Talk:Birthday problem

"Near matches" why so much trouble?

Can't you just divide a year in 52 weeks (52.17 or whatever) and use the same approach you do with 365 days? —Preceding unsigned comment added by 193.198.81.100 (talk) 21:31, 28 April 2009 (UTC)

"Near matches" - careless comment?

It is very careless to say that

"Thus in a family with six members, it is more likely than not that two members will have a birthday within a week of each other."

While i realise that it's tempting to make the group of six people a family, birthdays within a family are generally far from independent, especially if siblings are included!

Given the article it belongs to, i think the statement is out of place.

—The preceding unsigned comment was added by Rileen (talk • contribs) 12:51, 9 August 2006 (UTC).

Good point. Fixed. Thanks. --Keeves 18:33, 9 August 2006 (UTC)

Glad to see that - this was my first (which is why i also forgot to mention my name) teeny-weeny contribution to Wikipedia. Thanks! --Rileen, 11 August 2006

This is a valid statement within the context of the 'uniform birthday distribution' ideal set forth at the start of article. —The preceding unsigned comment was added by Qe2eqe (talk • contribs) 18:17, 17 April 2007 (UTC).

29 February

It makes the article significantly more complex to take into account 29 February, so most of it doesn't — except for bits of the first paragraph. I think the article should simply state early on that 29 February will be completely ignored, and then it should abide by that.

Does anyone object to this?

Ruakh 17:09, 13 August 2006 (UTC)

I had thought of that exact idea yesterday, and I'm glad you suggested it. I have now added such a paragraph, and I hope it will be noticable enough to bring this back-and-forth to an end. --Keeves 21:32, 13 August 2006 (UTC)

Hmm. Your paragraph is very clear, but I worry that it might be a bit much; after all, it's not really the central point of the article, but really just a caveat clarifying that the article is about an abstract mathematical concept and does not necessarily apply perfectly to real life. It seems like a single sentence could conceivably suffice. (Further, it's not true that "it is just as likely that a randomly-chosen person's birthday will be January 21 or October 5 or almost any other day of the calendar"; different times of year actually have fairly different birth rates.) Ruakh 03:12, 14 August 2006 (UTC)

A footnote should be more then sufficent to account for the leapyear birthdays and just have a further explaination thereof. —The preceding unsigned comment was added by 12.163.97.74 (talk • contribs).

Please see "Another Birthday Paradox" below.Cuddlyable3 (talk) 11:15, 3 October 2008 (UTC)

Mystery!

"23"! --nlitement ^[talk] 13:37, 30 September 2006 (UTC)

Approximation based on Taylor Series expansion

Isn't the expression obtained for p_bar(n) =approx= 1*e^-1/356*e^-2/365*...*e^-(n-1)/365 in the "Approximation" subsection based on the inequality "(1-x) < (e^-x)"? It doesn't seem to be based on the Taylor Series expansion of the exponential function; instead, it directly uses the exponential function. —The preceding unsigned comment was added by Wiki user 618 (talk • contribs) 03:13, 9 November 2006 (UTC).

But exp(-x) = 1 - x + x^2/2 - ... IS the Taylor expansion of exp(-x). Cut the expansion off after the term in x, and you get exp(-x) = 1-x. You can see that 1-x < exp(-x), by looking at the quadratic term of the Taylor expansion (the Taylor expansion is an alternating series). So you are basically saying the same thing, but the article actually shows why the inequality holds. —The preceding unsigned comment was added by 134.58.253.131 (talk • contribs) 11:30, 12 February 2007 (UTC).

Near matches?

Can we either cite or insert the math for the "near match" birthdays? The information is stated without proof. arctic 00:28, 27 November 2006 (UTC)

You're right to be concerned; I just wrote a quick Perl script to investigate this probabilistically, and it seems like the correct values for the table would be:

within k days	# people required
0	23
1	14
2	11
3	9
4	8
5	8 ← not 7, as the page currently has
7	7 ← not 6, as the page currently has

(Note: that was only probabilistic, but I used some basic techniques to ensure accurate results, so I'm fairly confident those are the correct values. As you say, though, a source or mathematical justification is quite necessary here. BTW, lest anyone be concerned that my random number generator might be biased — that's a valid concern, and it might well be, but if that were affecting the results, you'd typically expect greater synchronicity, hence lower values for the number of people necessary to ensure a 50% probability of a near match. The only bias that would produce higher values is a bias against synchronicity; that is, if the generator has a greater-than-random tendency against yielding recently-yielded numbers. I'll run some tests to ensure that that's not the case here, then comment back.)

Ruakh 01:49, 27 November 2006 (UTC)

Okay, I ran some tests, and it doesn't look like my random number generator has any bias against synchronicity. So, I think the table in the article is wrong. Ruakh 02:11, 27 November 2006 (UTC)

The table was indeed wrong, so I corrected it (and cited a source for the data). It's unfortunate that the error remained so long. --Sopoforic 03:16, 21 February 2007 (UTC)

Collision counting

This was useful to me testing a password generation programme, and I couldn't find anywhere else in Wikipedia that gave this information. The hash collision article could perhaps link directly to this section.

I suggest that the section's usefulness would be improved by the addition that for 1 << n << d, as is typical, the formula reduces to $n^{2}/2d$.

Also, knowing the standard deviation would be useful, as I'd like to assess how reasonable my observed number of collisions is. Jlittlenz 05:07, 8 December 2006 (UTC)

I am also interested in more detail on the counting issue. In particular, rather than just knowing the expected number of collisions, I'm wondering what is the distribution of the maximum number of collisions for any particular day across the year. To put it in hash terms, if you do get a hash collision, you typically use a tree or a linked list to store all the data with the same hash value. The question is, how many items do you have in your tree or linked list, in particular how long is the longest list going to be? It would be good to have a formula p(n,d,m) where n is the number of people/hash entries, d is the number of days per year/size of the hash table, and m is the maximum number of collisions. For example p(1,d,0) = 1 for all d, since there will be 0 collisions with only one person! Conversely p(1,d,m) = 0 for all m > 0, more generally p(n,d,m) = 0 for all m >= n—Preceding unsigned comment added by 213.123.216.147 (talk) 13:38, 18 October 2007 (UTC)

Forward vs. reverse?

