User talk:Handy2000

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The MHP[edit]

Hello Handy2000, Devlin is right. You are citing his words:

There are several ways to explain what is going on here. Here is what I think is the simplest account.
Suppose the doors are labeled A, B, and C. Let's assume the contestant initially picks door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C. Notice that he can always do this because he knows where the prize is located. (This piece of information is crucial, and is the key to the entire puzzle.) The contestant now has two relevant pieces of information:
1. The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3.
2. The prize is not behind door C.
Combining these two pieces of information yields the conclusion that the probability that the prize is behind door B is 2/3.
Hence the contestant would be wise to switch from the original choice of door A (probability of winning 1/3) to door B (probability 2/3).
[...]
(Oddly enough, people ... usually justify this by saying that after Monty has opened his door, the contestant faces a new and quite different decision, independent of the initial choice of door. This reasoning is fallacious, but I'll pass on pursuing this inconsistency here.)

If Monty opened his door randomly, then indeed his action does not help the contestant, for whom it makes no difference to switch or to stick. But Monty's action is not random. He knows where the prize is, and acts on that knowledge. That injects a crucial piece of information into the situation. Information that the wise contestant can take advantage of to improve his or her odds of winning the grand prize. By opening his door, Monty is saying to the contestant "There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3."

 

Initially the chances on each of the three doors are 1/3. So the odds on the door first selected by the guest will be 1/3. Devlin now says that it is most important that the host KNOWS where the car is. So his opening of a door showing a goat did NOT provide for a "new game" with two (still) closed doors of equal chance:

The one first selected by the contestant and the "other" closed host's door, just  "one of them" hiding the car, the other one hiding the goat  (with chances 1/3 : 1/3 - or 1/2 : 1/2).  That could occur only in the very "subset" of cases  that e.g. the host did NOT KNOW where the car was, and that he just "occasionally by chance" had opened  "a goat door". Just in that subset of cases.

But – as the host KNOWS where the car is, he can and will open a goat door in each and every case, and not just in a "subset of cases". Given that - in case he got two goats - he chooses which one of his two doors to open "unbiased" and "fully at random" (not giving away any special additional information on the actual location of the car)  you will have learned nothing that allows you to revise the odds on the door first selected by the guest. With other words: The odds on the car first selected by the guest still remain unchanged 1/3 under that condition.

As the odds on the car first selected by the guest are still 1/3, the two unchosen doors "together" have a chance of 2/3. Unchanged under that condition. The same as before the host has shown a goat. And you know already from the start that at least ONE of the two host's doors must hide a goat, as there is only one car. And the host just showed you a goat. That means that the odds on his OTHER still closed door did go up from original 1/3 to 2/3.

See also what I wrote in Arguments in reply @Ohiostandard

I wrote:

Will that table in your opinion help to show that the contestant will win by staying in 1/3 of cases, but will win by switching in 2/3 of cases,
no matter at all which door she should have chosen (1, 2 or 3), and no matter which other door whatsoever the host has opened, showing a goat:

(Supposed door A was chosen by contestant)
 

"Another" door opened by the host, showing a goat:



contestant's final decision:
Initial arrangement behind doors A, B, C
(probability)
Open Door A
(probability)
Open Door B
(probability)
Open Door C
(probability)
Joint
(probability)
Win by
staying
Win by
switching
Car Goat Goat (1/3) No Yes
(1/2
)
No 1/3 x 1/2 Yes
(1/6
)
No
No No Yes
(1/2
)
1/3 x 1/2 Yes
(1/6
)
No
Goat Car Goat (1/3) No No Yes
(1
)
1/3 x 1 No
Yes
(1/3
)
Goat Goat Car (1/3) No Yes
(1
)
No 1/3 x 1 No
Yes
(1/3
)
Note that the host has no choice when the contestant fails and chooses a goat,
because in that case the host is just bound to show the second goat.
Win by
staying
1/3
Win by
switching
2/3

See also "The Monty Hall puzzle".  Okay?  Regards,  Gerhardvalentin (talk) 13:02, 12 April 2011 (UTC)

