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July 12

Simultaneous equations

I would like to know if there is an analytic way to find all the solutions to this pair of simultaneous equations:
$y^{2}-x^{2}+1+2xy=0$
$y^{2}-x^{2}-1-2xy=0$
I can find some of the solutions by inspection: I can find the solutions where $x=y$ or $x=-y$ because then the $y^{2}-x^{2}$ term disappears and the two equations become exactly the same. When $x=y$ we have $2x^{2}+1=0$ which has solutions $x=y={\frac {i}{\sqrt {2}}}$ and $x=y={\frac {-i}{\sqrt {2}}}$. When $x=-y$ we have $2x^{2}-1=0$ which has solutions $x={\frac {1}{\sqrt {2}}},y={\frac {-1}{\sqrt {2}}}$ and $x={\frac {-1}{\sqrt {2}}},y={\frac {1}{\sqrt {2}}}$. Widener (talk) 09:02, 12 July 2011 (UTC)

Subtract one equation from the other and you get

$2+4xy=0$

$\Rightarrow y={\frac {-1}{2x}}$

Substituting this in either equation gives

${\frac {1}{4x^{2}}}-x^{2}=0$

$\Rightarrow x^{4}={\frac {1}{4}}$

$\Rightarrow x=\pm {\frac {1}{\sqrt {2}}}{\text{ , }}\pm {\frac {i}{\sqrt {2}}}$

So the four solutions you have found are the only solutions. Gandalf61 (talk) 09:23, 12 July 2011 (UTC)

Analog of Bezout's Identity with d=2, a,b,x,y odd

I have been generating pairs of consecutive numbers with certain properties (which are preserved under multiplication) by using Bézout's identity; I construct two numbers with partial properties, use Bezout's identity to generate consecutive candidate numbers based on the original two, then check the candidates to see if the extra factors have completed the properties.

I would now like to do a similar thing, but generate pairs of consecutive ODD numbers (differing by 2) with the same properties. I do not see how to adapt Bezout's identity to generate a difference of 2, without introducing a factor of two into every term. A second consideration is to keep the generated factors odd, so that I am generating consecutive odd numbers.

Does anyone have any ideas? Walt (talk) 11:34, 12 July 2011 (UTC)

Could you give us some more details please? You mention Bézout's identity and you mention that d = 2 while a, b, x and y are odd. Bézout's identity tells us that if gcd(a,b) = 2 then there exist integers x and y such that ax + by = 2. If I recall, you can use the Chinese remainder theorem to calculate x and y. Could you please explain a little more about the motivation, and about what these "partial" properties are? An explicit example would be great. — Fly by Night (talk) 00:02, 13 July 2011 (UTC)

Well, it's not really Bezout's identity because 2 is not the GCD of the two odd numbers (the GCD is 1). That's why I referred to it as an analogue. It is just a form of linear diophantine equation. From the section on linear diophantine equations (now the difference is called c): "It follows that there are also infinite solutions if c is a multiple of the greatest common divisor of a and b. If c is not a multiple of the greatest common divisor of a and b, then the Diophantine equation ax + by = c has no solutions." Of course, both this statement and some examples show that solutions exist. The problem is twofold: applying Bezout's identity and then multiplying through by two doesn't necessarily give the smallest solution, and the products ax and by are no longer odd.

The problem is finding integers with certain numbers of divisors. For example, 33 and 35 are the smallest consecutive odd numbers with 4 divisors each. I have an algorithm for generating tons of single numbers with this property, but generating all occurrences is a very inefficient way to find consecutive pairs. Bezout's identity provided a great tool when doing the problem for consecutive numbers (one even and one odd) since the difference of 1 satisfied the conditions of the identity. Walt (talk) 02:04, 13 July 2011 (UTC)

Probability problem; not your usual combinatorics problem

Hello, all. I've been working on this problem for some time, and am quite unsure of how to go about it properly, mainly due to my lack of knowledge in the topic. Anyway, enough with the irrelevant talk, here it is:

There are six boxes on a marry-go-round, each arranged so that they together form a circle on the merry-go-round that has been divided into six equal parts. Within the boxes are (let's use a colorful example) 1 boy, 2 girls, and 3 dogs. If the merry-go-round is spun once, and when it stops with a particular section (box) facing at a predetermined set, so that there are only six possible arrangements (think of the Wheel_of_Fortune_(U.S._game_show) and the mechanism used for that), then a box is opened and the item inside it is removed. Thus (here's the problem:), what are the chances that 1 boy, 1 girl, and 1 dog remain in their boxes after the merry-go-round has been spun 3 separate times?

The reason this isn't your normal combinatorics problem -- we can't just take the possible unordered, non-repeating set of three items -- is because there are only 3 spins permitted. In that case, one has to consider a whole new panoply of possibilities, which wouldn't be the case if the merry-go-round was spun just once with three boxes opened in some order.

Any ideas on how to tackle this properly? Numbility (talk) 20:19, 12 July 2011 (UTC)

My intuition tells me that your (presumed random) spins will treat each box alike, and so the problem can be considered to be the same as finding the chance that, given a random arrangement of one boy, two girls and three dogs, the first three will be, in some order, a girl and two dogs. Each of GDDxxx, DGDxxx and DDGxxx can arise in 6 ways (permuting BGD, which are the members of xxx), and so the total number of favourable ways is 18. Six objects of which two are alike and three are alike can be arranged in 60 ways, so your answer is 18/60 or 3/10.←86.155.185.195 (talk) 21:16, 12 July 2011 (UTC)

This was my initial guess, namely, computing the probability of the six possible arrangements (3dogs * 2girls * 1boy) divided by n!/r!(n-r)!, where r=3, the items chosen, and n=6, the total number of items. But as I said, the fact that there are three spins is a definite constraint that cannot be avoided. Thus, 6/20 = .3000. But I doubt that this is correct, which is why it's not your usual combinatorics problem. Numbility (talk) 23:16, 12 July 2011 (UTC)

On second thoughts, on each spin a particular box has 1 chance in 6 of being chosen. So the probability of the first three being, in order, GDD is (2/6)(3/6)(2/6) = 12/216. Likewise, the possibilities DGD and DDG also have probability 12/216. So the probability of some one of these events ocurring, and so leaving behind one each of boy, girl and dog, will be 3.12/216 = 1/6.←86.155.185.195 (talk) 21:33, 12 July 2011 (UTC)

Keep in mind that once a box has been opened it can still be a result in the following spins. It does not appear to me valid to successively multiply the probability of a box once it has been opened, because the kind of computation required will change. Thus, with each spin and an opening of a box, the problem will necessarily change and this has to be accounted for in the result. Numbility (talk) 23:16, 12 July 2011 (UTC)

The problem is small enough to just list the possible ways to be left with 1 of each. You can pick the same girl or dog 3 times in a row, which gives you 5 ways. You can pick a girl and a dog in 6 ways and each pair has 6 ways of being chosen (GGD, GDG, DGG, DDG, DGD, GDD). You can pick 3 pairs of dog and there are the same 6 ways for that pair to be chosen. There are 6 sets of 1 girl and 2 dogs, each of which can be chosen in 6 ways. So that should come out to 95/216 if I haven't done anything wrong. Maybe someone else has a more clever way to go about it though. Rckrone (talk) 01:50, 13 July 2011 (UTC)

I thought that, and used a pretty good method of finding out the bare probability by counting all of the possibilities by hand, as it were. However, since I don't know the formalisms involved, I'm not sure it's a valid result, which is why I'm trying to bounce this off you guys.

a=1 ; b=2 c=2 ; d=3 e=3 f=3
(Count the numbers next to the cases to obtain the result for each row.)
1st		{a,b,c,d,e}*3 [+3*2] {a,b,d,e,f}*2 [+2*2] {b,c,d,e,f}*1 [+1*2]				= 6 [+12] = 18
spin
result		One must take into account that the next spin may land on
		a previously emptied box. Thus add two self-returning paths here 
		for each unique case.
 
2nd		{a,b,c,d}*3 [+3] {a,b,d,e}*6 [+6] {a,d,e,f}*1 [+1] {b,c,d,e}*3 [+3] {b,d,e,f}*2 [+2]	= 15 [+15] = 30
spin
result		As above, but add one for one spin remaining.

3rd		{a,b,c}*1 >>>>{a,b,d}*6<<<< {a,d,e}*3 {d,e,f}*1 {b,c,d}*3 {b,d,e}*6			= 20
spin
result		"><" signifies the sought-for case (one of each item), of which there are six.

Therefore, 6/68 is the probability.

On that note, your calculation looks wrong to me. Anyone got better ideas? Numbility (talk) 03:49, 13 July 2011 (UTC)

Perhaps I am misunderstanding the problem or your answer, but I don't think you can pick the same girl three times in a row, because once a spin lands on her once, she is removed from her box and is not available to be chosen again. Also, if you happened to land on the same girl's box three times in a row with your three spins, you would be left with one boy, one girl, and three dogs in the boxes on the merry-go-round, which is not the desired outcome. —Bkell (talk) 03:06, 13 July 2011 (UTC)

It should be pretty clear that there are only six ways in which there can be 1 boy, 1 girl, and 1 dog remaining. The issue is figuring out all of the general cases in which these 6 occur. Numbility (talk) 03:57, 13 July 2011 (UTC)

I interpreted the question to mean you want at least 1 boy, 1 girl and, 1 dog left. If you want exactly those left you just consider the last case of picking 2 dogs and a girl which as I mentioned has 36 ways of happening, so 1/6. Rckrone (talk) 03:39, 13 July 2011 (UTC)

No, what are the chances that 1 boy, 1 dog, and 1 girl will remain after only three spins? Look at my previous answer above to see the "bare bones" of the problem. Numbility (talk) 03:53, 13 July 2011 (UTC)

If I'm understanding the problem correctly, your three spins must land on dogs twice and a girl once, not necessarily in that order. So, the three spins could land girl–dog–dog, dog–girl–dog, or dog–dog–girl. We just need to compute the probabilities of each of those three and add them together (since they are mutually exclusive). What is the probability of getting girl–dog–dog on the three spins? The probability of landing on a girl on the first spin is 2/6, or 1/3; after that, the probability of landing on a dog on the next spin is 3/6, or 1/2; and after that, since there are only two dogs remaining, the probability of landing on a dog on the last spin is 2/6, or 1/3. So the probability of getting girl–dog–dog is (1/3)×(1/2)×(1/3)=1/12. Similarly, in order to get dog–girl–dog, the first spin must land on a dog, which will happen with probability 3/6=1/2; the second must land on a girl, with probability 2/6=1/3; and the last spin must land on one of the two remaining dogs, with probability 2/6=1/3. So the probability of getting dog–girl–dog is again (1/2)×(1/3)×(1/3)=1/12. If you work out the probability of getting dog–dog–girl like this, you'll see that it is also 1/12. So the overall probability of a desirable outcome is (1/12)+(1/12)+(1/12)=3/12=1/4. —Bkell (talk) 02:59, 13 July 2011 (UTC)