Is there any reason for the notion that the "birthday problem" is computing p given n, while the "reverse birthday problem" is computing n given p (as implied in #Reverse problem)? Insofar as the two are separate problems, it seems to me that the ordinary birthday problem is computing the minimal n such that p > ½, so the "reverse birthday problem" would be computing p given n. —Ruakh_TALK 16:23, 8 December 2006 (UTC)

366 People isn't necessarily 100%

In the article, it claims, although it cannot actually be 100% unless there are at least 366 people.[1] However, the real number should be 367, because there are 366 days in a leap year. Someone could be born on February 29th, and then no one would have the same birthday out of 366 people. Correct me if I'm wrong, Thanks, RAmen, Demosthenes 21:48, 15 February 2007 (UTC)

Yes, you are correct, but check the footnote. We're assuming 365 days in a year for sake of simplicity in calculations. — Edward Z. Yang^(Talk) 22:48, 15 February 2007 (UTC)

Featured Article

I think this is an interesting subject, great article i think it's featured article quality comments? I dont know how to tag it for that status myself.. —The preceding unsigned comment was added by 83.49.105.6 (talk) 20:56, 12 March 2007 (UTC).

There are instructions at WP:FAC. Greeves ^{(talk • contribs)} 15:50, 31 March 2007 (UTC)

Conundrum, not paradox

Paradox? Don't you mean conundrum. a paradox is something I thought to be impossible!! And this is not. —The preceding unsigned comment was added by 131.111.8.98 (talk • contribs) 00:05, 18 April 2007 (UTC).

To quote the third sentence of the article:

This is not a paradox in the sense of leading to a logical contradiction; it is called a paradox because mathematical truth contradicts naive intuition: most people estimate that the chance is much lower than 50%.

—Ruakh_TALK 01:49, 18 April 2007 (UTC)

I have only come across this before referred to as the 'Birthday Surprise' which is a better description since it is not a paradox. You should mention that it is also known as 'Birthday Surprise' (as it has in note 4 and the Klamkin and Newman reference).62.255.240.100 12:18, 15 May 2007 (UTC)

Look up "paradox" in a dictionary. The first definition usually is something like, "A true statment that seems not to make sense at first glance." Then, somewhere further down the list of definitions, you'll find the one that says, "a mathematical proposition that can neither be proven, nor disproven." —Preceding unsigned comment added by 192.55.12.36 (talk) 20:08, 29 February 2008 (UTC)

My IB Math Studies Project

Hi people, I have already contributed in the past to this article, and having done my math studies project on this topic, I would like to contribute with my testings.

By the way why is my name always deleted from the references?? I contributed to this article —The preceding unsigned comment was added by 90.0.168.169 (talk) 19:15, 30 April 2007 (UTC).

Hi,

Thanks for your contributions!

Please note that the purpose of the "references" section is to show the sources from which Wikipedians take their information, not to show the Wikipedians who took their information from those sources.

Unless you can cite a specific claim in the article that comes from your high school math project and not from the other references, it looks (and is) really awkward to list a contributor's high school math project in the references. And if there is a claim that comes from your high school math project and not from the other references, then we should consider removing that claim until such time as you get your research published in a peer-reviewed journal; verifiability from reputable sources is a fundamental element of Wikipedia. (This is not in any way to criticize your math project; I'm familiar with the IB program, and am sure that your project was very impressive by high school standards. Kudos to you. But that doesn't make it appropriate as a reference here. After all, just because something was an IB project, that doesn't mean that it garnered a good grade, or that it would stand up to professional scrutiny. IB projects are first and foremost a learning experience.)

—Ruakh_TALK 01:23, 1 May 2007 (UTC)

Hi again As I was doing my math project, ergo in the past year and a half, I periodically updated this article with what I found with my researches. Unfortunately I didn't know that my project would be sent to IB to be verified. This means that they may think I have copied, even though it is the exact contrary :( That's why I would like my name to appear. Anyways thanks for the answer —The preceding unsigned comment was added by 90.0.46.7 (talk • contribs) 12:32, 1 May 2007 (UTC).

Oh, scary. Best of luck with that. :-/ —Ruakh_TALK 15:02, 1 May 2007 (UTC)

So is it ok to put me in the references? —The preceding unsigned comment was added by 90.0.46.7 (talk • contribs) 16:08, 1 May 2007 (UTC).

I'm sorry, but that's still not what references are for. :-/ —Ruakh_TALK 16:41, 1 May 2007 (UTC)

And In the External link can I put back the test I did with the players? —The preceding unsigned comment was added by 90.0.46.7 (talk • contribs) 20:36, 1 May 2007 (UTC).

Nonsense.

i didn't believe that these calculations are correct so i made an experiment: i rounded up 3 groups of people. every group constisted of 50 people so there SHOULD be a 97% chance that 2 of them celebrate their birthday on the same day (in each group).

guess what: there weren't 2 guys with the same birthday in none of the groups. in fact there weren't even any guys in these 150 persons who had their birthday on the same day.

mathematically, the chance of that happening is close to 0 (like 0,0x%). so the result of the experiment would be far more than highly unlikely. still, it proves that this calculation (like everything in maths) is just another pathetic attempt of human beings to define the nature of our environment. —The preceding unsigned comment was added by Sonandzon (talk • contribs) 21:26, 5 May 2007 (UTC).

Re: Nonsense

Unsigned, juvenile, biased, and disparaging commentary will be (and has been, here) deleted. The validity of basic statistical and mathematical methodology is beyond reproach. Lesotho 23:09, 5 May 2007 (UTC)

Re: who you talking to? —The preceding unsigned comment was added by 90.4.17.239 (talk • contribs) 12:38, 8 May 2007 (UTC).

Sonandzon, I think this is an interesting study. If what you found is repeatable, it shows that although the mathematics is correct, the assumptions behind it are not. Why don't you do a serious experiment and publish your findings in a peer reviewed conference or journal? -Pgan002 00:26, 22 June 2007 (UTC)

Re: Nonsense: You are obviously lying. If you would have done what you claim you would have found lots of matching pairs.Likebox 14:38, 18 October 2007 (UTC)

Please don't feed trolls. (In this case it's not a big deal, as (s)he's presumably moved on, but it's a bad habit to get into. If you don't know what I'm talking about, please read Wikipedia:What is a troll?.) —Ruakh_TALK 03:25, 19 October 2007 (UTC)

Wrong

I disagree with the mathematics on which this article is based.

The probability that in a group of two, both share the same birthday = 1/366.

The probablility that in a group of three, one pair share the same birthday (three people make 3 pairs = (1/366) + (365/366)*(1/365) + (365/366)*(364/365)*(1/364) = 3/366

The probability that in a group of four (i.e. 6 pairs) that one pair share the same birthday is (1/366) + (365/366)*(1/365) + (365/366)*(364/365)*(1/364) + (365/366)*(364/365)*(363/364)*(1/363) + (365/366)*(364/365)*(363/364)*(362/363)*(1/362) + (365/366)*(364/365)*(363/364)*(362/363)*(361/362)*(1/361) = 6/366

From this it can be seen that for a group of $n$ people, there will be $n(n-1)/2$ pairs, and that the probability that one pair share the same birthday will thus be $(n(n-1)/2)/366$.

