My Math Forum Can someone explain the Monty Hall problem to me?

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October 13th, 2015, 08:34 PM   #11
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Quote:
 Originally Posted by skipjack that's incorrect. If you picked correctly, you lose by switching, and if you picked incorrectly you win by switching, regardless of the value of q. Hence the probability that you win by switching is independent of the value of q.
q=1/2.

Last edited by skipjack; October 14th, 2015 at 02:49 AM.

October 14th, 2015, 05:09 AM   #12
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Quote:
 Originally Posted by skipjack That's incorrect. If you picked correctly, you lose by switching, and if you picked incorrectly you win by switching, regardless of the value of Q. Hence the probability that you win by switching is independent of the value of Q.
Did you read it, or try to understand it? You are trying to use only prior probabilities, and I'm talking about probabilities conditioned on the evidence you have. And it is objections based on what one thinks is the correct solution, without expressing it rigorously, that lead to controversy in such problems.

Use the labels I described before: C, R, and L. Then:
• There is a 1/3 prior probability that the car is behind L, and Monty must open R. So P(Open=R & Car=L)=1/3.
• There is a 1/3 prior probability that the car is behind R, and Monty must open L. So P(Open=L & Car=R)=1/3.
• There is a 1/3 prior probability that the car is behind C, and a Q probability he would open L given that the car is behind C. So P(Open=L & Car=C)=P(Open=L|Car=C)*P(Car=C)=Q/3.
• There is a 1/3 prior probability that the car is behind C, and a (1-Q) probability he would open R given that the car is behind C. So P(Open=R & Car=C)=P(Open=R|Car=C)*P(Car=C)=(1-Q)/3.

The conditional probability that the car is behind R, given that Monty opened L, is now:

P(Car=R|Open=L) = P(Car=R & Open=L)/[P(Car=c & Open=L)+P(Car=R & Open=L)]
= (1/3)/[(Q/3+1/3)
= 1/(1+Q).

Similarly, P(Car=L|Open=R) = 1/[1+(1-Q)] = 1/(2-Q).

Now, when we aren't told how to determine Q, we must assume a value to get an answer. And the only value e can assume is Q=1/2, because we can't assume he favors R over L or L over R. Note that P(Car=R|Open=L)=P(Car=L|Open=R)=2/3 in this case.
Quote:
 Originally Posted by mathman q=1/2.
Assumed to be, yes.

But I fear we have lost our original questioner.

Last edited by JeffJo; October 14th, 2015 at 05:11 AM.

 October 14th, 2015, 07:07 AM #13 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 I did read it, but your explanation calculates that 1/(1 + Q) is the probability that the car is behind R, given that Monty opened L. That's different from your original assertion, which was that your probability of winning by switching is 1/(1 + Q). Both 1/(1 + Q) and 1/(2 - Q) are at least 1/2. Hence switching is an optimal policy, regardless of the value of Q, whether that value is known, or which door Monty opens. The contestant can reason as follows: if I never switch, my chance of winning is 1/3, and so if I always switch, my chance of winning is 2/3. This means that calculating conditional probabilities is unnecessary. The contestant is simply uninterested in his conditional chance of winning in a particular game, given the circumstances of that game, as there is no policy better than "always switch".
 October 14th, 2015, 08:45 AM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra The problem is that we don't know whether Monty has made a choice or not. What we do know is the probability that he made a choice ($\newcommand{\pr}[1]{\Pr\left({#1}\right)} \pr{\text{choice}} = \frac13$) and the probability that we win by switching given that he made a choice ($\pr{\text{Car$\ne$C | choice}} = 0$). We also know the probability that he didn't make a choice ($\pr{\text{no choice}} = \frac23$) and the probability that we win by switching given that he didn't make a choice ($\pr{\text{Car$\ne$C | no choice}} = 1$). Thus we have $$\pr{\text{Car \ne C | choice}} = {\pr{\text{Car \ne C and choice}} \over \pr{\text{choice}}} \implies \pr{\text{Car \ne C and choice}} = \pr{\text{Car \ne C | choice}} \cdot \pr{\text{choice}} = 0 \cdot \tfrac13 = 0 \\ \pr{\text{Car \ne C | no choice}} = {\pr{\text{Car \ne C and no choice}} \over \pr{\text{no choice}}} \implies \pr{\text{Car \ne C and no choice}} = \pr{\text{Car \ne C | no choice}} \cdot \pr{\text{no choice}} = 1 \cdot \tfrac23 = 0 \\[12pt] \pr{\text{Car \ne C }} = \pr{\text{Car \ne C and choice}} + \pr{\text{Car \ne C and no choice}} = 0 + \tfrac23 = \tfrac23$$ Both are valid approaches, working on different assumptions. JeffJo's unwritten assumption is that we know something about Monty's approach to choosing. The version set out above makes no such assumption. If we happen to know the value of $q$, it makes sense to use it, but without it we are better off focussing on what we know. Of course, it should come as no surprise that the methods agree in the case where we pick $q=\frac12$ because both cases postulate that his actual choice is immaterial, that we gain no information from the choice, that the situation is symmetric. In essence, JeffJo's solution introduces a variable upon which we have no information, so at the point we introduce it we already know what value we will pick for it and that it will be the symmetric solution and that as such any asymmetry in the problem will give us no useful information.
October 14th, 2015, 10:50 AM   #15
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Quote:
 Originally Posted by skipjack I did read it, but your explanation calculates that 1/(1 + Q) is the probability that the car is behind R, given that Monty opened L. That's different from your original assertion, which was that your probability of winning by switching is 1/(1 + Q).
Ummmm, you'd switch to door R? So the two are supposed to be the same?