No, it's not that they "must" land that way. They can land any which way they please, even on an empty box, so the calculation has to take that into account. Numbility (talk) 03:53, 13 July 2011 (UTC)

Bkell means that they must land that way in order to reach the outcome you want: Exactly 1 boy, 1 girl, and 1 dog remain. However, Bkell incorrectly wrote (1/3)×(1/2)×(1/3)=1/12. But (1/3)×(1/2)×(1/3)=1/18. Then the total probability of success becomes 3/18 = 1/6, as 86.155.185.195 and Rckrone have already said. PrimeHunter (talk) 04:28, 13 July 2011 (UTC)

Oh, yep, I can't multiply. Thanks. —Bkell (talk) 12:06, 13 July 2011 (UTC)

@PrimeHunter (or anyone), so how do I relate my crude analysis, which gave me 6/68 or 2/34, to this 1/6 probability? Where am I wrong? Numbility (talk) 15:16, 13 July 2011 (UTC)

I think I may be beginning to see the light, and I hope you guys will help me on this. Looking at my table of scenarios, I've come to the realization that I should discard the undesired outcomes. Thus, at the first spin, I have only 5 of 6 total cases that I want, at the second, I have only 9 of 15 that I want, and at the third only 6 of 20. At this stage, I take the fractions of each ordinal spin-result and multiply them: (5/6)(9/15)(6/20) = 60/1800 which reduces to 3/20. This answer is surprisingly close to 1/6 and seems to make a lot more sense than 1/6. What say you all? Numbility (talk) 15:55, 13 July 2011 (UTC)

I don't understand your analysis. Can you explain it a little more thoroughly, in words? —Bkell (talk) 19:58, 13 July 2011 (UTC)

Okay, after I've reread what you've written, let me say a little bit about what I think you're doing. You are correct when you say that your first spin will lead toward a desirable outcome 5/6 of the time (the other 1/6 of the time the first spin will land on a boy, and then certainly the desired outcome will not occur, no matter what happens in the next two spins). However, the probability that your second spin will continue toward a desirable outcome depends on what exactly happened in the first spin. If the first spin landed on a girl, then the second spin must land on a dog to lead toward a desirable outcome, and the probability of that happening is 3/6. On the other hand, if the first spin landed on a dog, then the second spin could land either on another dog or on a girl and still lead toward a desirable outcome, so the probability is 4/6. So you can't just multiply the whole 5/6 from the first spin by the probability of "success" in the second spin (where "success" means "leading toward a desirable outcome"), because that second probability is not independent of what happened in the first spin. Instead, you should split up that 5/6 from the first spin like this: With probability 3/6, the first spin will land on a dog, and then, if that happens, the probability of success in the second spin is 4/6; the probability of both of these things happening in the first two spins is (3/6)×(4/6)=1/3. With probability 2/6, the first spin will land on a girl, and then, if that happens, the probability of success in the second spin is 3/6; the probability of both of these things happening in the first two spins is (2/6)×(3/6)=1/6. Then add the probabilities of these two mutually exclusive events to get the total probability of success through the first two spins: (1/3)+(1/6)=1/2. Now, to finish this analysis, you must split up the 4/6 probability of success in the second spin following a dog in the first spin, for the same reason: After the first spin lands on a dog, and a "successful" second spin, the probability of success in the third spin depends on what exactly happened in the second spin (dog or girl). This analysis is what I was doing above. —Bkell (talk) 20:13, 13 July 2011 (UTC)

I guess, in this particular problem, after the first two spins are "dog" and "success" (i.e., either "dog–dog" or "dog–girl"), the probability of success in the third spin happens to be the same (2/6) whether the second spin was a dog or a girl. In general, though, when dealing with conditional probabilities like this (where the probability of an event B is different depending on whether or not some other event A happened), you need to consider different cases for the event A: one line of reasoning for the case when A occurs, and another for the case when A does not occur. This is called the law of total probability. —Bkell (talk) 20:41, 13 July 2011 (UTC)

Maybe I'm dense (others seem to have quickly agreed that the answer is 1/6, but I'm not sure), but I'm having a hard time seeing how the logic of conditional probability applies to this problem. Perhaps a simplified version of the problem would help. Going back to the basics, let's say I have four cards: 2 spades, 1 heart, 1 diamond. After the deck has been shuffled, a card is withdrawn and a blank is inserted in its place, and the deck is shuffled again. What is the probability of there being 1 spade and 1 heart after 2 shuffles and withdrawals? So we have:

a=1, b=2, c=3, d=3
{a=heart, b=diamond, c=d=spade}
1st	(one blank	{a,b,c}*2	{b,c,d}*1	{a,c,d}*1			=	4 cases
	for all cases)	(good)		(bad)		(good)

2nd	(two blanks	{b,c}*2		{a,b}*1		{c,d}*1		>>{a,c}*2<<		=	6 cases
	for all cases)	(bad)		(bad)		(bad)		(good)

1st has 3 good cases out of 4, thus, 3/4; 2nd has 2 good cases out of 6, thus, 1/3. The total probability
is (3/4)(1/3)= 3/12 = 1/4.

What is the flaw with this analysis as we consider the "good" cases in the numerator over the total cases in the denominator? It should be clear that in the good cases, the result we're after is a possibility therein, and so we can multiply the good cases and divide this by the product of individual ordinal cases. This problem is fundamentally the same as my earlier problem with a few simplifications (fewer cases, so hopefully less confusion). Thanks for your time. Numbility (talk) 03:17, 14 July 2011 (UTC)

In addition, if you look closely, we don't interpret the 2nd row as being "conditional on" the first result, nor do we consider the 1st row as being "conditional on" the random arrangement of the cards at the initial condition. It seems pretty clear to me that in the merry-go-round version, even a 3rd row of results aren't "conditional on" what happened in the 2nd row, which means I can divide the product of good cases by the product of all cases that exist. So 3/20 must be the answer, not 1/6. Numbility (talk) 03:37, 14 July 2011 (UTC)

The answer can't be 3/20. Here's why. There are 6 possible outcomes for the first spin (one for each box). Then, for each of those possible outcomes (and independently of the first outcome), there are 6 possible outcomes for the second spin. Likewise there are 6 possible outcomes for the third spin. This gives 6×6×6=216 possible outcomes for the three spins together. All of these 216 outcomes are equally likely. So the probability that you get a desired outcome has got to be x/216 for some value of x. No such fraction can simplify to 3/20, because 20 has a factor of 5 and 216 doesn't. —Bkell (talk) 04:01, 14 July 2011 (UTC)

I count 15 unique cases at the second spin and 20 at the 3rd. So there are a total of 1,800 unique cases within the "space" of the merry-go-round, not 216. Numbility (talk) 04:06, 14 July 2011 (UTC)

Here are the 216 possible ways the three spins could turn out. I've stereotypically named the boy Johnny, the girls Mary and Susan, and the dogs Fido, Rover, and Spot. If a spin lands on someone's box who was already chosen in a previous spin, that box is empty.