Hence the probability that a single pair in a group of $n$ people will share the same birthday will be at least 50% when there are 133 pairs, because 133=366/2. This occurs when $n=19$ (approximate to nearest whole).

The probability that a single pair in a group of $n$ people will share the same birthday is 100% when there are 366 or more pairs of people: this occurs when $n=27$ (approximate to nearest whole). --Justificatus 15:47, 6 August 2007 (UTC)

The article explicitly skips leap years, but it sounds like you're calculating based on one. Mmernex 16:02, 6 August 2007 (UTC)

You're close, but a bit off. The problem is, it's possible for three people to all share a birthday, and you're counting each case of this as though it were three cases of two people sharing a birthday, when for our purposes it's actually equivalent to just one case of two people sharing a birthday. (And, analogously with larger groups.) In more general terms, remember that P ( A ∨ B ) = P ( A ) + P ( B ) − P ( A ∧ B ); hence, P ( A ∨ B ) ≠ P ( A ) + P ( B ) (except in the special case that A and B are mutually exclusive). —Ruakh_TALK 18:39, 6 August 2007 (UTC)

There may be something more deeply wrong with Justificatus' argument than you realize. Read that last paragraph again. It says that the probability of a shared birthday in a random group of 27 people is 100%. That is to say, it's guaranteed. That is to say, there is no possible way to choose 27 people such that each of them has a different birthday.

Justificatus "disagrees" with the math on this page. But, this is not some ivory tower abstraction that we're talking about here, this should be basic common sense. 192.55.12.36 (talk) 20:53, 29 February 2008 (UTC)

I agree with all of your comment except the "more deeply wrong" part. Superficially, the problem with his argument is that its result is nonsensical; but I think "your result is wrong because it contradicts this obvious fact: ___" is less helpful than "your reasoning is flawed because you're making this erroneous assumption: ___". Hence, I tried to explain why his reasoning was flawed, rather than how I knew it was flawed. —Ruakh_TALK 23:06, 19 April 2008 (UTC)

365 different birthdays -- what am I missing?

I wish to take issue with the statement, "After having met people with n different birthdays (n < 365), the chance that the next person you meet has a colliding birthday is (365-n)/365."

Suppose I am just getting started. I have met one person, and recorded their birthday. The number of different birthdays n is then 1.

The quoted text asserts that there are 364 chances out of 365 that the second person I meet will have the same birthday as the first one. This seems counter-intuitive. Likewise, if my list of birthdates is long, and I have n = 364, then the equation predicts that there is only one chance in 365 that the next person I meet has a birthday that is already on my list.

In summary, the equation says that at the beginning when n is small, I will have a large number of collisions, and as the size of my list of birthdays grows, the likelihood of a collision will decrease. --Jamesglong 18:14, 13 August 2007 (UTC)

Heh, good catch. I'm not sure how to fix this, though; in general, that section is assuming some things I'm not sure about. I think for now I'll label it "disputed", and we can work on fixing it up. —Ruakh_TALK 18:49, 13 August 2007 (UTC)

I think that (365-n)/365 is the probability that the next person you meet will have a non-colliding birthday. After you've collected one birthday, there is a good chance (364 in 365) that the next person will have a different birthday. Once you've collected 365 birthdays, you cannot collect any new birthdays and the probability is zero.--143.215.153.96 19:53, 14 August 2007 (UTC)

Coupon collector's problem

I didn't find a page for this problem on wikipeida, altough it is related to this article. There are N coupons on a desk, and you have to pick from them randomly (without taking the picked one from the desk). What is the expected amount of picks, until you have picked al of them at least once.

Google found the answer N*(ln N + Euler's constant) here —Preceding unsigned comment added by 81.183.27.21 (talk) 17:20, August 24, 2007 (UTC)

Pigeonhole principle

The pigeon hole principle says that there MUST be at least two people with the same birthdays if you have 367 or more people. Imagine 366 people (allowing for leap years), with one birthday on each of the 366 days of the (leap) year. When you introduce the 367th. person, this persons birthday can only happen on one of the already occupied birthdays. —Preceding unsigned comment added by 80.2.196.26 (talk) 18:22, 14 September 2007 (UTC)

Please read the first footnote. :-) —Ruakh_TALK 21:43, 14 September 2007 (UTC) —Preceding unsigned comment added by Ruakh (talk • contribs)

How about sharing the same birthday date?

I think one of the earlier posts discussing a test with groups of 50 may have been looking for the same birthday date (month/day/year) as opposed to simply the same day (month/day). Can you compute the odds for the same date? I suppose the distribution of people would impact the odds - but, can you make estimates based on assumptions for your groups ("even distribution of ages 20-60", or "distribution mirroring earth's population at each age")? Now that seems like a fun excercise. And one I'm not capable of completing.  :-(

69.134.204.75 20:58, 14 September 2007 (UTC)

Well, the "even distribution of ages 20–60" version is not hard: it's the same as the normal birthday problem, but assuming 14,975 distinct days instead of 365. With that assumption, it would take 145 people before you have a 50% chance that two had the same DOB — still amazingly few, IMHO. The latter version would obviously requiring knowing that distribution, which I do not. (And the math would be significantly more complicated, too. I'm actually not sure exactly how one could go about that, other than a brute-force or probabilistic method.) —Ruakh_TALK 21:53, 14 September 2007 (UTC) —Preceding unsigned comment added by Ruakh (talk • contribs)

GA Review

This article does not meet the Good Article criteria at this time, and will not be listed. Currently, I'd probably grade it as somewhere on the border of Start-class and B-class. The main issues are some major organizational and prose issues, as well as insufficient reference citations. I am also unclear whether the items mentioned under 'references' were really used to back up anything in the article at all, or are, in fact, just extra books that some might want to use to look up additional information about the subject. In which case, the section should be renamed to further reading.

There's a considerable amount of text in the 'notes' section. Why isn't this used in the article itself. These are clearly not footnotes! It would help editors to review WP:CITE for guidelines on how to properly use inline citations to cite material in an article.

The section header titles are fairly long, and not very clear and concise. It would be advisable to shorten and simplify these so that they are easier to understand when reading the table of contents. Some sections might be reorganized and merged with others as well, and check that all second, third, and fourth level subsection headers are absolutely necessary.