Very kind of you to help me out. Of course I do not doubt that on the average you win the car 2/3 of the times by switching. It is the reasoning of Devlin I think to be incorrect. You speak of odds that do or do not change. But in the article one does not speak off odds or changing odds. So it seems you give your own argument, and do not explain the arguments of Devlin. For instance it says in the article (with my comment inserted): "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3 (Yes, but each single door has chance 1/3). I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner (Yes, originally; and each of the other doors also has probability 1/3). I have not changed that (See, each door had and has chance 1/3 on the car). But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3 (No, as we've seen, all three doors have probability 1/3 on the car - random distribution).'" I think Devlin confuses probability and conditional probability. Handy2000 (talk) 21:08, 12 April 2011 (UTC)
Devlin just tries to explain "what is going on here", he does not present a conditional probability calculus. He says:
"There are several ways to explain what is going on here. Here is what I think is the simplest account."  And he just tries to help the contestant to understand the actual situation, emphasizing that the host KNOWS about the actual location of the three objects. And I try to add some remarks:
If Monty opened his door RANDOMLY, then indeed his action does not help the contestant, for whom it makes no difference to switch or to stick. But Monty's action is not random. He knows where the prize is, and acts on that knowledge. That injects a crucial piece of information into the situation. Information that the wise contestant can take advantage of to improve his or her odds of winning the grand prize. By opening his door, Monty is saying to the contestant "There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3."
Originally the probability for the three doors to hide the car is 1/3+1/3+1/3=1.
Then the contestant first selects one door, and that still does not alter this given fact, probability did still remain 1/3 + (1/3+1/3)=1.
But please take notice that the contestant now has definitively and irrevocably divided  those three doors into  two groups:  Into the "group" of the one door that she just has selected (probability 1/3) and the "other group" of the two unselected host's doors (1/3+1/3). If you have a closer look on those two unselected (host's) doors (only 1 car is in the show) you can see that at least one door of that group  must  hide a goat. You do not know which one, but the host knows. So at least one of those two host's doors definitely has a chance of Zero. That's a fact.
As both host's doors together have a probability to hide the car of 2/3  (3/3-the first selected door 1/3), the "other" host's door now must have a chance of 2/3.
In other words: In 2 out of 3 this "other" door will hide the car, and in 1 out of 3 the "other" door will hide the second goat, "we" just do not know which one. This is true as soon as the contestant has made her first choice, i.e. even BEFORE the host has opened a door:  0/3+2/3=2/3  resp.  2/3+0/3=2/3.
Now let's distinguish the two possible scenarios:
First scenario: In 1 out of 3 the contestant has first selected the car, and the host got two goats (0/3+0/3), and he will be going to open now one of his two goat-hiding doors. In this scenario staying will win, because his "other" door is 0/3 also.
But in 2 out of 3 the contestant has first selected one of the two goats, and the host got the car and only ONE goat (resp. only ONE goat and the car), that means that he is bound to open that only single door of his that hides his only goat. He can do this because he knows which one of his two doors hides the car and which one hides the goat, and soon he will open one door.
But - as I said before - at least ONE of his two doors hides a goat (0/3), and before the host opens a door, even YOU know about this given fact. And in this latter scenario the "other" door hides the car (3/3), and the host KNOWS which one is which one. So in opening of his only one door that hides a goat, he is showing you that his "other" door must hide the car in this scenario.
Devlin tries to explain what is going on. Regards, Gerhardvalentin (talk) 23:59, 12 April 2011 (UTC)

Thank you very much, for your efforts. But (!) the point is that Devlin's explanation seems not to be a logically correct explanation. Let us call the probabilities on the car pA, pB and pC for the three doors A, B and C. Then pA=pB=pC=1/3. This never changes, whatever happens in the game. If the contestant chooses door A, indeed is pB+pC=2/3, but never these chances change into 0 and 2/3. There is no way to come to such a conclusion. So, when the host opens door C, the chance pC (on the average) on the car there is still 1/3. But the conditional probability, given the door hides a goat is (of course) 0. That's where we think Devlin is mistaken. Regards, Handy2000 (talk) 09:38, 13 April 2011 (UTC)

As I said before, he does not present mathematical probability theorems and calculi for students of his class here, he just tries to help anyone (grandma and grandson) to understand "what is going on". So just call (0 and 2/3) "conditional probability", if you like. No problem. Gerhardvalentin (talk) 10:30, 13 April 2011 (UTC)
Devlin's reasoning only requires the observation of an obvious and intuitive symmetry inherent in the problem to be correct. Martin Hogbin (talk) 14:48, 5 May 2011 (UTC)
I didn't notice your remark here. Our problem is Devin does not refer to any symmetry. And even so, how does symmetry help? As I wrote above: pB+pC=2/3, due to pB=pC=1/3. How do we continue? Handy2000 (talk) 07:19, 8 May 2011 (UTC)
I think Gerhard has the answer. In everyday speech probabilities can change to reflect the revealing of new information. For example pC becomes zero after the host opens door C to reveal a goat. Now, you professor might want to insist that pC does not change but that new conditional probability is created when the host opens door C, and I am sure he is correct. If you want to add in the word 'conditional' as Gerhard suggests, I too have no objection.
The question we need to ask ourselves to arrive at the solution is, 'How does the revealing of a goat behind door C affect the the (now conditional if you like) probability that the car is behind door A?'. It is intuitively obvious that it does not affect it at all. The host must reveal a goat and, as the situation is symmetrical with respect to door number, the specific door he opens can make no difference. Thus no information, that you were not already aware of, about the probability of the the car being behind door A has been revealed and the (now conditional if you like) probability has not changed.
No doubt your professor will have told you about a contrived version of this problem in which the host has a preference for one of the doors. You might like to discuss with him how this would have any effect on the probabilities when using the subjective Bayesian definition of probability, if the player is not aware of this preference.
Maybe your professor is just trying to get his students to pass their exams (which is not in itself a bad thing) rather than to think about the meaning of probability too deeply. Martin Hogbin (talk) 09:33, 8 May 2011 (UTC)

Not edited before?[edit]

You seem to be very interested and informed about this topic. I am surprised that you have not commented before. Martin Hogbin (talk) 14:51, 5 May 2011 (UTC)

I do not quite follow you. I discuss this topic with some fellow students, and we're interested in the different views. Professor Richard Gill explained several things to me, and as far as I understand, he also admits the solution of Devlin given in the article is wrong. Rick Block told me Devlin himself seems to have changed his original solution. Handy2000 (talk) 15:42, 5 May 2011 (UTC)

My point was just that you seem remarkably familiar with the article, and other editors for someone whose first contribution was on 5 May 2011. Martin Hogbin (talk) 15:52, 5 May 2011 (UTC)