216 possible outcomes of the three spins

1. Spins land on Johnny, Johnny (now empty), Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover, Spot. BAD.
2. Spins land on Johnny, Johnny (now empty), Mary. Those remaining in boxes after 3 spins: Susan, Fido, Rover, Spot. BAD.
3. Spins land on Johnny, Johnny (now empty), Susan. Those remaining in boxes after 3 spins: Mary, Fido, Rover, Spot. BAD.
4. Spins land on Johnny, Johnny (now empty), Fido. Those remaining in boxes after 3 spins: Mary, Susan, Rover, Spot. BAD.
5. Spins land on Johnny, Johnny (now empty), Rover. Those remaining in boxes after 3 spins: Mary, Susan, Fido, Spot. BAD.
6. Spins land on Johnny, Johnny (now empty), Spot. Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover. BAD.
7. Spins land on Johnny, Mary, Johnny (now empty). Those remaining in boxes after 3 spins: Susan, Fido, Rover, Spot. BAD.
8. Spins land on Johnny, Mary, Mary (now empty). Those remaining in boxes after 3 spins: Susan, Fido, Rover, Spot. BAD.
9. Spins land on Johnny, Mary, Susan. Those remaining in boxes after 3 spins: Fido, Rover, Spot. BAD.
10. Spins land on Johnny, Mary, Fido. Those remaining in boxes after 3 spins: Susan, Rover, Spot. BAD.
11. Spins land on Johnny, Mary, Rover. Those remaining in boxes after 3 spins: Susan, Fido, Spot. BAD.
12. Spins land on Johnny, Mary, Spot. Those remaining in boxes after 3 spins: Susan, Fido, Rover. BAD.
13. Spins land on Johnny, Susan, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Fido, Rover, Spot. BAD.
14. Spins land on Johnny, Susan, Mary. Those remaining in boxes after 3 spins: Fido, Rover, Spot. BAD.
15. Spins land on Johnny, Susan, Susan (now empty). Those remaining in boxes after 3 spins: Mary, Fido, Rover, Spot. BAD.
16. Spins land on Johnny, Susan, Fido. Those remaining in boxes after 3 spins: Mary, Rover, Spot. BAD.
17. Spins land on Johnny, Susan, Rover. Those remaining in boxes after 3 spins: Mary, Fido, Spot. BAD.
18. Spins land on Johnny, Susan, Spot. Those remaining in boxes after 3 spins: Mary, Fido, Rover. BAD.
19. Spins land on Johnny, Fido, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Rover, Spot. BAD.
20. Spins land on Johnny, Fido, Mary. Those remaining in boxes after 3 spins: Susan, Rover, Spot. BAD.
21. Spins land on Johnny, Fido, Susan. Those remaining in boxes after 3 spins: Mary, Rover, Spot. BAD.
22. Spins land on Johnny, Fido, Fido (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Rover, Spot. BAD.
23. Spins land on Johnny, Fido, Rover. Those remaining in boxes after 3 spins: Mary, Susan, Spot. BAD.
24. Spins land on Johnny, Fido, Spot. Those remaining in boxes after 3 spins: Mary, Susan, Rover. BAD.
25. Spins land on Johnny, Rover, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Spot. BAD.
26. Spins land on Johnny, Rover, Mary. Those remaining in boxes after 3 spins: Susan, Fido, Spot. BAD.
27. Spins land on Johnny, Rover, Susan. Those remaining in boxes after 3 spins: Mary, Fido, Spot. BAD.
28. Spins land on Johnny, Rover, Fido. Those remaining in boxes after 3 spins: Mary, Susan, Spot. BAD.
29. Spins land on Johnny, Rover, Rover (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Spot. BAD.
30. Spins land on Johnny, Rover, Spot. Those remaining in boxes after 3 spins: Mary, Susan, Fido. BAD.
31. Spins land on Johnny, Spot, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover. BAD.
32. Spins land on Johnny, Spot, Mary. Those remaining in boxes after 3 spins: Susan, Fido, Rover. BAD.
33. Spins land on Johnny, Spot, Susan. Those remaining in boxes after 3 spins: Mary, Fido, Rover. BAD.
34. Spins land on Johnny, Spot, Fido. Those remaining in boxes after 3 spins: Mary, Susan, Rover. BAD.
35. Spins land on Johnny, Spot, Rover. Those remaining in boxes after 3 spins: Mary, Susan, Fido. BAD.
36. Spins land on Johnny, Spot, Spot (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover. BAD.
37. Spins land on Mary, Johnny, Johnny (now empty). Those remaining in boxes after 3 spins: Susan, Fido, Rover, Spot. BAD.
38. Spins land on Mary, Johnny, Mary (now empty). Those remaining in boxes after 3 spins: Susan, Fido, Rover, Spot. BAD.
39. Spins land on Mary, Johnny, Susan. Those remaining in boxes after 3 spins: Fido, Rover, Spot. BAD.
40. Spins land on Mary, Johnny, Fido. Those remaining in boxes after 3 spins: Susan, Rover, Spot. BAD.
41. Spins land on Mary, Johnny, Rover. Those remaining in boxes after 3 spins: Susan, Fido, Spot. BAD.
42. Spins land on Mary, Johnny, Spot. Those remaining in boxes after 3 spins: Susan, Fido, Rover. BAD.
43. Spins land on Mary, Mary (now empty), Johnny. Those remaining in boxes after 3 spins: Susan, Fido, Rover, Spot. BAD.
44. Spins land on Mary, Mary (now empty), Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover, Spot. BAD.
45. Spins land on Mary, Mary (now empty), Susan. Those remaining in boxes after 3 spins: Johnny, Fido, Rover, Spot. BAD.
46. Spins land on Mary, Mary (now empty), Fido. Those remaining in boxes after 3 spins: Johnny, Susan, Rover, Spot. BAD.
47. Spins land on Mary, Mary (now empty), Rover. Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Spot. BAD.
48. Spins land on Mary, Mary (now empty), Spot. Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover. BAD.
49. Spins land on Mary, Susan, Johnny. Those remaining in boxes after 3 spins: Fido, Rover, Spot. BAD.
50. Spins land on Mary, Susan, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Fido, Rover, Spot. BAD.
51. Spins land on Mary, Susan, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Fido, Rover, Spot. BAD.
52. Spins land on Mary, Susan, Fido. Those remaining in boxes after 3 spins: Johnny, Rover, Spot. BAD.
53. Spins land on Mary, Susan, Rover. Those remaining in boxes after 3 spins: Johnny, Fido, Spot. BAD.
54. Spins land on Mary, Susan, Spot. Those remaining in boxes after 3 spins: Johnny, Fido, Rover. BAD.
55. Spins land on Mary, Fido, Johnny. Those remaining in boxes after 3 spins: Susan, Rover, Spot. BAD.
56. Spins land on Mary, Fido, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Rover, Spot. BAD.
57. Spins land on Mary, Fido, Susan. Those remaining in boxes after 3 spins: Johnny, Rover, Spot. BAD.
58. Spins land on Mary, Fido, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Rover, Spot. BAD.
59. Spins land on Mary, Fido, Rover. Those remaining in boxes after 3 spins: Johnny, Susan, Spot. GOOD.
60. Spins land on Mary, Fido, Spot. Those remaining in boxes after 3 spins: Johnny, Susan, Rover. GOOD.
61. Spins land on Mary, Rover, Johnny. Those remaining in boxes after 3 spins: Susan, Fido, Spot. BAD.
62. Spins land on Mary, Rover, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Spot. BAD.
63. Spins land on Mary, Rover, Susan. Those remaining in boxes after 3 spins: Johnny, Fido, Spot. BAD.
64. Spins land on Mary, Rover, Fido. Those remaining in boxes after 3 spins: Johnny, Susan, Spot. GOOD.
65. Spins land on Mary, Rover, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Spot. BAD.
66. Spins land on Mary, Rover, Spot. Those remaining in boxes after 3 spins: Johnny, Susan, Fido. GOOD.
67. Spins land on Mary, Spot, Johnny. Those remaining in boxes after 3 spins: Susan, Fido, Rover. BAD.
68. Spins land on Mary, Spot, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover. BAD.
69. Spins land on Mary, Spot, Susan. Those remaining in boxes after 3 spins: Johnny, Fido, Rover. BAD.
70. Spins land on Mary, Spot, Fido. Those remaining in boxes after 3 spins: Johnny, Susan, Rover. GOOD.
71. Spins land on Mary, Spot, Rover. Those remaining in boxes after 3 spins: Johnny, Susan, Fido. GOOD.
72. Spins land on Mary, Spot, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover. BAD.
73. Spins land on Susan, Johnny, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Fido, Rover, Spot. BAD.
74. Spins land on Susan, Johnny, Mary. Those remaining in boxes after 3 spins: Fido, Rover, Spot. BAD.
75. Spins land on Susan, Johnny, Susan (now empty). Those remaining in boxes after 3 spins: Mary, Fido, Rover, Spot. BAD.
76. Spins land on Susan, Johnny, Fido. Those remaining in boxes after 3 spins: Mary, Rover, Spot. BAD.
77. Spins land on Susan, Johnny, Rover. Those remaining in boxes after 3 spins: Mary, Fido, Spot. BAD.
78. Spins land on Susan, Johnny, Spot. Those remaining in boxes after 3 spins: Mary, Fido, Rover. BAD.
79. Spins land on Susan, Mary, Johnny. Those remaining in boxes after 3 spins: Fido, Rover, Spot. BAD.
80. Spins land on Susan, Mary, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Fido, Rover, Spot. BAD.
81. Spins land on Susan, Mary, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Fido, Rover, Spot. BAD.
82. Spins land on Susan, Mary, Fido. Those remaining in boxes after 3 spins: Johnny, Rover, Spot. BAD.
83. Spins land on Susan, Mary, Rover. Those remaining in boxes after 3 spins: Johnny, Fido, Spot. BAD.
84. Spins land on Susan, Mary, Spot. Those remaining in boxes after 3 spins: Johnny, Fido, Rover. BAD.
85. Spins land on Susan, Susan (now empty), Johnny. Those remaining in boxes after 3 spins: Mary, Fido, Rover, Spot. BAD.
86. Spins land on Susan, Susan (now empty), Mary. Those remaining in boxes after 3 spins: Johnny, Fido, Rover, Spot. BAD.
87. Spins land on Susan, Susan (now empty), Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover, Spot. BAD.
88. Spins land on Susan, Susan (now empty), Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Rover, Spot. BAD.
89. Spins land on Susan, Susan (now empty), Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Spot. BAD.
90. Spins land on Susan, Susan (now empty), Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover. BAD.
91. Spins land on Susan, Fido, Johnny. Those remaining in boxes after 3 spins: Mary, Rover, Spot. BAD.
92. Spins land on Susan, Fido, Mary. Those remaining in boxes after 3 spins: Johnny, Rover, Spot. BAD.
93. Spins land on Susan, Fido, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Rover, Spot. BAD.
94. Spins land on Susan, Fido, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Rover, Spot. BAD.
95. Spins land on Susan, Fido, Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Spot. GOOD.
96. Spins land on Susan, Fido, Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Rover. GOOD.
97. Spins land on Susan, Rover, Johnny. Those remaining in boxes after 3 spins: Mary, Fido, Spot. BAD.
98. Spins land on Susan, Rover, Mary. Those remaining in boxes after 3 spins: Johnny, Fido, Spot. BAD.
99. Spins land on Susan, Rover, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Spot. BAD.
100. Spins land on Susan, Rover, Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Spot. GOOD.
101. Spins land on Susan, Rover, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Spot. BAD.
102. Spins land on Susan, Rover, Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Fido. GOOD.
103. Spins land on Susan, Spot, Johnny. Those remaining in boxes after 3 spins: Mary, Fido, Rover. BAD.
104. Spins land on Susan, Spot, Mary. Those remaining in boxes after 3 spins: Johnny, Fido, Rover. BAD.
105. Spins land on Susan, Spot, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover. BAD.
106. Spins land on Susan, Spot, Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Rover. GOOD.
107. Spins land on Susan, Spot, Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Fido. GOOD.
108. Spins land on Susan, Spot, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover. BAD.
109. Spins land on Fido, Johnny, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Rover, Spot. BAD.
110. Spins land on Fido, Johnny, Mary. Those remaining in boxes after 3 spins: Susan, Rover, Spot. BAD.
111. Spins land on Fido, Johnny, Susan. Those remaining in boxes after 3 spins: Mary, Rover, Spot. BAD.
112. Spins land on Fido, Johnny, Fido (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Rover, Spot. BAD.
113. Spins land on Fido, Johnny, Rover. Those remaining in boxes after 3 spins: Mary, Susan, Spot. BAD.
114. Spins land on Fido, Johnny, Spot. Those remaining in boxes after 3 spins: Mary, Susan, Rover. BAD.
115. Spins land on Fido, Mary, Johnny. Those remaining in boxes after 3 spins: Susan, Rover, Spot. BAD.
116. Spins land on Fido, Mary, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Rover, Spot. BAD.
117. Spins land on Fido, Mary, Susan. Those remaining in boxes after 3 spins: Johnny, Rover, Spot. BAD.
118. Spins land on Fido, Mary, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Rover, Spot. BAD.
119. Spins land on Fido, Mary, Rover. Those remaining in boxes after 3 spins: Johnny, Susan, Spot. GOOD.
120. Spins land on Fido, Mary, Spot. Those remaining in boxes after 3 spins: Johnny, Susan, Rover. GOOD.
121. Spins land on Fido, Susan, Johnny. Those remaining in boxes after 3 spins: Mary, Rover, Spot. BAD.
122. Spins land on Fido, Susan, Mary. Those remaining in boxes after 3 spins: Johnny, Rover, Spot. BAD.
123. Spins land on Fido, Susan, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Rover, Spot. BAD.
124. Spins land on Fido, Susan, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Rover, Spot. BAD.
125. Spins land on Fido, Susan, Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Spot. GOOD.
126. Spins land on Fido, Susan, Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Rover. GOOD.
127. Spins land on Fido, Fido (now empty), Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Rover, Spot. BAD.
128. Spins land on Fido, Fido (now empty), Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Rover, Spot. BAD.
129. Spins land on Fido, Fido (now empty), Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Rover, Spot. BAD.
130. Spins land on Fido, Fido (now empty), Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover, Spot. BAD.
131. Spins land on Fido, Fido (now empty), Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Spot. BAD.
132. Spins land on Fido, Fido (now empty), Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover. BAD.
133. Spins land on Fido, Rover, Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Spot. BAD.
134. Spins land on Fido, Rover, Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Spot. GOOD.
135. Spins land on Fido, Rover, Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Spot. GOOD.
136. Spins land on Fido, Rover, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Spot. BAD.
137. Spins land on Fido, Rover, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Spot. BAD.
138. Spins land on Fido, Rover, Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Susan. BAD.
139. Spins land on Fido, Spot, Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Rover. BAD.
140. Spins land on Fido, Spot, Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Rover. GOOD.
141. Spins land on Fido, Spot, Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Rover. GOOD.
142. Spins land on Fido, Spot, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover. BAD.
143. Spins land on Fido, Spot, Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Susan. BAD.
144. Spins land on Fido, Spot, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover. BAD.
145. Spins land on Rover, Johnny, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Spot. BAD.
146. Spins land on Rover, Johnny, Mary. Those remaining in boxes after 3 spins: Susan, Fido, Spot. BAD.
147. Spins land on Rover, Johnny, Susan. Those remaining in boxes after 3 spins: Mary, Fido, Spot. BAD.
148. Spins land on Rover, Johnny, Fido. Those remaining in boxes after 3 spins: Mary, Susan, Spot. BAD.
149. Spins land on Rover, Johnny, Rover (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Spot. BAD.
150. Spins land on Rover, Johnny, Spot. Those remaining in boxes after 3 spins: Mary, Susan, Fido. BAD.
151. Spins land on Rover, Mary, Johnny. Those remaining in boxes after 3 spins: Susan, Fido, Spot. BAD.
152. Spins land on Rover, Mary, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Spot. BAD.
153. Spins land on Rover, Mary, Susan. Those remaining in boxes after 3 spins: Johnny, Fido, Spot. BAD.
154. Spins land on Rover, Mary, Fido. Those remaining in boxes after 3 spins: Johnny, Susan, Spot. GOOD.
155. Spins land on Rover, Mary, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Spot. BAD.
156. Spins land on Rover, Mary, Spot. Those remaining in boxes after 3 spins: Johnny, Susan, Fido. GOOD.
157. Spins land on Rover, Susan, Johnny. Those remaining in boxes after 3 spins: Mary, Fido, Spot. BAD.
158. Spins land on Rover, Susan, Mary. Those remaining in boxes after 3 spins: Johnny, Fido, Spot. BAD.
159. Spins land on Rover, Susan, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Spot. BAD.
160. Spins land on Rover, Susan, Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Spot. GOOD.
161. Spins land on Rover, Susan, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Spot. BAD.
162. Spins land on Rover, Susan, Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Fido. GOOD.
163. Spins land on Rover, Fido, Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Spot. BAD.
164. Spins land on Rover, Fido, Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Spot. GOOD.
165. Spins land on Rover, Fido, Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Spot. GOOD.
166. Spins land on Rover, Fido, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Spot. BAD.
167. Spins land on Rover, Fido, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Spot. BAD.
168. Spins land on Rover, Fido, Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Susan. BAD.
169. Spins land on Rover, Rover (now empty), Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Fido, Spot. BAD.
170. Spins land on Rover, Rover (now empty), Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Spot. BAD.
171. Spins land on Rover, Rover (now empty), Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Spot. BAD.
172. Spins land on Rover, Rover (now empty), Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Spot. BAD.
173. Spins land on Rover, Rover (now empty), Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido, Spot. BAD.
174. Spins land on Rover, Rover (now empty), Spot. Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido. BAD.
175. Spins land on Rover, Spot, Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Fido. BAD.
176. Spins land on Rover, Spot, Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Fido. GOOD.
177. Spins land on Rover, Spot, Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Fido. GOOD.
178. Spins land on Rover, Spot, Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Susan. BAD.
179. Spins land on Rover, Spot, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido. BAD.
180. Spins land on Rover, Spot, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido. BAD.
181. Spins land on Spot, Johnny, Johnny (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover. BAD.
182. Spins land on Spot, Johnny, Mary. Those remaining in boxes after 3 spins: Susan, Fido, Rover. BAD.
183. Spins land on Spot, Johnny, Susan. Those remaining in boxes after 3 spins: Mary, Fido, Rover. BAD.
184. Spins land on Spot, Johnny, Fido. Those remaining in boxes after 3 spins: Mary, Susan, Rover. BAD.
185. Spins land on Spot, Johnny, Rover. Those remaining in boxes after 3 spins: Mary, Susan, Fido. BAD.
186. Spins land on Spot, Johnny, Spot (now empty). Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover. BAD.
187. Spins land on Spot, Mary, Johnny. Those remaining in boxes after 3 spins: Susan, Fido, Rover. BAD.
188. Spins land on Spot, Mary, Mary (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover. BAD.
189. Spins land on Spot, Mary, Susan. Those remaining in boxes after 3 spins: Johnny, Fido, Rover. BAD.
190. Spins land on Spot, Mary, Fido. Those remaining in boxes after 3 spins: Johnny, Susan, Rover. GOOD.
191. Spins land on Spot, Mary, Rover. Those remaining in boxes after 3 spins: Johnny, Susan, Fido. GOOD.
192. Spins land on Spot, Mary, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover. BAD.
193. Spins land on Spot, Susan, Johnny. Those remaining in boxes after 3 spins: Mary, Fido, Rover. BAD.
194. Spins land on Spot, Susan, Mary. Those remaining in boxes after 3 spins: Johnny, Fido, Rover. BAD.
195. Spins land on Spot, Susan, Susan (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover. BAD.
196. Spins land on Spot, Susan, Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Rover. GOOD.
197. Spins land on Spot, Susan, Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Fido. GOOD.
198. Spins land on Spot, Susan, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover. BAD.
199. Spins land on Spot, Fido, Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Rover. BAD.
200. Spins land on Spot, Fido, Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Rover. GOOD.
201. Spins land on Spot, Fido, Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Rover. GOOD.
202. Spins land on Spot, Fido, Fido (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover. BAD.
203. Spins land on Spot, Fido, Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Susan. BAD.
204. Spins land on Spot, Fido, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover. BAD.
205. Spins land on Spot, Rover, Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Fido. BAD.
206. Spins land on Spot, Rover, Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Fido. GOOD.
207. Spins land on Spot, Rover, Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Fido. GOOD.
208. Spins land on Spot, Rover, Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Susan. BAD.
209. Spins land on Spot, Rover, Rover (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido. BAD.
210. Spins land on Spot, Rover, Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido. BAD.
211. Spins land on Spot, Spot (now empty), Johnny. Those remaining in boxes after 3 spins: Mary, Susan, Fido, Rover. BAD.
212. Spins land on Spot, Spot (now empty), Mary. Those remaining in boxes after 3 spins: Johnny, Susan, Fido, Rover. BAD.
213. Spins land on Spot, Spot (now empty), Susan. Those remaining in boxes after 3 spins: Johnny, Mary, Fido, Rover. BAD.
214. Spins land on Spot, Spot (now empty), Fido. Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Rover. BAD.
215. Spins land on Spot, Spot (now empty), Rover. Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido. BAD.
216. Spins land on Spot, Spot (now empty), Spot (now empty). Those remaining in boxes after 3 spins: Johnny, Mary, Susan, Fido, Rover. BAD.