Many sections seem overly dependent on some very complex equations with very little text to actually describe the significance and meaning of these equations. This is likely to scare off many readers. Equations should also have a citation, so that we can verify that what is being included is clearly not original research. The citation should not appear next to the equation itself, it should appear in the sentence preceding and referring to the equation, usually immediately after a colon, such as in "the probability that two birthdays coincide can be attributed to the following equation:

The lead section actually does a reasonably good job of introducing the topic, but could better summarize the article. It might help to review WP:LEAD for tips on this. Also, the 'birthday attack' is mentioned in the lead, and briefly in the text, but there is no source for this information, and the text is pretty vague.

Hopefully, this will help editors improve the article. Good luck! Dr. Cash 04:59, 2 October 2007 (UTC)

footnote86.42.66.94 (talk) 00:59, 10 January 2008 (UTC)

Thanks! :-) —Ruakh_TALK 14:50, 2 October 2007 (UTC)

Knapsack Problem?

The formulation of the knapsack problem on this page is quite different from the knapsack problem with which I am familiar, which is as it is described in its own Wikipedia page. A bit of research reveals that what is described here appears to be the partition problem. The partition problem is a special case of the subset-sum problem, which is itself a special case of the knapsack problem. Thus, I suppose the characterization in this article is not strictly incorrect, but it is certainly misleading. I apologize if this discussion ends up at the wrong place; this is my first contribution to Wikipedia and I am not quite sure what I am doing. —Preceding unsigned comment added by 216.164.23.210 (talk) 03:24, 26 October 2007 (UTC)

It ended up in exactly the right place! It's a good idea to sign your talk entries (but not article changes) - writing four tildes (~) or clicking the signature button over the edit window will do that. If you are not logged in, it will just add your IP number and a timestamp (click "Show preview" below the edit window to see the effect).

I've changed the article to say "a variant of the knapsack problem"; I think it's correct this way, though it could still be improved upon. Go ahead, be bold, and do so, if you see a way!--Niels Ø (noe) 08:12, 26 October 2007 (UTC)

Solution for "near matches"

Hi,

Can anyone point me to a solution for the "near matches" version (i.e. birhdays within one week of each other)?

Thanks, nyenyec ☎ 17:28, 10 November 2007 (UTC)

Ok. Its actually rather trivial, if you just say Born the same week, and make the number 52. If you want within one week, The problem becomes born the same forghtnight. ( 26 ). —Preceding unsigned comment added by 99.33.92.113 (talk) 10:55, 17 November 2009 (UTC)

Understanding the problem, not

The section #Understanding the problem is rather confusing to me, and I'm not a total math idiot, although I am a math idiot. In particular, I can't make much sense of the following paragraphs:

The actual birthday problem asks whether any of the 23 people have a matching birthday with any of the others — not one in particular. (See "#Same birthday as you" below for an analysis of this much less surprising alternative problem.)

• The section #Same birthday as you appears to contradict this, and even if not, it doesn't help understand the problem.

Since in every group of 23 people there are 23*22/2=253 pairs, which is more than half of the number of days in the year, the chance that one of these pairs has a matching birthday is not small. For 28 people, the number of pairs exceeds the number of days, and the probability of matching is considerably greater.

• "not small"? How great exactly?

• 20*19/2=190 pairs is also more than half of the number of days in the year. Why then is exactly 23 used? Should be explained.

• 28? Where does that number come into play?

• "considerably greater"? How great exactly? How is this relevant to understanding the problem?

I dorftrottel I talk I 03:22, December 3, 2007

Both the sentences you quote are takes at making the computational result, following from a rather complicated analysis, less counter-intuitive. While, in a group of 23, the chances that someone (anyone in the group) has the same birthday as a previously identified person are low, the chances that someone (anyone in the group) has the same birthday as someone else (anyone else in the group) are higher. The second sentence makes this somewhat quantitative, though still less quantitative (and less complicated) than the exact calculation. If you in a list of 23 persons compare the birthday of the first person on the list to the others, you have 22 chances of success, but if you compare each and everyone to the others, you have 253 chances. (Why 253? Well, you can compare each of the 23 to each of the other 22, yielding 23*22 comparisons, but then you'll have made all comparisons twice (comparing A to B and comparing B to A), so we divide by two, 23*22/2=253.) Each of these many chances is small, because there are 365 days in a year. In a group of 28, there are 28*27/2=378, which exceeds 365. It is thus not so surprising that you have a fair chance of at least one success.

I think this is a good point to make in the article; possibly it could and should be made more clear. If you understand it now, you may be the right one to improve on it!--Niels Ø (noe) 08:27, 3 December 2007 (UTC)

Thanks a lot for the explanations! I'm not confident enough (yet) to edit the article, but I'll do some more reading and get back to it later. I dorftrottel I talk I 07:45, December 4, 2007

This explanation can also cause misunderstandings.

In a list of 23 persons, if you compare the birthday of the first person on the list to the others, you have 22 chances of success, but if you compare each to the others, you have 253 chances. This is because in a group of 23 people there are 23*22/2=253 pairs, which is more than half of the number of days in the year. So the chance that one of these pairs has a matching birthday is not small.

The numbers are correct, but arguing that the number of pairs is more than half of the number of days in the year would also fit for just 22 people! Furthermore some readers might be inclined to conclude that if the number of pairs exceeds the number of days we would have a sure hit. Maybe it should be pointed out in the main article that this isn't so. --Stevemiller (talk) 17:55, 11 March 2008 (UTC)

If you have 253 pairs of dates (people), it doesn't intuitively make sense that the odds are better than 50% that one of those pairs matches considering there are 132,860 possible pairs of dates which do not match, and only 365 that do match. 71.108.195.111 (talk) 07:12, 20 July 2008 (UTC)

Reordering and clarification

It seems to me the Halmos argument section should come before the approximations, since it follows from the discussion in the first paragraph of "Understanding the problem".

It's also not clear how the Poisson approximation is derived. What does the Binomial distribution have to do with the problem at hand? —Preceding unsigned comment added by Justin.mauger (talk • contribs) 19:53, 4 January 2008 (UTC)

Extremely confusing, text butchered?

