Monty Hall Problem... For Newbies

[dlorde]

There is a dispute over what the odds in a given situation would be, so people are discussing it.

Fair enough, although it seems to me that this stuff is pretty simple once the problem is fully and unambiguously specified. Given the earlier discussion, there seems to be a high probability that some of these disagreements are a result of having different interpretations of an ambiguous or incompletely specified problem.

Just sayin'

 

[Meed]

I think people are getting a little overly hung up on the word "discarded". All it means is for a question "What are the odds of X given Y?", that when we run a simulation which can generate either Y or not-Y, we are only interested in the Y cases, so we don't bother with the rest. It's just a way of fitting a simulation to a conditional probability question.

If, alternatively, you answer the question by calculating the odds directly instead of running a simulation, then there is no "discarding". We could calculate an answer to the question "In the random Monty, what are the odds that sticking will win, given that Monty has revealed a goat?" by using the conditional probability formula:

P(B|A)* = P(A and B) / P(A)

P(A) = P(Monty reveals a goat) = 2/3
P(B) = P(Win by sticking) = 1/3
P(A and B) = 1/3

P(B|A) = (1/3) / (2/3) = 1/2

Translation: The probability of winning by sticking, given that Monty has revealed a goat in the Random Monty, is 1/2.

*"P(B|A)" denotes "the probability of B, given A".

 

[dlorde]

I think people are getting a little overly hung up on the word "discarded".

If that was referring to me, I used it simply to emphasise that, in the modified MH problem being discussed, only a subset of all trials was being considered, so that a specific set of initial choices was not being considered, i.e. compared to the standard MH problem, some trials were being discarded.

Not sure how that equates to being 'hung up' on the word

 

[tsig]

The mythbusters tested this and found out that it pays to switch but most people won't switch.


When presented with the Monty Hall Problem, people would be more likely to win if they changed their decision.

confirmed

They built a small-scale simulator to do 50 trials each, with Adam always switching his choice and Jamie never switching. Adam won far more often than Jamie did, and Jamie explained the reason: because the player has a 2/3 probability of choosing a losing door at first, switching turns the odds in his favor.


http://mythbustersresults.com/wheel-of-mythfortune

 

[Meed]

If that was referring to me, I used it simply to emphasise that, in the modified MH problem being discussed, only a subset of all trials was being considered, so that a specific set of initial choices was not being considered, i.e. compared to the standard MH problem, some trials were being discarded.

Not sure how that equates to being 'hung up' on the word

It's just that statements such as

"If you start discarding trials on various criteria, you can make the odds pretty much what you like. So what?"
"The discarding is what directly changes the odds"
"What Monty does only creates the need to discard."
"Only because the games in which he reveals the car are all discarded."

are vaguely confusing to me.

Like, say I did an empirical analysis supporting the conclusion "when it rains it pours". Maybe I found that 90% of the time it rained, it also poured. If someone responded "only because you didn't include the days in which it didn't rain at all in your analysis", I would be confused. Not that the response would be incorrect exactly, I just wouldn't understand why it was being said.

I guess it really doesn't matter. *shrug*

 

[dlorde]

Like, say I did an empirical analysis supporting the conclusion "when it rains it pours". Maybe I found that 90% of the time it rained, it also poured. If someone responded "only because you didn't include the days in which it didn't rain at all in your analysis", I would be confused. Not that the response would be incorrect exactly, I just wouldn't understand why it was being said.

The difference is the context. In the context of the thread - the standard MH problem, the modified version had caused some confusion. I just thought it worth pointing out the significant change in the problem. YMMV.

 

[Meed]

Fair enough.

 

[JohnnyG]

Translation: The probability of winning by sticking, given that Monty has revealed a goat in the Random Monty, is 1/2.

The probability of that happening in the Random Monty is 2/3, making overall odds of winning 1/3 if you stick. Those are the same odds of winning as in the original MH problem where the odds of revealing a goat are 1 because it is specified that he knows where they are. The odds changed to 1/2 because you discarded 1/3 of the possible outcomes, when you factor them back in the odds return to 1/3.

 

[Meed]

The probability of that happening in the Random Monty is 2/3, making overall odds of winning 1/3 if you stick.

Indeed. That's why we start with P(A) = P(Monty reveals a goat) = 2/3 , P(B) = P(Win by sticking) = 1/3, and P(A and B) = 1/3.

If we ask the same conditional probability question when Monty is following the standard rules, then it's P(A) = P(Monty reveals a goat) = 1 , P(B) = P(Win by sticking) = 1/3, and P(A and B) = 1/3.

Those are the same odds of winning as in the original MH problem where the odds of revealing a goat are 1 because it is specified that he knows where they are. The odds changed to 1/2 because you discarded 1/3 of the possible outcomes, when you factor them back in the odds return to 1/3.

I suppose one could think of it this way, but personally I find it rather muddled to speak of "discarding" outside the context of a simulation. I just see it as a change in the value of P(A).

 

[JohnnyG]

Indeed. That's why we start with P(A) = P(Monty reveals a goat) = 2/3 , P(B) = P(Win by sticking) = 1/3, and P(A and B) = 1/3.