I don't have time just this minute to improve the program I wrote to list these so that it labels them as good or bad outcomes, but if you count them you'll see that 36 of these result in a good outcome, so the probability is 36/216, or 1/6. —Bkell (talk) 04:43, 14 July 2011 (UTC)

Okay, now that I have a few minutes, I've improved my little program so that it more clearly states, for each of the 216 equally possible outcomes, where the spins land, who is remaining in their boxes after the three spins, and whether this outcome is good (exactly one boy, one girl, and one dog remain) or bad (otherwise). If you count up the GOOD rows, you will see that there are 36 of them. —Bkell (talk) 06:25, 14 July 2011 (UTC)

Your answer of 1/4 happens to be correct for this problem, I believe, but I don't see why your analysis makes sense. For the first draw, you have {a,b,c}*2; I assume this is shorthand for {a,b,c} and {a,b,d} because those two cases are equivalent. Likewise, for the second draw, you have {b,c}*2, which I assume means {b,c} and {b,d}, and you have {a,c}*2, which I take to mean {a,c} and {a,d}. You say, for the second draw, "6 cases for all cases"—but that's not true; you don't have the same six possibilities for all cases. If the first draw resulted in {a,b,c}, for example, you can't get {b,d} after the second draw, because you've removed d from the deck. So {b,c} should count only once in that case. The same goes for {a,c}, and {c,d} can't happen at all—and you forgot the possibility that after the second draw you might be left with {a,b,c} because you drew the blank. So, if the first draw leaves you with {a,b,c}, then there are only four possibilities, not six, for what you are left with after the second draw: {a,b}, {a,c}, {b,c}, and {a,b,c}. One of these four, {a,c}, is a good outcome. The probability of getting to {a,c} in this manner equals the probability of getting {a,b,c} after the first draw (which is 1/4) multiplied by the probability of then getting {a,c} in the second draw (which is 1/4), so the probability of getting {a,c} in this way is 1/16. Exactly equivalent is the case where you get {a,b,d} after the first draw and {a,d} after the second; that gives another 1/16. Then there's a whole other possibility to consider: If you get {a,c,d} on the first draw, then your possibilities after the second draw are {a,c}, {a,d}, {c,d}, and {a,c,d}. Out of those four, two are good: {a,c} and {a,d}. So the probability of getting a good result down this path is the probability of getting {a,c,d} on the first draw, which is 1/4, multiplied by the probability of getting either {a,c} or {a,d} on the second, which is 1/2; this gives 1/8. Then 1/16+1/16+1/8=1/4. —Bkell (talk) 03:47, 14 July 2011 (UTC)

Great response; you've interpreted my shorthand correctly. First, I know that the good cases lead to each other, that's the point I'm trying to make by saying they're "good". On paper, I draw lines connecting the cases that can lead to each other, but my means of doing this here are limited. Further, I didn't forget about the blanks, they're implied by the notes in each row. In any case, we can't exclude some of the cases merely due to the outcome of one particular dealing, so I don't understand how there couldn't be 6 possible cases at the 2nd shuffling (not counting that we might have drawn a blank and remained at the first row). Furthermore, all unique cases have been listed, even if at the 2nd withdrawal I get a blank card, since by default such a case is described in the first row and has already been counted. This last point is very significant, and I hope you realize that we can't count cases that have already been counted in our computation, which would result in inflation. I think it is a unique coincidence we got exactly the same result, but I think my method is correct. Numbility (talk) 03:58, 14 July 2011 (UTC)

Okay, here's a table of the possible outcomes of the two draws in the simplified problem. Let's say the cards we start with are A♠, 2♠, A♥, A♦.

		Second draw
		A♠	2♠	A♥	A♦	blank
First draw	A♠	(impossible)	BAD	BAD	GOOD	BAD
	2♠	BAD	(impossible)	BAD	GOOD	BAD
	A♥	BAD	BAD	(impossible)	BAD	BAD
	A♦	GOOD	GOOD	BAD	(impossible)	BAD

Out of the 16 possible outcomes, 4 of them are good. This gives a probability of 1/4.

If we call a first draw of A♥ a failure, because it cannot lead to a good outcome, and the other three possible first draws successes, then the probability of a successful first draw is 3/4; we both agree on this. Your method gives a calculation of (3/4)×(1/3). This would make sense if the probability of success on the second draw was a constant 1/3, independent of the result of the first draw (given that the first draw was a success). But you can see that this probability is not constant: The first two rows of the table above have just one good outcome each, while the third row has two good outcomes. And there are four possible outcomes in each row, so a fraction of 1/3 doesn't even make sense.