Something must've been removed, since you go straight on to the approximations without actually stating the problem mathematically. Am I missing something? —Preceding unsigned comment added by 85.228.97.83 (talk) 21:54, 7 January 2008 (UTC)

It did indeed get butchered, by this anonymous edit and failure to properly revert afterwards. I believe I've just repaired it. —Keenan Pepper 03:33, 24 January 2008 (UTC)

At one point in the section you restored, the text says "It is easier to first calculate the probability p(n) that all n birthdays are different." Why are we bothering with an assumption that all birthdays are different? As unlikely as it is, they could all be the same day. They each could be any day. This section should be deleted if it isn't seen to be useful to the reader. At the very least, it should be made more plain that this "easier" method is just a step toward understanding the actual problem. Binksternet (talk) 04:18, 24 January 2008 (UTC)

Um, I don't think you understand what it says. It's not assuming that all birthdays are different. If you assumed all birthdays are different, then no two birthdays could be the same, so the probability of any two birthdays being the same is zero, and there's nothing to calculate. It's calculating the probability that all the birthdays are different, and calling this probability ${\bar {p}}(n)$. Then the probability that any two birthdays are the same is $p(n)=1-{\bar {p}}(n)$, because "all birthdays are different" is the logical negation of "at least two birthdays are the same". This is not an "easier" method and does not make any simplifying assumptions. It calculates the exact answer. —Keenan Pepper 05:20, 24 January 2008 (UTC)

It seems this issue has been solved, and the text is no longer very confusing, so i'm removing the tag. If someone still finds it hard to read, please add the tag back. --Storkk (talk) 11:58, 1 February 2008 (UTC)

Pool of possible birthdays stays 365

Some of the article text and prose references assumed that if you document one person's birthday then for the next person you document you could expect their birthday to come from a pool of birthdays reduced by one. However, the pool of possible birthdays is always the same (in the simplified problem format presented here.) Every single new person you see will always have 365 (actually 365.24219) possible birthdays. Binksternet (talk) 23:26, 21 January 2008 (UTC)

The pool of birthdays that you need to hit in order not to have a match is reduced by 1. --Storkk (talk) 13:34, 3 February 2008 (UTC)

Why? Binksternet (talk) 17:28, 3 February 2008 (UTC)

Because you're trying to not have a match, meaning that if you were to hit the same birthday as someone else, you would have a match, so there's one less possibility each time. For example if the first person was born on Jan 1, in order to not have a match you'd need to be born on a day other than Jan 1, which, assuming we're not talking about leap years, is 364 possible days.131.247.152.4 (talk) 14:57, 5 May 2008 (UTC)

Seasonal peaks and troughs

This whole thing works only in theory, which the article does not address. There are seasonal peaks and troughs that vary geographically, which alter the real-life probabilities discussed in this article. For example, there is a peak of birthdays in April/May in Canada, and a trough in December/January. So the probability of any given birthday falling on a particular date is not 1/365. Am I wrong here? 76.71.33.251 (talk) 02:33, 20 February 2008 (UTC)

Nevermind... I see that the article indeed addresses this. Thanks! 76.71.33.251 (talk) 02:39, 20 February 2008 (UTC)

Partition Problem problem

The text for this section says "...If there are only two or three weights, the answer is very clearly no..." But surely weights of 1,3 & 4 grams will balance - 1,3 vs 4 ? --83.105.33.91 (talk) 12:16, 17 March 2008 (UTC)

You don't understand. It says each weight is an integer number of grams randomly chosen between one gram and one million grams. The question is whether it's possible for a majority of the cases. I'll try to clarify the article. —Keenan Pepper 02:00, 18 March 2008 (UTC)

Birthday Pair Problem

I am comparing this article to page 322 in Applied Cryptography by Bruce Schneier and either the book and my math are wrong or the following line is in correct.

"...23 people there are 23*22/2=253 pairs"

It should read "...23 people there are 23*22=506 pairs." Perhaps the division by 2 is present due to pair AB being equivalent to pair BA. If that is the case the article should be more specific because mathematically there are 506 and not 253 pairs. —Preceding unsigned comment added by Rabunbike (talk • contribs) 18:27, 26 March 2008 (UTC)

You have 506 ordered pairs and 253 pairs without a need to order. You are not doing the test twice on each pair, but once on each pair of people. That is why 253 pairs (no order) are correct for this problem. -- 87.162.75.104 (talk) 10:26, 19 April 2008 (UTC)

I agree to the fact that the division by 2 should be explained briefly in the article to help understand the calculation. Florentnicoulaud (talk) 15:11, 17 July 2008 (UTC)

Presidents of the United States

We have thus far had 42 Presidents of the United States and have yet to have a matching birthday pair. The odds that two of our Presidents will share a birthday is rapidly(relatively speaking) approaching 1.0, is it not? —Preceding unsigned comment added by NCJRB (talk • contribs) 01:34, 5 May 2008 (UTC)

That's simply not true. James Polk and Warren G. Harding share the birthday November 2. It is probably fairly unlikely that there would be only one such match (other than the obvious fact that Grover Cleveland shares a birthday with himself), but still there's one. There are also considerably more matching pairs in their death days.Eebster the Great (talk) 03:04, 7 May 2008 (UTC)

Same birthday as you

In the graph next to the same birthday as you section, shouldn't q(n) = 1 when n = 365 (assuming n is the number of people other than you) or n = 366 (assuming n is the number of people including you) ? —Preceding unsigned comment added by 131.247.152.4 (talk) 15:00, 5 May 2008 (UTC)

No, it seemed that way to me at first, too, but that would only be the case if nobody else shared a birthday. You can imagine a group of you and 200 million other people, but all 200 million of the other people were born January 1 and you were born June 30. However, it's rather surprising how slowly that probability q(n) grows.Eebster the Great (talk) 03:04, 7 May 2008 (UTC)

Thought I had a grasp on this but I don't.

Alright, I read the section that said 23 different people means 253 different pairs, which made sense to me. However, I was doing the calculations upwards, and you end up with more than 365 different pairs in the with only 28 or so people range. So why isn't the probability a hundred percent at that point? —Preceding unsigned comment added by 69.62.140.50 (talk • contribs) 03:17, 8 May 2008 (UTC)

You're comparing apples and oranges — number of pairs with number of days. It's possible to have a group of 365 people, one having each possible birthday (not counting 29 February); in that case you'd have 66,430 pairs of people, but each pair having two distinct birthdays. The "253 different pairs" thing is a bit of an intuitive hint that even with only 23 people, there are more possibilities than it might seem at first, but it's not a valid way to determine what the actual probability is of a match. (The problem is that the pairs aren't independent; if A and B share a birthday, and B and C share a birthday, then A and C must share a birthday, and conversely, if A and B don't share a birthday, and B and C don't share a birthday, then there's a slightly elevated chance that A and C do share a birthday.) I think this part of the article needs to be tweaked a bit, as you're not the first person to come out of it with that same question. —Ruakh_TALK 04:08, 8 May 2008 (UTC)

same birthday as you

Regarding this edit [1]. I think the old explanation was better. "Note that this number is significantly higher than 365/2 = 182.5: the reason is that it is likely that there are some birthday matches among the other people in the room." That makes sense because the reason you need so many people is that their birthdays overlap, and take up fewer days than people. Cretog8 (talk) 19:43, 28 June 2008 (UTC)

Notability?