If we ask the same conditional probability question when Monty is following the standard rules, then it's P(A) = P(Monty reveals a goat) = 1 , P(B) = P(Win by sticking) = 1/3, and P(A and B) = 1/3.



I suppose one could think of it this way, but personally I find it rather muddled to speak of "discarding" outside the context of a simulation. I just see it as a change in the value of P(A).

I think the highlighted value should be 1/2.

I think they are two different ways of looking at the same thing. We are essentially saying that comparing the Random Monty P(B) = 1/2 to the Standard Monty P(A and B) = 1/3 is rather meaningless because it only considers a subset of all available outcomes.

 

[ynot]

If anyone believes that discarding the games when random Monty inadvertently reveals the car doesn’t affect the odds then why discard? Make it that the contestant automatically wins the car. After all, its Monty’s mistake, not the contestants.

It simply doesn’t make any sense that the game would be played in a way that Monty could inadvertently reveal the car before the contestant has had a chance to guess where it is. It therefore doesn’t make any sense to discuss a game that would never be played in practice. Unless you’re a math nerd.

 

[Modified]

If anyone believes that discarding the games when random Monty inadvertently reveals the car doesn’t affect the odds then why discard? Make it that the contestant automatically wins the car. After all, its Monty’s mistake, not the contestants.

Because the interesting question is whether the contestant should switch. Cases where they have no choice are not relevant to that question.

 

[ynot]

Because the interesting question is whether the contestant should switch. Cases where they have no choice are not relevant to that question.

Perhaps I should have added a smiley to let you know I wasn’t being that serious.

As mentioned earlier, the rules in the OP clearly define that Monty knows which door the car is behind and always reveals a goat, so there’s never a game played where he could or would inadvertently reveal the car. This thread is about that standard Monty game so why is it being derailed with a non-standard scenario?

 

[ynot]

Because the interesting question is whether the contestant should switch. Cases where they have no choice are not relevant to that question.

Not only not relevant but also silly. If the interesting question is whether the contestant should switch why then create a silly scenario in which their switching choice is often negated?

 

[Meed]

I think the highlighted value should be 1/2.

No, P(B) is 1/3 in both cases. It's the probability that our initial pick was correct when we have no other information. It isn't actually used in the conditional probability formula, but it's useful for determining what the value of P(A and B) is. P(A and B) is the joint probability that our initial pick was correct AND Monty reveals a goat. It's also 1/3 in both cases, because Monty will always reveal a goat when we've picked the car in both the Standard Monty and the Random/Ignorant Monty. The only thing different between the two situations is P(A). That is, the probability of Monty revealing a goat.

I think they are two different ways of looking at the same thing. We are essentially saying that comparing the Random Monty P(B) = 1/2 to the Standard Monty P(A and B) = 1/3 is rather meaningless because it only considers a subset of all available outcomes.

I don't know what you mean by "comparing the Random Monty P(B) to the Standard Monty P(A and B)". The relevant comparison would be to compare P(B|A) for the Random Monty with P(B|A) for the Standard Monty.

And I've no idea what you mean by saying it's "rather meaningless because it only considers a subset of all available outcomes". Conditional probabilities are far from meaningless. Pretty much the entire field of statistics and Bayesian data analysis are dependent on conditional probabilities.

 

[Modified]

Not only not relevant but also silly. If the interesting question is whether the contestant should switch why then create a silly scenario in which their switching choice is often negated?

Because it can help in understanding the original question, as several people have already pointed out.

 

[ynot]

Because it can help in understanding the original question, as several people have already pointed out.

If people can’t understand the relatively simple standard Monty game then I don’t see how making it more complicated and convoluted is going make things any easier for them.

 

[Modified]

If people can’t understand the relatively simple standard Monty game then I don’t see how making it more complicated and convoluted is going make things any easier for them.

It can because they may be essentially mistaking one game for the other when computing the odds.

 

[hgc]

New rule: In some future year, when someone starts another new thread on this topic, please phrase the problem thusly:

A contestant on The Price is Right is told by Monte Hall that behind Doors 1, 2 and 3 are two goats and one car. THIS IS TRUE. IT IS ALWAYS TWO GOATS AND ONE CAR. The contestant, trying to win the car, guesses a door and tells Monte. Before revealing what is behind the door selected, Monte opens one of the other doors, to reveal a goat, and gives the contestant the choice to switch. MONTE ALWAYS REVEALS THE GOAT. HE KNOWS WHAT'S BEHIND EACH DOOR, SINCE IF HE DIDN'T HE WOULDN'T KNOW WHICH DOOR TO REVEAL. THERE IS NEVER A VARIATION. HE ALWAYS REVEALS A GOAT. Should the contestant stay or switch?

That should clear things up. Or not.

ETA: Permitted variations -- curtains instead of doors or mules instead of goats, but not both

 

[Roboramma]

New rule: In some future year, when someone starts another new thread on this topic, please phrase the problem thusly:





That should clear things up. Or not.

You forgot to mention that if there are two goat doors for Monty to choose from, he chooses randomly between them.

Also, what's wrong with discussing other variations of the problem? Regarding the original problem, we've already finished the discussion: there's no one left posting in the thread who is unclear about the odds of that scenario.

But some people found some other interesting things to talk about. What's wrong with that?

I found the two coin flip problem interesting, for instance.