The reason conditional probability comes into play here is that, given two events A and B, the probability of both A and B occurring is P(A)×P(B|A), that is, the probability of A, multiplied by the probability of B given A. It is not simply P(A)×P(B), unless the two events are independent, in which case P(B|A)=P(B). In these problems the two events you are trying to use are not independent: The probability of being left with {a,c} after the second draw is different depending on whether or not you were left with {a,b,d} after the first draw, for example.

(By the way, you are misusing the word "unique" as it is used in mathematics. "Unique" means "only one exists"—so, for example, we may talk about "the unique solution" to an equation, if that equation has one and only one solution. So it doesn't make sense to talk about "1,800 unique cases"—if there are 1,800 of them, they are not unique! You mean "1,800 distinct cases".) —Bkell (talk) 07:01, 14 July 2011 (UTC)

Perhaps this is the core of the problem: Your multiplication for counting possibilities isn't quite right. Let's consider a simple counting problem. Suppose I choose to write either "ABC" or "CDE" on a card, and then I choose one of the letters on the card to color red. How many possibilities are there for the final result on the card?

Clearly there are six: I have two choices for the word I write on the card, and then, for each of those two choices, I have three choices for the letter I color red. Two times three is six. This is called the rule of product. We can list out the six possibilities: ABC, ABC, ABC, CDE, CDE, and CDE.

Your method of counting in the probability problems above, however, seems to proceed like this: After the first step, there are two possibilities for what I have written on the card (ABC or CDE); this is true. After the second step, there are five possibilities for the letter colored red (A, B, C, D, or E); this is also true, in a way. Therefore, this method concludes, there must be 2×5=10 possible results. Of course, you can see that this is incorrect. The reason is that some of the possibilities in the second step (coloring a letter red) are impossible for some of the possibilities in the first step (choosing ABC or CDE): If I choose ABC in the first step, then I cannot possibly choose D or E to color red in the second step. So 2×5 counts too many.

Likewise, in the merry-go-round problem, you say that after the first step there are six possibilities (GGDDD, BGDDD×2, BGGDD×3), which is correct, and then you say that after the second step there are 30 possibilities. Now, I count 21 possibilities after the second step (one way to have GGDDD remaining, two ways to have BGDDD, three ways to have BGGDD, two ways to have GDDD, three ways to have GGDD, one way to have BDDD, six ways to have BGDD, and three ways to have BGGD), but either way, we do not have all 30 or 21 or however many possibilities for each one of the six possibilities in the first step (for instance, the possibility BGDD after the second step is impossible if the outcome of the first step was GGDDD), so multiplying these numbers together is incorrect. —Bkell (talk) 07:30, 14 July 2011 (UTC)

Thanks for your detailed and thorough responses; this really helps me get to the logic of your thinking, which I'm striving to see as valid. I truly appreciate the time you're taking here to explain things. There are a few things I'd like to mention, however: in my original card problem, there are only 4 cards, namely, 2 spades, 1 heart, and 1 diamond - you mistakenly put 5 cards, so your interpretation of the problem doesn't pertain to the problem I posed; furthermore, I have been following the rule of product every step of the way, which is why I know, at the end of the day, as in the case of the card problem, I know there are only 2 ways the desired result can appear (because there are 2 spades and 1 heart), and as in the case of the merry-go-round problem, I know there are only 6 ways the desired result can appear (3 dogs, 2 girls, and 1 boy, the product of which is 6); and, yes, my knowledge of technical terms in mathematics isn't precise, and I never attempted to use "unique" in the sense that would be appropriate here, as I only expect my meaning to get across through lay-terms, which I succeeded in doing, but thanks for pointing it out to me. Let me try to illustrate my thinking in a modified version of the card problem, and we'll just go from there:

There are 2 spades and 2 hearts. After the deck has been shuffled, a card is withdrawn, then a blank is inserted in its place, and the deck is shuffled once more. What is the probability that there will be one of each card remaining after 2 shufflings and withdrawals?

a=1, b=1; c=2, d=2 (4 items, 2 of 2 kinds)
(a=b=spade; c=d=heart)
1st	(one blank	{a,b,c}*2cases	{a,c,d}*2cases			= 4cases
	for all cases)	good		good

2nd	(two blanks	{a,b}*1case	{a,c}*4cases	{c,d}*1case	= 6cases
	for all cases)	bad		good		bad

Thus, there are 4 (rule of product and addition - there is 2 of one kind, and two different cases of this, so 2+2=4)
out of 4 cases that are good at the 1st withdrawal, so the probability is 1; there are 4 (rule of product -
there are 2 of each kind, so 2*2=4) out of 6 cases that are good, so the probability is 2/3 of obtaining 1 of each card.

Wherein lies the flaw with this mode of analysis? Isn't the probability of this occurring 2/3? If not, why exactly? Point out the step(s) which are erroneous to me and I'll try to see if the underlying logic of the criticism is correct or intuitively obvious.

In anticipation of your form of analysis, I'll try to emulate it: at the first step we have 4 in 4 chances of making a desired withdrawal, so 1/1, then, at the second, we have 2 in 3 chances of making a desired withdrawal, and the two chances together make: (1/1)(2/3) = 2/3, so the chance of making the desired outcome is 2/3. Have I followed you correctly? If so, I'm not sure why my method appears appealing to me, but I need to know why it's wrong. Numbility (talk) 14:49, 14 July 2011 (UTC)

One error in your analysis is that you have omitted the cases where the second draw draws a blank and replaces it with a blank, so the final deck is {blank, S, H, H} or {blank, S, S, H}. The probability of the final deck being {blank, blank, S, H} is 1/2. Gandalf61 (talk) 15:38, 14 July 2011 (UTC)

Yes, I see that, and hadn't yet posted about that, waiting for Bkell's verdict. Without loss of generalization, I can say then about the merry-go-round is that at the first spin, I have 5/6 good cases, then the second 4/6 or 2/3 good cases, and the third 2/6 or 1/3 cases; therefore, the probability that 1 dog, 1 boy, and 1 girl will remain is: 10/54 or 5/27. Correct? Where am I failing to get the supposedly correct 1/6? Numbility (talk) 17:07, 14 July 2011 (UTC)

I do not have five cards. I have four cards: A♠, 2♠, A♥, A♦. I also have a column in my table for a blank, but only for the second draw, because your problem stated, "a card is withdrawn and a blank is inserted in its place." Please note that I have labeled some entries in my table as "(impossible)", because according to the problem as you gave it, it is impossible to draw, say, A♠ twice in a row. So the table is really a 4×4 table—one cell in each row is not actually there. —Bkell (talk) 05:42, 15 July 2011 (UTC)

I don't know how to answer your questions any more clearly. Your method does not give the correct answer for the merry-go-round problem, and that should be convincing proof that it is not valid. Several people have explained why the answer is 1/6. I went so far as to explicitly list out all 216 possible outcomes and label them as good or bad, and you can see by brute-force counting that the probability is 36/216=1/6. If you still don't believe that the probability is 1/6, I don't know how to make it clearer to you. If you won't accept this complete listing of all possible outcomes, and a count of how many of them are "good", which is basically the fundamental definition of probability, then what will you accept as proof that the answer is 1/6?

For your two-spade/two-heart problem, the answer is 1/2, not 2/3. Here is another table, listing all possible outcomes, to show that:

		Second draw
		A♠	2♠	A♥	2♥	blank
First draw	A♠	(impossible)	BAD	GOOD	GOOD	BAD
	2♠	BAD	(impossible)	GOOD	GOOD	BAD
	A♥	GOOD	GOOD	(impossible)	BAD	BAD
	2♥	GOOD	GOOD	BAD	(impossible)	BAD

Here's how to read that table. If the first draw is A♠ (corresponding to the first row), then the second draw can be either 2♠, A♥, 2♥, or blank. Draws of A♥ and 2♥ lead to a desired outcome; draws of 2♠ or blank lead to an undesired outcome. The other rows should be interpreted similarly. I do not have "five cards" here—I have five columns because, overall, there are five things you might get on the second draw (A♠, 2♠, A♥, 2♥, or blank), so I have a column for each of them. Of course, for any one given card for the first draw, only four of these five are actually possible, so one column in each row is marked "(impossible)". [If you prefer, you can remove the "blank" column and reinterpret the intersection of, say, A♠ and A♠ to mean that the second draw actually produced a blank, thereby changing all of the "(impossible)" cells to "BAD" cells.] Looking at this table, it is clear that out of the 16 (not 20) possible outcomes, 8 of them are good, so the probability is 8/16=1/2. If your analysis gives 2/3 as the answer, then the analysis cannot be valid.

Gandalf61 has already pointed out a flaw in your analysis, namely, that you apparently failed to include the possibility of getting a blank in the second draw. If you enforce a rule that the blank must not be drawn in the second draw (perhaps by not putting the blank into the deck at all), then the probability is indeed 2/3. We can see that with another table listing all possible outcomes:

		Second draw
		A♠	2♠	A♥	2♥
First draw	A♠	(impossible)	BAD	GOOD	GOOD
	2♠	BAD	(impossible)	GOOD	GOOD
	A♥	GOOD	GOOD	(impossible)	BAD
	2♥	GOOD	GOOD	BAD	(impossible)

I have removed the "blank" column because now we are not adding a blank card after the first draw. In this table, out of the 12 (not 16) possible outcomes, 8 of them are good, so the probability of a good outcome is 8/12=2/3.

If you would like me to attempt to pinpoint specific flaws in your reasoning, please explain your analysis much more thoroughly, in words rather than in the tables and symbols you've been giving, and do so for one of the problems we've discussed rather than introducing a new one. Also, give as clear and precise a justification as possible for every step of your analysis. Remember, in mathematics, reasoning is not "correct until proven wrong"—it is "incorrect until completely justified from earlier established truths." —Bkell (talk) 06:13, 15 July 2011 (UTC)

Let me repeat again an important point about the rule of product. The rule of product says that, if there are m ways to perform one task, and, for every one of those m ways, there are n ways to perform a second task, then there are m×n ways to perform the two tasks together. The phrase "for every one of those m ways" is crucial here. If some of the m ways of performing the first task correspond to more or fewer than n ways to perform the second task, then the rule of product cannot be used.

Rephrased in terms of probabilities, the rule of product says that if one experiment has probability p of success, and, for any successful outcome of the first experiment, a second experiment has probability q of success, then the probability that both experiments are successful is pq. Again, the phrase "for any successful outcome of the first experiment" is crucial here.