Perhaps I'm being an idiot but in what way is the topic of this article notable? It seems like just a randomly chosen application of elementary probability theory. 196.41.124.8 (talk) 18:42, 10 July 2008 (UTC)

It might not be important, but it is notable. It's very popular as an illustration of probability concepts. Do a web search, and you'll probably be surprised by how much it comes up. Cretog8 (talk) 19:00, 10 July 2008 (UTC)

Clearer definition of "birthday"?

I scanned through the comments, but didn't see my question addressed (sorry if I missed it), but here it is, in two parts:

1. Is it likely that people unfamiliar with this particular "problem"/concept will read this article? And if the answer is "yes," then

2. Do you think that such a person will wonder whether "birthday" means just the day and the month, or the whole day/month/year set?

The reason I ask this question is because it was the very first question I asked when I began to read the article, and I found that, since that information wasn't specified right at the beginning of the article, I wasn't able to get really absorbed in the potential fantasticness of the rest of the narrative related to the phenomenon.

I'm just saying. So can someone just pop the answer in the beginning of the article somewhere? Real quick? XO -- Sugarbat (talk) 02:21, 13 July 2008 (UTC)

If you click the link for birthday, you get, rather unambiguously, the appropriate definition. But it might be worth clarifying. The simplest i could think of would be to add one word, "annual", before "birthday" in the first sentence. --pfunk42 (talk) 02:20, 14 July 2008 (UTC)

Actually, I have exactly the same problem with the opening definition of "birthday" in its own article -- and you reiterate that problem in your last sentence. :) And as a matter of fact, only part of that definition (of "birthday") is correct - since "anniversary," by definition, implies that there has been a period of time since a particular event -- a period marked by the recurrence of a point in time (subsequent to the event) that matches a similarly named point in time at which the event occurred (i.e., one-week anniversary [Tuesday/Tuesday], two-minute anniversary [120 seconds/120 seconds], 100th-century anniversary [C/C or M/M], etc.). However, "birthday" can mean either the event *or* the "anniversary" of that event (someone's birth). Ergo a birthday can be literally the day, month, and year you were born (the actual date of your birth), or just the day and the month (the "anniversary" of your birth). That's why this article, here, is confusing, and it's also why the "birthday" definition in that article is confusing. I don't think anyone's complained, there, just because, there, I don't think differentiating matters as much as it does in this article/with regard to this concept. Right? Sugarbat (talk) 04:51, 19 July 2008 (UTC)

"Two-week anniversary" is a misnomer. An anniversary, by definition, recurs annually. Notice the letters "ann-" at the beginnings of both words: "anniversary", "annual". That's not a coincidence. Michael Hardy (talk) 11:41, 19 July 2008 (UTC)

That's interesting, but doesn't quite apply to the topic. Also, although you're (partly) right about the literal etymology, you're a little off w/r/t the conventional use/definition of the word "anniversary" in English. That's one reason it's called "English" and not "Latin."  ;) Sugarbat (talk) 01:47, 31 July 2008 (UTC)

Another Birthday Paradox

In the operetta The Pirates of Penzance the birthday on February 29th of Frederic the pirate apprentice leads to this exchange about a resulting birthday paradox:

FRED. How quaint the ways of Paradox!

At common sense she gaily mocks!

Though counting in the usual way,

Years twenty-one I've been alive,

Yet, reckoning by my natal day,

I am a little boy of five!

RUTH and KING. He is a little boy of five! Ha! ha! ha!

ALL. A paradox, a paradox,

A most ingenious paradox!

This fictional birthday paradox is different from the "birthday paradox" named in the article and is certainly notable. Cuddlyable3 (talk) 21:59, 11 August 2008 (UTC)

Partition Problem Error

The section on the partition problem seems to be clearly incorrect, or at least to use a confusing metaphor. It states that the problem involves moving weights on a scale between the pans in order to balance the scale, and claims that "if there are two or three weights, the answer is clearly no" but this is not true; if a 5 and 10 weight are on the left pan, and a 5 weight is on the right pan, you can move the 5 from the left to the right and balance the scale with three weights. Is there some detail missing that would exclude this case? —Preceding unsigned comment added by Hyphz (talk • contribs) 11:19, 8 September 2008 (UTC)

Paragraph removed

I removed this paragraph:

It is easier to figure the probability that the birthdays will be different, such as: with one person they have 365 opportunities to have a different birthday. The second person only has 364 possibilities to have a different birthday than the first person. The third person has 363 days, and so on. Thus when the group reaches 366 a clash is inevitable — all the days will have been used up (except for leap years of course).

First off, how can the author know what's easier for a specific reader? Secondly, the number 366 amply takes care of leap years. More important, though, is the way that this is presented: it's not explained up front that the goal of this 'easier' mental exercise is to have a 'hit' between any two persons out of a collection of people in a group rather than the question of whether somebody in the room shares, say, your birthday. The exercise isn't set up adequately. Binksternet (talk) 12:57, 6 December 2008 (UTC)

Extra parentetheses?

Why do we need the extra parentheses around the scientific numeral? Is not the order of operations sufficient to just put (100 − 3×10−129)%, for example?— trlkly 20:16, 22 December 2008 (UTC)

Suggestion for addition to the "Generalizations" section

Hi -

I have a suggestion for anyone who is particularly probability-inclined - if anyone would like to add a section on how to calculate the probability of having not just a pair of birthdays, but a triplet of birthdays, a quadruplet of birthdays, and so on, on the same day, I would be very interested in reading about it. I could not figure out how to do this on my own as I realized that the probabilities for each unique set of birthdays are unfortunately not all independent.

The function would look something like this:

P(n,k) = probability that out of n people, at least one set of k people share a birthday

This whole article is simply about P(n,2), and I would be interested to see a generalization of that if possible. —Preceding unsigned comment added by Face Kicker (talk • contribs) 08:02, 25 December 2008 (UTC)

Error in the calculations

I can't believe no one saw this by now. There's a slight mistake in one of the equations. —Preceding unsigned comment added by 89.123.197.192 (talk) 00:15, 13 February 2009 (UTC)

Do tell us about it. Cuddlyable3 (talk) 22:30, 12 June 2009 (UTC)

I believe that he scribbled it in the margin. ( Proof left to the reader ). —Preceding unsigned comment added by 99.33.92.113 (talk) 10:58, 17 November 2009 (UTC)

Calculating the Probability

This section should explain the reasoning behind the calculation more fully and more clearly. It looks as if either it has been reproduced without it being understood or that it has come from someone who is so familiar with it that he forgets that most of us really do need an explanation........There is a better treatment here http://www.curiousmath.com/index.php?name=News&file=article&sid=78 ....... .mikeL assisted by sinebot, bless it —Preceding unsigned comment added by 92.238.234.172 (talk) 08:36, 1 June 2009 (UTC)

Software for Birthday Paradox Cases

There is software for this type of calculations, especially the most inclusive cases of probability of collisions. The results are instantaneous for numerical values up to 18 digits wide. Two programs are available now:

Collisions.exe

BirthdayParadox.exe

They are presented on the page of the most thorough analysis of the Birthday Paradox and probability of collisions:

The Birthday Paradox: Combinatorics, Probability of Duplication, Coincidences, Collisions, Roulette, Social Security Numbers, Genetic Code, DNA Sequences.