Quote:
 Both 1/(1 + Q) and 1/(2 - Q) are at least 1/2. Hence switching is an optimal policy, regardless of the value of Q, whether that value is known, or which door Monty opens.
Yes, the best strategy is to switch. I never said anything about that.

The explanations given (if you could switch to two doors, you would, wouldn't you?) are not. They leave out an important factor. Yes, it does average out to the same over the long run, but in a single game - which is what the question is about - it is not the same.

Quote:
 Originally Posted by v8archie The problem is that we don't know whether Monty has made a choice or not. What we do know is the probability that he made a choice ... Both are valid approaches, working on different assumptions.
No, they are not both valid approaches. You also know what the choice was, if he made a choice. Ignoring information is never correct, even if it works out to the same answer. Because there are other problems - and you know the one I mean - where they don't.

Quote:
 JeffJo's unwritten assumption is that we know something about Monty's approach to choosing.
No, my assumption is that we know nothing about it. Your approach here is essentially the same, but a naive version of it. But in that other problem you assume you do know something about it.

Quote:
 In essence, JeffJo's solution introduces a variable upon which we have no information...
No, it recognizes a random variable that MUST exist in the problem, and uses the fact that we have no information about it to estimate the probability distribution for it. Please stop misrepresenting that.

A similar example is that we don't know how the car's door was selected. This is a random variable upon which we have no information. If we ignore it, and use your methods, we would get (using the usual door numbers) P(Car=3)=0 and P(Car=1)=P(Car=2)=1/2. That would change the answer.

What we actually do is recognize that this selection is a random variable, that it has three equivalent values, and that outcomes with one of those values are eliminated.

October 14th, 2015, 06:18 PM   #16
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Quote:
 Originally Posted by JeffJo Ummmm, you'd switch to door R? So the two are supposed to be the same?
Can you clarify? There is always only one way to switch. What two are you referring to? I don't recall stating that any two things are the same or are supposed to be the same.

Quote:
 Originally Posted by JeffJo Yes, the best strategy is to switch.
If you always switch, what you called "an important factor" is not really important, as it never changes your decision to switch.

Quote:
 Originally Posted by JeffJo I never said anything about that.
You did - you earlier stated "Since there are twice as many cases left where the car is behind R, than cases where it is behind C, you should switch."

 October 14th, 2015, 07:41 PM #17 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 This thing is beginning to look/sound like the missing dollar puzzle/joke. Jeff, your replies are beginning to sound like you no longer care about the solution, but more interested in getting the last word.
October 15th, 2015, 06:06 AM   #18
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Quote:
 Originally Posted by skipjack Can you clarify? There is always only one way to switch. What two are you referring to?
The two cases you compared when you said "That's different from your original assertion." The two times I said switching has probability 1/(1+Q) of winning. One where I explicitly said "to door R," and one where I said "assume Monty opens L," so the door you would switch to is R.

And note that people who don't think switching matters think the probability that either decision wins is 1/2. That fits your description "Both [conditional probabilities] are at least 1/2."

+++++
My point is not that a properly-handled Q affects your decision to switch. It is that Q's existence is what makes it advantageous to switch. People who think switching can't matter are implicitly assuming P(Open=L|Car=C)=P(Open=R|Car=C)=1, so "Both 1/(1 + Q) and 1/(2 - Q) are at exactly 1/2."

My point is they they sense there is something fishy about the argument "you'd switch to both doors, wouldn't you?" even if they do not know what that something fishy is. That something is that the argument reduces a conditional probability to an unconditional probability by ignoring Q. Since it is Q that makes the answer what it is, you need to make them understand it or they won't believe you.

Quote:
 Originally Posted by Denis Jeff, your replies are beginning to sound like you no longer care about the solution, but more interested in getting the last word.
I'm less concerned with the last word, than with getting this point across to people who keep explaining the Monty Hall Problem is a way that has an established history of not convincing anybody.

October 15th, 2015, 06:23 AM   #19
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Quote:
 Originally Posted by JeffJo I'm less concerned with the last word, than with getting this point across to people who keep explaining the Monty Hall Problem is a way that has an established history of not convincing anybody.
Well, it such doth reduceth the price of groceries,
then Go Man Go!!!

 July 22nd, 2016, 10:00 AM #20 Newbie   Joined: Jul 2016 From: Israel Posts: 2 Thanks: 0 This video explains the monty hall problem I think the best, but, it also has a 3d simulation for downloading in the description for free. This simulation proves this problem once and forever!

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