Let's consider your analysis of the two-spades/two-hearts problem, ignoring the failure to consider getting a blank on the second draw. (In other words, I am going to consider your analysis as written; I am not going to insert a blank card into the deck after the first draw, so that the second draw will never produce a blank.) Your analysis seems to be flawed in its use of the rule of product. Your two experiments are the first draw and the second draw. The probability of success in the first draw is, as you say, 1. However, you then appear to claim that, for any successful outcome of the first draw, there are six possibilities for the second draw. This is false. Instead, for any successful outcome of the first draw, there are only three possibilities for the second draw, as follows:

• If the first draw is A♠, then the three possibilities for the second draw are 2♠, A♥, and 2♥. Two of these (A♥ and 2♥) are good, so the probability of success in the second experiment is 2/3.
• If the first draw is 2♠, then the three possibilities for the second draw are A♠, A♥, and 2♥. Two of these (A♥ and 2♥) are good, so the probability of success in the second experiment is 2/3.
• If the first draw is A♥, then the three possibilities for the second draw are A♠, 2♠, and 2♥. Two of these (A♠ and 2♠) are good, so the probability of success in the second experiment is 2/3.
• If the first draw is 2♥, then the three possibilities for the second draw are A♠, 2♠, and A♥. Two of these (A♠ and 2♠) are good, so the probability of success in the second experiment is 2/3.

Since, for any successful outcome of the first draw, the probability of success in the second draw is 2/3 (not 4/6, even though these are numerically equal), here is where we can use the rule of product to get that the final probability is 1×(2/3)=2/3.

The flaw in your analysis seems to be that you are considering all possibilities after the second draw and including all of them with each one of the possibilities after the first draw—even pairings that don't make sense (like pairing {A♠, 2♠, A♥} after the first draw with {A♠, 2♥} after the second draw—this pairing is impossible, because you can't have 2♥ remaining after the second draw if you didn't have it remaining after the first draw). When you consider only pairings that make sense, you will see that every possibility of the first draw should have three, not six, possibilities for the second draw. —Bkell (talk) 06:58, 15 July 2011 (UTC)

I would like to apologize for misreading a part of your note: I mistakenly thought you'd considered 5 cards, which is obviously wrong. Anyway, considering everything and the overwhelming evidence, I consider this issue resolved and the answer to the original problem 1/6. Thanks! Numbility (talk) 13:16, 15 July 2011 (UTC)

July 13

Dominion probability

What's the probability of getting a 5-2 copper split in the two opening hands of Dominion, in either order? (The deck starts with 7 coppers and 3 estates, and you draw 5 cards per hand, discarding all 5 after your turn.) --134.10.113.106 (talk) 00:25, 13 July 2011 (UTC)

There are 10 choose 5 = 252 possible first hands, and of those, 7 choose 5 = 21 are all copper. So 8.33% probability of your first hand being all copper. The same likelihood of your second hand being all copper, and since they're disjoint possibilities, they add to get 16.66% probability.--Antendren (talk) 01:17, 13 July 2011 (UTC)

Non-associativity of addition with infinite series

I've seen examples illustrating the non-associativity of addition with infinite series, but they all illustrate the property for divergent series only. Are there any examples of conditionally-convergent illustrating this property also?--Leon (talk) 07:17, 13 July 2011 (UTC)

No. Associating terms in a series is effectively selecting a sub-sequence of the sequence of partial sums. For any convergent series (even just conditionally convergent), the sequence of partial sums converges, and thus any sub-sequence will converge to the same value.--Antendren (talk) 07:28, 13 July 2011 (UTC)

It may be worth emphasizing that conditionally convergent series of reals can be rearranged (by permuting the terms) to achieve any real value. This is the Riemann rearrangement theorem. 173.75.158.107 (talk) 11:45, 13 July 2011 (UTC)

OK, and commutativity: the addition of terms of any conditionally convergent series cannot be commutative, owing to the Riemann rearrangement theorem; is this true of divergent series also? To cut to the chase, can we say the following about the arrangement of terms in series?

• Unconditionally convergent: always commutative and associative.
• Conditionally convergent: never commutative, always associative.
• Divergent: never commutative, not always (but sometimes) associative.

(if my reasoning is both incorrect and unclear, I think that since a conditionally convergent series, which is always associative, can be rearranged into a divergent one by a permutation, that some divergent series may inherit that property) --Leon (talk) 14:17, 13 July 2011 (UTC)

Depends what you mean by a divergent series being associative or commutative. For example, consider the series 1 + 1 + 1 + 1 + 1 + 1 + ... No matter how you associate terms or rearrange terms, the series will still diverge to positive infinity. You might call this series associative and commutative (of course, some would call this series convergent).--Antendren (talk) 21:08, 13 July 2011 (UTC)

Given that by associating terms of this series to get 1 + 2 + 3 + 4 + … or 1 + 2 + 4 + 8 + …, is it really correct to say that it is associative?--Leon (talk) 13:22, 14 July 2011 (UTC)

Well, that depends on if you say those series also sum to positive infinity, or if you are using a summability method to assign a finite value to them. If the former, it's not a counterexample to associativity. If the latter, we're getting into an area I don't know much about, so I can't speak on it.--Antendren (talk) 02:51, 15 July 2011 (UTC)

Cute way to evaluate a series

The series $\sum _{n=0}^{\infty }{\frac {n+1}{2^{n}}}$ can be evaluated by arranging the dyadic fractions into a triangular array:

${\begin{matrix}1&{\frac {1}{2}}&{\frac {1}{4}}&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&{\frac {1}{2}}&{\frac {1}{4}}&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&&{\frac {1}{4}}&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&&&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&&&&\cdots &\cdots \end{matrix}}$

The sum down the $n$th column is $(n+1)/2^{n}$, so the value of the series is the sum of the entries in the array. The sum of the $n$th row is $1/2^{n-1}$, so the sum of the series is $\sum _{n=0}^{\infty }{\frac {1}{2^{n-1}}}=4$.

This trick works for any series of the form $\sum (n+1)x^{n}$. Although we can certainly evaluate this by less elementary methods, this is the most elegant way I've seen to compute the series. My question is, presumably this trick is well-known. Does it have a home in some article in Wikipedia? (Maybe in the article about the series itself, but where would that be?) Sławomir Biały (talk) 12:00, 13 July 2011 (UTC)

More generally, assuming everything converges

$\sum _{n=0}^{\infty }\left(\sum _{i=0}^{\infty }f_{i}\right)g_{n}=\sum _{i=0}^{\infty }f_{i}\left(\sum _{n=i}^{\infty }g_{n}\right).$

This is basically changing the "order of summation", a discrete version of order of integration (calculus). I found this at wolfram.com which seems to cover it, look under the heading "Double infinite summation", but I can't find it here. Maybe this will help refine the search a bit.--RDBury (talk) 13:49, 13 July 2011 (UTC)

Yes it's easy to justify the interchange in order of summation. (For instance, we could clobber it with Tonelli's theorem.) The question is whether writing this single series as a double series is a well-known/notable trick. Sławomir Biały (talk) 14:36, 13 July 2011 (UTC)

Just thinking that a more general term would be easier to find. Another generalization is

$\sum _{n=0}^{\infty }a_{n}x^{n}={\frac {1}{1-x}}\left(a_{0}+\sum _{n=1}^{\infty }(a_{n}-a_{n-1})x^{n}\right)$

which is like step 1 of a binomial transform. The full binomial transform would give you a finite sum in your example. I'm sure people like Euler and Gauss used this kind of thing all the time, they may have though it too trivial to write out though. It's exactly the kind of thing you need to work with hypergeometric series for example.

Hitting it with Tonelli's theorem seems like hunting for rabbits with nuclear tipped cruise missiles.Naraht (talk) 19:10, 13 July 2011 (UTC)

That was Sławomir's point. — Fly by Night (talk) 19:26, 13 July 2011 (UTC)

If you generalize the technique you get

$\sum _{k=0}^{\infty }f_{k}g_{k}=$

${\begin{matrix}f_{0}g_{0}&+f_{0}g_{1}&+f_{0}g_{2}&+f_{0}g_{3}&+f_{0}g_{4}&\cdots \\&+(f_{1}-f_{0})g_{1}&+(f_{1}-f_{0})g_{2}&+(f_{1}-f_{0})g_{3}&+(f_{1}-f_{0})g_{4}&\cdots \\&&+(f_{2}-f_{1})g_{2}&+(f_{2}-f_{1})g_{3}&+(f_{2}-f_{1})g_{4}&\cdots \\&&&+(f_{3}-f_{2})g_{3}&+(f_{3}-f_{2})g_{4}&\cdots \\&&&&\cdots &\cdots \end{matrix}}$

$=f_{0}\sum _{k=0}^{\infty }g_{k}+(f_{1}-f_{0})\sum _{k=1}^{\infty }g_{k}+(f_{2}-f_{1})\sum _{k=2}^{\infty }g_{k}+\dots$

which is the first equation in Summation by parts#Newton series, putting n=infinity. So the answer I'm going with for the judges is "summation by parts". Not sure where Tonelli's theorem comes in with that idea though.--RDBury (talk) 21:42, 13 July 2011 (UTC)

Nice. Sławomir Biały (talk) 22:09, 13 July 2011 (UTC)

Need name for a distribution

Hi. The function $f(x)=\exp(x)/(1+\exp(x))^{2}$ defines a probability density function. Does anyone know a name for it? Robinh (talk) 21:36, 13 July 2011 (UTC)

Try logistic distribution.--RDBury (talk) 21:57, 13 July 2011 (UTC)

That's it! Easy if you know. Best wishes, Robinh (talk) 23:30, 13 July 2011 (UTC)

Resolved

quotation

how can this possibly be right:
"All the taxes paid over a lifetime by the average American are spent by the government in less than a second." That would mean after 1 second, there would have to be a one-lifetime deficit, and a year has 31 million seconds. So, even if a lifetime is a paltry "20 years", that would be 20 * 30 = 600 million times that average annual earnings in defiicit. If the annual earnings is even $10,000 then that would mean there would be a 6 trillion dollar deficit. In fact the deficit is only 40billion (a huge difference of over 100-fold, especially considering that a lifetime is surely more than 20 years and the average salary surely more than 10k per annum). so how can that quotation possibly be right? 188.222.102.201 (talk) 23:12, 13 July 2011 (UTC)

"Spending" here probably means budget spending, not deficit spending. The annual budget is about three and a half trillion, or ten billion per day, four hundred million per hour, seven million per minute, one hundred thousand per second. A person paying $2500 tax per year would total one hundred thousand in a 40 year career. So it seems about right to me.--RDBury (talk) 23:37, 13 July 2011 (UTC)

I think you're confusing average with total, because your calculation doesn't take into account the size of the population. As you say, assuming a 20 year lifetime, the quote says the government spends 600 million lifetimes in 20 years. But there are 300 million people in the US, so we generate 300 million lifetimes in 20 years. That's only a 2:1 ratio of spending to earning. Since lifetimes are closer to 80 years, the quote actually gives us a ratio of roughly 8:1.--Antendren (talk) 23:43, 13 July 2011 (UTC)

Working lifetimes are not so long. —Tamfang (talk) 22:51, 15 July 2011 (UTC)

Rewriting an absolutely convergent series

Related to the above question, an absolutely convergent series is said to also be "unconditionally convergent", ie, satisfy the property that any permutation of the terms of the series will still yield the same sum. But this isn't the only way to rewrite the sum. What if you take a series and convert it to an arbitrary well-ordered sum, with the obvious requirement that there be a bijection between the terms of the well-ordered sum and the original series? Is that guaranteed to converge to the same value as the original series? --COVIZAPIBETEFOKY (talk) 23:32, 13 July 2011 (UTC)

Yes. You can start by proving it for series of all non-negative terms, by induction on the order-type of the sum: if the well-ordered sum converges to something less than the original series, grab finitely many terms from the original series which sum to more than the well-ordered sum, deriving a contradiction. If the well-ordered sum converges to something greater than the original series, necessarily the order-type is a successor ordinal $\alpha +1$. But then since finite sums are commutative, you can move the last summand to the front of the series, and get a well-ordered sum of order-type $\alpha$ which converges to more than the sum of the original series, contradicting the inductive hypothesis.