There is also the complementary probability: The Reversed Birthday Paradox. The same software performs the most accurate calculations for such situations as well. The reverse collisions probability: Calculate the number of persons (elements) when the probability is known. In the Wiki article, such calculations are only approximations.

Parpaluck (talk) 17:58, 4 June 2009 (UTC)

Probability of duplication

The following statement in the Wiki article is flawed:

"It is easier to first calculate the probability p(n) that all n birthdays are different. If n > 365, by the pigeonhole principle this probability is 0."

The Birthday Paradox is a tiny element of the probability of duplication or probability of collisions. If the set has N elements (e.g. N = 365 birthdays), then for number of elements > 365 the probability is 1 (or 100%) that there is duplication.

Let's take the case of dice rolling. Number of elements N = 6 (6 numbered faces from 1 to 6).

Throw 2 dice at a time. The probability of duplication (e.g. 1-1 or 6-6) is 16.66% (6 / 36).

Total number of sets: 6 ^ 2 = 36

Number of sets WITHOUT duplicates: 30

Number of sets with duplicates: 6

Throw 6 dice at a time. The probability of duplication (e.g. 1-1-?-?-?-? or ?-6-?-?-?-6) is 98.46% (45936 / 46656).

Total number of sets: 6 ^ 6 = 46656

Number of sets WITHOUT duplicates: 720

Number of sets with duplicates: 45936

Throw 7 dice at a time. The probability of duplication (e.g. ?-?-?-?-1-?-1 or 6-?-?-?-?-6-?) is 100% (279936 / 279936).

Total number of sets: 6 ^ 6 = 46656

Number of sets WITHOUT duplicates: 0

Number of sets with duplicates: 279936.

Parpaluck (talk) 18:05, 4 June 2009 (UTC)

Obviously, but how does that make the statement "the probability p(n) that all n birthdays are different ... if n > 365 ... is 0" flawed? Note, that's the probability of there not being duplication. —JAO • T • C 19:58, 12 June 2009 (UTC)

If n>365 then the probability that all n birthdays are different is 0 (zero means impossible) BECAUSE the probability of duplication is 1 (or 100% means certain). Parpaluck has not shown any flaw, only that the same thing can be stated in two ways. Cuddlyable3 (talk) 22:08, 12 June 2009 (UTC)

Horrendous Mathematics on Top of Plagiarism

The Birthday Paradox article at Wikipedia must be rewritten immediately. It is one of those materials that make Wikipedia the laughing stock in the common sense world. Forget about intelligentsia or academia or the virtual world of intelligent and cultured humans! It is things like this article or materials as those dedicated to factorials or combinatorics that make all normally intelligent and knowledgeable persons run away from Wikipedia. Some might only laugh. Most of them, however, have the worst of feelings regarding this laudable attempt to make all knowledge available to the commoners. Each and every one of us is a commoner in most fields of knowledge — nobody should have hard feelings in that regard.

The opening of the Birthday paradox article is blatant plagiarism:

"In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 366…"

The sentences are taken, without quotation, from Warren Weaver's famous book Lady Luck (page 132).

Then, the formulae that follow represent an exercise in idiotic mathematics. Alas, if the author(s) would have plagiarized correctly (!) the formula in Warren Weaver's book!!

I urge the Wikipedia editors make immediate changes to this article. They might want also to contact the author of this benchmark article:

The Birthday Paradox: Combinatorics, Probability of Duplication, Coincidences, Collisions, Roulette, Social Security Numbers, Genetic Code, DNA Sequences.

It's for the sake of intelligence, as in human reasoning!

Parpaluck (talk) 19:47, 4 June 2009 (UTC)

While pages 132–135 of Weaver's book deal with this problem, neither of these sentences are anywhere to be found, not even partially, on these pages, at least in not the version available here. —JAO • T • C 20:13, 12 June 2009 (UTC)

I've just looked at page 132 of Warren Weaver's book Lady Luck and that sentence is not there. It's not surprising that the same ideas are expressed, since the problem has been well known for a long time. Weaver's book was published in 1963. I don't actually know whether the problem was well-known before Weaver's book came out but I'd be a bit surprised if it was not. As a mathematician I'm failing to find the "idiotic mathematics" that "Parpaluck" says is here. As for the articles on factorials and combinatorics, could Parpaluck please be specific and post to Wikipedia talk:WikiProject Mathematics, saying what things are objectionable? Michael Hardy (talk) 21:52, 12 June 2009 (UTC)

The sentences were reworded! Looked like more to disguise the "source of inspiration". The keyword is 23 — as is in 23 persons for a 50-50 chance that at least two share the same birthday.

But the worst part is, really, the MATHEMATICS. You still don't get it!

1) Let's take … Warren Weaver's favorite example: n = 23. Your formula is clumsy:

[365 x 364 x (365 – 23 + 1)] / (365 ^ 23) then leads to a formula of combinations! The best, in words:

The Probability_Of_Collisions (Coincidences, Birthday Paradox) = Number_Of_Duplicate_Sets (M, N) / Exponents (M, N) where Number_of_Duplicate_Sets (M, N) = Exponents (M, N) – Arrangements (M, N)

Simply, the formula is:

Probability_Of_Collisions (Coincidences, Birthday Paradox) = 1 –{[365 x 364 x (365 – 23 + 1)] / (365 ^ 23)}

2) "It is easier to first calculate the probability p(n) that all n birthdays are different. If n > 365, by the pigeonhole principle this probability is 0."