Next, you can basically repeat the proof that absolute convergence implies unconditional convergence: grab finitely many terms that bring the sum of the original series within epsilon, bound the remainder of the well-ordered sum by the sum of the absolute values. I suppose this requires first showing that finite rearrangement does not change well-ordered sums, and that a well-ordered sum is bounded by the sum of its absolute values, but those are simple enough.--Antendren (talk) 00:23, 14 July 2011 (UTC)

Thanks, that mostly makes sense. I'm not clear on why the order type of a sum which converges to something greater than the original series must be a successor. --COVIZAPIBETEFOKY (talk) 01:11, 14 July 2011 (UTC)

Sorry, it's based on the inductive hypothesis again. Since the sum at a limit ordinal is just the supremum of the sums at previous ordinals, if it happened at a limit ordinal, there must be some previous ordinal where the partial sum is greater than the sum of the original series. But the partial sum at this ordinal corresponds to the rearrangement of a subseries of the original series, and the sum of a subseries (of non-negative terms) is at most the sum of the original series, so this contradicts the inductive hypothesis.--Antendren (talk) 01:29, 14 July 2011 (UTC)

Another way of putting this is to compare the well-ordered sum with the Lebesgue integral (treating your sequence as a function from a discrete measure space where every point has mass 1). You can prove by induction on α that the well-ordered sum up to α equals the integral of the function restricted to ordinals less than α, using additivity of the integral at successors and dominated convergence at limits. Algebraist 08:53, 14 July 2011 (UTC)

Thanks a lot! --COVIZAPIBETEFOKY (talk) 13:03, 14 July 2011 (UTC)

Resolved

July 14

Constrained minimization

Suppose a plant produces $q$ engines each week using $K$ assembly machines and $L$ workers.

The number of engines produced each week is

$q=3K^{1/2}L^{1/2}$

Each assembly machine rents for $r$ dollars per week and each worker costs $w$ dollars per week.

Part one of the question is, how much does it cost to produce $q$ engines, if the plant is cost-minimizing?

I am assuming that this is a constrained maximization problem where we need to maximize $-rK-wL$ subject to $q=3K^{1/2}L^{1/2}$

I'm trying to solve this with the Lagrangian $G=-rK-wL-\lambda (q-3K^{1/2}L^{1/2})$

that is, with

$G_{K}=-r-\lambda ({\frac {3}{2}}L^{1/2}K^{-1/2})=0$

$G_{L}=-w-\lambda ({\frac {3}{2}}L^{-1/2}K^{1/2})=0$

$G_{\lambda }=q-3K^{1/2}L^{1/2}=0$

I think I worked out from the first two conditions that $\lambda =-w\left({\frac {2}{3}}K^{-1/2}L^{1/2}\right)=-r\left({\frac {2}{3}}K^{1/2}L^{-1/2}\right)$ and so $L={\frac {r}{w}}K$

Then from the last condition $q=3K^{1/2}L^{1/2}=3K\left({\frac {r}{w}}\right)^{1/2}$

I get $K={\frac {q}{3}}\left({\frac {w}{r}}\right)^{1/2}$ and $L={\frac {q}{3}}\left({\frac {r}{w}}\right)^{1/2}$

So the overall cost function is

$cost=rK+wL={\frac {2q}{3}}w^{1/2}r^{1/2}$

Have I done this correctly, or is there another way to do this?

I am also asked what are the average and marginal costs for producing $q$ engines.

For the average cost I presume you just divide the total cost by $q$, while for the marginal cost you take the derivative of the total cost with respect to $q$

In this case the average and marginal costs are the same. Does this indicate constant returns to scale?

118.208.40.147 (talk) 02:28, 14 July 2011 (UTC)

The easiest way to solve this is to simply substitute $L={\frac {q^{2}}{9K}}$ in $rK+wL$ and find the unconstrained extremum of this single-variable function. Using this I got the same result as you, so your derivation is probably correct. And yes, this result indicates that the cost is linear. -- Meni Rosenfeld (talk) 09:00, 14 July 2011 (UTC)

Matrix square root

I have read the Square root of a matrix article.

I have a real square positive-definite but possibly non-symmetric matrix M, and I seek a (preferably real) matrix S such that transpose(S)*S=M (the * indicates matrix product). I don't have to be able to construct S, it suffices to know that it exists (does it?).

Cholesky decomposition will not do, as the matrix M is not guaranteed to be symmetric.

A square root based on diagonalization won't help either, as that would give S*S=M and not transpose(S)*S=M

The Square root of a matrix article writes further

"In linear algebra and operator theory, given a bounded positive semidefinite operator (a non-negative operator) T on a complex Hilbert space, B is a square root of T if T = B* B, where B* denotes the Hermitian adjoint of B. According to the spectral theorem, the continuous functional calculus can be applied to obtain an operator T^½ such that T^½ is itself positive and (T^½)² = T. The operator T^½ is the unique non-negative square root of T. "

This seems very close to what I'm looking for, but, alas, "citation needed". I know next to nothing about continuous functional calculus and that article isn't very helpful.

213.49.89.115 (talk) 17:47, 14 July 2011 (UTC)

If M is non-symmetric then there is no such matrix S. Let M be a non-symmetric matrix, and assume that there exists S such that M = S^TS. Notice that S^TS is symmetric because (S^TS)^T = S^TS^TT = S^TS. If M = S^TS then M must also be symmetric; which is a contradiction. — Fly by Night (talk) 01:47, 15 July 2011 (UTC)

Thanks! (and my apologies for the stupid-in-hindsight question :) ) 213.49.89.115 (talk) 17:10, 15 July 2011 (UTC)

It's not a stupid question at all. It's just that you needed to add the assumption that M is symmetric. In fact, you need det(M) to be non-negative (which it is because M is positive definite, and so all of its eigenvalues are positive). Moreover, you need all of the entries on the leading diagonal to be non-negative too. These are all necessary conditions, but I'm not sure if they are sufficient. The bottom line is that M being positive definite is not enough for you to be able to solve M = S^TS. Even if M is positive definite, symmetric, and has non-negative entries along the leading diagonal; I'm not sure that that's enough. I'd be interested to see what progress you make. — Fly by Night (talk) 22:31, 15 July 2011 (UTC)

July 15

Where do we find linear equations?

The article about them says " rise quite naturally when modeling many phenomena,". What are these "many phenomena"?

One example is Hooke's law, another is the relationship between time and velocity under a constant force. Basically just open a science textbook at random and start looking. I tried a physics textbook and found the relationship between area and the force of pressure in a body of fluid.--RDBury (talk) 13:06, 15 July 2011 (UTC)

A huge number of situations where you have oscillations about an equilibrium point are well described by linear equations when the amplitude of oscillation is small. These include waves, vibrations, the movement of a pendulum or spring, etc. Looie496 (talk) 17:46, 15 July 2011 (UTC)

Another couple of examples: Ohm's law gives a linear relationship between voltage and current, assuming a constant resistance, and the hydrostatic pressure of an incompressible fluid, such as water pressure in the ocean, is a linear function of depth. —Bkell (talk) 20:22, 16 July 2011 (UTC)

Typically everything behaves linearly for small deviations when away from special points. That's practically the first assumption any scientist makes in any discipline. Singularity theory is the study of the next more complicated situation in general. Dmcq (talk) 20:37, 16 July 2011 (UTC)

Linear equations give first order approximations to any system. That is why they are so ubiquitous. The truth is that no physical theory is totally correct, and that they are all approximations. Every known physical theory breaks down outside of some set of parameters. Look at Newton's Laws. These can be viewed as (very, very good) approximations to what actually happens; but when things get really, really small, Newton's Laws fall apart. The truth is that linear equations turn up in every system: as a linear approximation to that system. — Fly by Night (talk) 23:03, 16 July 2011 (UTC)

That's a little bit of an overstatement. Lots of important systems (particularly in biology) show limit-cycle oscillations, which are not usefully modeled by linear equations. Looie496 (talk) 23:39, 16 July 2011 (UTC)

I didn't comment on the validity of the approximations. — Fly by Night (talk) 02:56, 17 July 2011 (UTC)

:-) Looie496 (talk) 17:00, 17 July 2011 (UTC)

What are these number tringles called?

When looking at Floyd's triangle on Wikipedia, i mentally replaced each counting number n with the nth odd number and found the kth row added to k^3. (This is easy to prove)

• 1
• 3 5
• 7 9 11
• 13 15 17 19
• 21 23 25 27 29
• ....

But related is

• 1
• 5 9 13
• 17 21 25 33 37
• 41 45 49 53 57 61 65

....

and

• 1
• 7 13 19 25
• 31 37 43 49 55 61 67
• ....

and generally

• 1
• 2m+1 4m+1 6m+1 ... 2(m)(m+1)+1
• ..... ... 2(m)(3m+2)+1
• ...

All of these triangles have nice cube sums on the rows. These aren't hard to prove either. I'm certain this is wellknown but I haven't found a name for them. In any case, they seem at least as interesting as floyd's triangle, so i'd like to get them on Wikipedia.Thanks, Rich Peterson199.33.32.40 (talk) 18:41, 15 July 2011 (UTC)

See [1] for the first traingle. We've got most of that material in Squared triangular number but with different emphasis. I searched the OEIS for the others but didn't turn up anything. The OEIS turns up most sequences that have been published.--RDBury (talk) 20:26, 15 July 2011 (UTC)

July 16

are numbers sets?