The Birthday Paradox is a tiny element of the probability of duplication or probability of collisions. If the set has N elements (e.g. N = 365 birthdays), then for number of elements > 365 the probability is 1 (or 100%) that there is duplication. The birthday paradox is less a problem of probability. In fact, it is better analyzed by the mathematics of sets. The die has 6 numbered faces. It is very easily provable that more than 6 dice can be rolled. Read the case above regarding a number of 7 dice. You can roll 100 dice, if you want to. Is that impossible? Look at the results. Can you find a series of throws that consists only of UNIQUE face values? NOT! Just look at 7-die series. Everything shows at least one duplicate. Matter of fact, most throws show a larger number of duplicates (more than just two dice). Then, how can be the probability of duplication 0? Have you ever thrown 7 dice and all seven shown only unique numbered faces???

Parpaluck (talk) 19:36, 17 June 2009 (UTC)

= OK. Mea culpa … perhaps! "This probability…." refers to the probability for sets with unique elements only. Probably other readers too think of "This probability…." to refer to the probability of the birthday paradox. Again, that's why I say that the birthday paradox is better understood via the mathematics of sets. More deeply, there are no formulae, but algorithms. Thus, the cases of absurdity for N > M can be avoided. The formula of arrangements (to calculate the number of sets with unique elements only) leads to an absurdity. You can't arrange 6 dice taken 7 at a time!

Parpaluck (talk) 19:46, 17 June 2009 (UTC)

Mistake in partition problem

The section on the partition problem mentions N weights with integer gram masses randomly chosen from [1, 1000000]. The last paragraph states that "the distribution of the sum of weights is approximately Gaussian, with a peak at 1,000,000 N and width $\scriptstyle 1,000,000{\sqrt {N}}$ ...". However, the peak (and mean) should actually be at $\scriptstyle {{{1000001N} \over 2}=500000.5N}$, and the width (standard deviation) should be $\scriptstyle {{{3{\sqrt {37037037037N}}} \over 2}\approx 288675.1{\sqrt {N}}}$. Here's my work:
The distribution of one weight's mass (in grams) is discrete uniform over [1, n], n = 1000000, so has
${\begin{aligned}\mu _{\text{one}}&={{n+1} \over 2}\\&={{\left(1000000\right)+1} \over 2}\\&={1000001 \over 2}\\\end{aligned}}$
and
${\begin{aligned}\sigma _{\text{one}}^{2}&={{n^{2}-1} \over 12}\\&={{\left(1000000\right)^{2}-1} \over 12}\\&={{1000000000000-1} \over 12}\\&={999999999999 \over 12}\\&={333333333333 \over 4}\\\end{aligned}}$
By the central limit theorem, the average mass (in grams) of N such weights would approximately follow a normal distribution with
${\begin{aligned}\mu _{\text{avg}}&=\mu _{\text{one}}\\\end{aligned}}$
and
${\begin{aligned}\sigma _{\text{avg}}^{2}&={\sigma _{\text{one}}^{2} \over N}\\\end{aligned}}$
so the sum of the masses (in grams) of N weights would approximately follow a normal distribution with
${\begin{aligned}\mu _{\text{sum}}&=N\mu _{\text{avg}}\\&=N\left(\mu _{\text{one}}\right)\\&=N\left({1000001 \over 2}\right)\\&={1000001N \over 2}\\&=500000.5N\end{aligned}}$
and
${\begin{aligned}\sigma _{\text{sum}}^{2}&=N^{2}\sigma _{\text{avg}}^{2}\\&=N^{2}\left({\sigma _{\text{one}}^{2} \over N}\right)\\&=N\sigma _{\text{one}}^{2}\\&=N\left({333333333333 \over 4}\right)\\&={333333333333N \over 4}\\\sigma _{\text{sum}}&={\sqrt {333333333333N \over 4}}\\&={3{\sqrt {37037037037N}} \over 2}\\&\approx 288675.1{\sqrt {N}}\\\end{aligned}}$
If I'm mistaken, please let me know. If not, could someone please figure out what consequences this has? Is the answer to the problem still N = 23? While I could take a crack at figuring that out myself, my stats is a bit rusty, so I'd at least like to have some confirmation that my calculations here are correct before going forward. Klparrot (talk) 00:11, 8 August 2009 (UTC)

reads first person

This article at times read directly to the reader. words such as "we" and "you" should be removed. I would do this except someone is curently editing the article. Thanks in advance ♠ B.s.n. ♥R.N.contribs 02:43, 4 October 2009 (UTC)

Why the Shirky Reference?

Why the reference to Clay Shirky? Is it just cruft from some earlier version? —Preceding unsigned comment added by 128.255.45.53 (talk) 21:29, 24 October 2009 (UTC)

Birthday solution

• I checked birth dates in List of Prime Ministers of Australia. 50% probability failed. There are thousands of articles about 'Living People'. It is easy to debunk this birthday problem practically.

• I recalled my school life. No birth date match with hundreds of classmates. In office, no match with birth date of almost 1000 employees. I had access to database of around 100,000 people and I found only 1 match(other than my birth date). This sounds improbable, almost impossible, but somehow it happened.

• The article does not mention minimum population to make this theory work. In village(or group) with population 365(each having unique birth date) surely this birthday problem will not arise. So how much population this village should have? Thanks! Rāmā (talk) 00:23, 19 November 2009 (UTC)

OK, OK, OK... I was checking match of birth dates of two prime ministers in List of Prime Ministers of Australia. 50% probability failed. Now after waking up, I checked whether my birth date match with birth date of any PM. And I must be honest, it does. It didn't happened in my entire life. And it happened in first randomely chosen list. Observation cause wave function collapse. Rāmā (talk) 10:59, 19 November 2009 (UTC)

Timeline-

• I was trying to sleep while thinking about 'birthday problem'. Somehow I came to the conclusion that 'birthday problem' means that if I toss coin in the air wishing it to be 'head' then there is almost 94% possibily that coin will fall on ground with showing 'head'. So 'birthday problem' is hoax!
• I came back on wikipedia, checked category 'living people' but birthdates on few random article is missing.
• How about checking 'list of prime ministers of India'? But that list repeat prime ministers twice in office. I need minimum 23 Prime Ministers.
• I applied commonsense that as most of the wikipedians are from developed countries, hence list about PMs of developed countries will be more perfect. I chose Australia i.e List of Prime Ministers of Australia.
• I took casual look at list. No match. Applying theory of probability, 99.99% chances are that in very first attempt no one will not find match.
• I declared on this talk page that birthday problem is wrong and went back to sleep.
• After waking up, I thought how about cross-checking my own birth date with list List of Prime Ministers of Australia? I found match!
• I recalled something link to 'list of US presidents' in 'birth day' article. I searched article and found that it is NOT list of US presidents. It is same list i.e List of Prime Ministers of Australia which I chose randomely and there is birth day match of two prime ministers.
• I don't know what to talk. Thanks! Rāmā (talk) 13:37, 19 November 2009 (UTC)