In either naive or axiomatic set theory, are numbers themselves considered sets? And if so, what elements do they contain? Aacehm (talk) 17:24, 16 July 2011 (UTC)

Natural numbers are often defined as sets of all natural numbers less than them. Zero is the empty set ($\emptyset$), 1 is the set containing 0 ($\{0\}=\{\emptyset \}$, 2 is the set containing zero and 1 ($\{0,1\}=\{\emptyset ,\{\emptyset \}\}$), etc.. --Tango (talk) 18:05, 16 July 2011 (UTC)

I can almost remember from 25 years ago what purpose this definition served. Does anyone else? 157.22.42.3 (talk) 00:09, 17 July 2011 (UTC)

I believe the definition comes from Principia Mathematica (or perhaps Frege); the aim was to construct mathematics from the smallest possible set of basic concepts and axioms. Looie496 (talk) 00:47, 17 July 2011 (UTC)

The natural numbers are defined using the Peano axioms (or variations upon them). Once you have a set of axioms, it can be helpful to find a model for them. It has become common to try and define all mathematics in terms of ZF set theory, and the model I described is a natural way to model the Peano axioms in ZF set theory. One thing finding such a model does is prove that the axioms are consistent within the theory - if they weren't, then you would have a big problem. --Tango (talk) 02:37, 17 July 2011 (UTC)

So It's possible to write the number 3 as something like this: {{},{{}},{{},{{}}}}? Aacehm (talk) 15:36, 17 July 2011 (UTC)

That's 2, (the empty set is 0), but yes, the idea is right. Looie496 (talk) 16:59, 17 July 2011 (UTC)

No 3 in that interpretation,{} is the empty set and corresponds to 0, and {{}} which is the set containing the empty set corresponds to 1. Dmcq (talk) 18:03, 17 July 2011 (UTC)

clothing with clothoids

I've lately become interested in Cornu spirals. Am I right in supposing that they can be determined by four points? How about replacing one or more with a tangent (not osculating) line or circle? (I don't mind if a solution is not unique, so long as the set of solutions is finite.) Where might I look for procedures? —Tamfang (talk) 00:12, 16 July 2011 (UTC)

Do you mean at most four points in general position? A straight line and a circle are both Euler spirals. Any two distinct points determine a unique straight line, while three non-linear points uniquely determine a circle. — Fly by Night (talk) 01:02, 16 July 2011 (UTC)

I don't know anything about these curves, but I imagine what Tamfang is saying is that given any four distinct points there is exactly one Cornu spiral that passes through all four. (Whether or not this is true, I don't know.) Sure, a straight line is determined by two distinct points, as long as you also have the information that it's a straight line; but if I just give you two distinct points on a Cornu spiral, you don't know whether the spiral is a straight line between them or a different shape of curve. Presumably, if I'm correctly interpreting what Tamfang is saying, you'd need to be given four collinear points to be sure that the curve is a straight line and not something else. Now, that actually seems to be false to me: I can pick out four collinear points on the curve in the illustration in the Euler spiral article. —Bkell (talk) 04:17, 16 July 2011 (UTC)

Yeah, that's true. — Fly by Night (talk) 04:53, 16 July 2011 (UTC)

Ah but can you pick out four arbitrarily spaced collinear points? Anyway, as I said, I'll be satisfied with a finite number of solutions. —Tamfang (talk) 06:24, 16 July 2011 (UTC)

It seems that Euler spirals have five degrees of freedom. According to Kimia, B. B.; Frankel, I.; Popescu, A. M. (2003). "Euler Spiral for Shape Completion" (PDF). International Journal of Computer Vision. 54 (1/2/3): 159–182. you need only specify a point p (two numbers), the angle of the tangent at p (one number), the curvature at p (one number), and the rate of change of the curvature at p (one number). Take a look at Proposition 1 at the bottom right of page 164, and the top left of page 165. For example, you could specify two points, say p and q, (giving four numbers) and then specify the direction of the tangent line at either p or q (one number). If you want an Euler spiral that passes through four fixed points then you are imposing eight conditions. That means that if you had a three parameter family of a quadruple of points, then you'd expect only one quadruple in the family to have an Euler spiral passing through each of the points. — Fly by Night (talk) 18:41, 16 July 2011 (UTC)

Every Euler/Cornu spiral has the same shape, differing from the "unit" spiral by rotation (one number), scale (one number) and translation (two numbers). The fifth apparent degree of freedom is consumed in positioning p along the curve. With four points, four of the eight degrees of freedom are thus consumed. Otherwise you wouldn't always be able to draw a line (two DoF) through two points (four numbers), or a circle (three DoF) through three points (six numbers).

Incidentally, today I sat down to compute the integral of squared derivative of curvature in Knuth's cubic, which is misstated in the article (p.162, eqn 1); but the expression soon became big enough that I felt more likely than not to make a blunder in the algebra .... someday I'll buy Mathematica. —Tamfang (talk) 09:46, 17 July 2011 (UTC)

I see what you mean. There is some quotient involved. For example, you can specify a unique line by a base point and an angle, but that same line is given by any base point on the line and the same angle. The space of unoriented lines in the plane is parametrised by the projective plane minus a point, i.e. is homeomorphic to the Möbius band; which is locally two dimensional. The paper says that you need five numbers to give an Euler spiral, $(x,y,\theta ,\kappa ,\kappa ').$ But there is some kind of quotient. For example, if we reverse the orientation then $(x,y,\theta ,\kappa ,\kappa ')\sim (x,y,\theta ,-\kappa ,-\kappa ').$ This seems like a non-trivial problem. Consider the space all initial conditions, which will be $\mathbb {R} ^{5}$, but then identify all points that give the same spiral (as a point-set). Like I said all the points $(x,y,\theta ,\kappa ,\kappa ')\sim (x,y,\theta ,-\kappa ,-\kappa ').$ But then we can "slide" $(x,y)$ along the curve, which changes $(\theta ,\kappa ,\kappa ')$, and all such points need to be identified. I recommend the book Olver, P. J. (2009). Equivalence, Invariants and Symmetry. Cambridge University Press. ISBN 0521101042., and especially the chapter about Jet and Contact transformations. — Fly by Night (talk) 20:32, 17 July 2011 (UTC)

P.S. What is the correct statement of Knuth's cubic, and how does the paper misstate it? — Fly by Night (talk) 21:41, 17 July 2011 (UTC)

$3s^{2}-2s^{3}+r_{0}e^{i\theta _{0}}s(1-s)^{2}-r_{2}e^{-i\theta _{2}}s^{2}(1-s)$ : a 'minus' was moved from within an exponent to immediately after it, splitting the last term into two. —Tamfang (talk) 04:45, 18 July 2011 (UTC)

1 radian as a pure number

The article Radian per second says that 2π rad/s = 1 Hz. Since the SI unit Hz is equal to (1/s), could we not divide each side by (1/s) and get 2π rad = 1 $\Rightarrow$ 1 rad = ${\frac {1}{2\pi }}$ which is a pure number not carrying any units? Widener (talk) 00:54, 16 July 2011 (UTC)

A radian is a pure number already, just like the 1 in the numerator of hz. — Carl (CBM · talk) 00:56, 16 July 2011 (UTC)

But is that pure number that the radian represents indeed the number ${\frac {1}{2\pi }}$?Widener (talk) 01:29, 16 July 2011 (UTC)

Yes, if you take the "1" in the numerator to mean "1 cycle". The identity you started with (2π rad/sec = 1 hz) is only sane when you read the "1 hz" to mean "1 cycle hz", that is, "1 cycle per second". You can do all sorts of crazy things with dimensionless units if you don't take care, for example (using the fact that radians are dimensionless):

2π rad/s = 1 hz = (1 rad) hz = 1 rad/s

This is what the article on radians per second is warning about when it talks about the importance of distinguishing angular frequency and ordinary frequency, both of which are dimensionally measured in hertz but which measure different things. Someone else may have a different way of explaining why it's nonsense. — Carl (CBM · talk) 01:52, 16 July 2011 (UTC)

I prefer not to use the term "radian" at all. The angle you are talking about is simply "1" and a full circle is just "2π". Angles don't have units, they are just numbers - they are defined as the ratio of two lengths (radius and arc length), which is just a number. The only time you might need to use the word "radian" is to make it clear that you aren't working in degrees, but that's almost never the case in maths beyond school-level since degrees are a completely arbitrary and useless concept from a mathematical point of view. --Tango (talk) 18:10, 16 July 2011 (UTC)

I respectfully disagree with Tango. The radian and the turn are both perfectly natural units of measurement of angle. See Angle#Units. The turn per second has the name hertz, but the radian per second has got no such special name. The rpm is also a commonly used unit for frequency. Angular frequency and ordinary frequency is the same concept, but measured in different units. Wavelengths are usually per wave rather than per radian. The unit of the planck constant is Js, meaning joule per hertz rather than in joule per radian per second. We cannot close our eyes to the fact that the radian is not the only unit of angle around. Bo Jacoby (talk) 19:22, 16 July 2011 (UTC).

Ok, turns can be useful. I stand by my statement that degrees aren't, though. Radians are still more natural mathematically than turns - a lot of useful formulae need an extra 2π in them if you work in turns rather than radians (some, of course, allow you to get rid of a 2π, but far fewer as far I as am aware). Waves don't really involve angles, but rather things that are analogous to them, so whether we throw a 2π into the analogy or not is really a matter of choice (basically, a choice between using k or ω). As for the Planck constant, a lot of physicists prefer the "reduced Planck constant" (denoted ħ), which is in radians. --Tango (talk) 23:32, 16 July 2011 (UTC)

When the unit is omitted we pay the price of having several notations for the same concept, ν and ω for frequency, h and ħ for planck constant. E=hν and E=ħω. The "reduced Planck constant" is not reduced but merely measured in another unit. You don't drive slower by measuring your speed in miles per hour rather than kilometers per hour, and an oscillation is not faster by being measured in rad per second rather than in turn by second. So, avoid confusion by saying radian, if that is what you mean, or turn or period or cycle or wave or tau (2π), if that is what you mean. The degree is probably here to stay for yet a while, even if we don't like it very much. Bo Jacoby (talk) 05:29, 17 July 2011 (UTC).

Taylor convergence

I'm working on code that uses a Taylor series to approximate $\int _{0}^{x}e^{f(t)}dt$, where f is a polynomial. The series coefficients $a_{n}$ are generated as follows:

$a_{0}=0$

$g\leftarrow e^{f(0)}$ (a constant polynomial)

$n\leftarrow 1$

loop

$a_{n}=g(0)$

$n\leftarrow n+1$

$g\leftarrow (g^{\prime }+gf^{\prime })/n$

endloop

I know the code is correct (though I could have mistranslated it to math notation here!). I've been arbitrarily using 100 terms. To speed it up, I want to estimate the remainder:

$L=|t_{max}|$

$k=|f^{\prime }(0)|$

$M=kL^{1+m}$ where m is the degree of f′

loop

...

$L\leftarrow ML/n$

if k*L < tolerance: break

...

endloop

Am I on the right track? —Tamfang (talk) 15:01, 16 July 2011 (UTC)

I think this is only reasonable if x is small. Otherwise think about what will happen if $f^{\prime }(0)=0$. Looie496 (talk) 17:48, 16 July 2011 (UTC)

Oops, I think I meant k to be the highest-ranking coefficient of f', rather than its constant term. —Tamfang (talk) 17:56, 16 July 2011 (UTC)

... I used $M=f_{a}^{\prime }(t_{max})$, where $f_{a}$ is f with each coefficient replaced by its absolute value. Didn't work. :( —Tamfang (talk) 19:46, 17 July 2011 (UTC)

And then I saw my blunder and changed M to $t_{max}f_{a}^{\prime }(t_{max})$. Now it works beautifully.

July 17

Time and angles, Similar Increments?

Why are the increments for angles and for time similar in name and quantity? eg. Angles - 60 seconds = 1 minute, 60 minutes = 1 degree

                    and
   Time   - 60 seconds = 1 minute, 60 minutes = 1 hour