Adaptation to Monty Hall Problem

[Dusseldorf]

We are familiar with the monty hall problem or we can check wikipedia for an outline. We agree that the probability, after the host intentionally reveals the door with a goat, is 1/3 for your original choice and 2/3 for the door you can switch to.

My question is this. Suppose now that Monty will behave randomly as follows:

If your original choice was correct, with 2/3 probability Monty will say "I think you should keep your first choice".

If you original choice was incorrect, with 1/3 probability he will say "I think you should keep your first choice".

Otherwise he will say nothing. How does this affect the probability distribution.

Thanks any replies. Supposedly I am teaching a lesson on the monty hall problem.

[MickeyHoldem]

You will pick the correct door 1/3 of the time and Monty will tell you your correct 2/3 of the time... 1/3 * 2/3 = 2/9

You will pick the wrong door 2/3 of the time and Monty will tell you your correct 1/3 of the time... 2/3 * 1/3 = 2/9

Looks like you have an equal chance of Monty saying your right no matter what, therefore you can ignore Monty, and continue to listen to the little voice inside that says "switch".

[jason1990]

I get that if he tells you to stay, then there's a 50% chance the prize is behind the door you picked; if he tells you nothing, there's only a 20% chance the prize is behind the door you picked.

Without loss of generality, suppose you initially picked door 1 and that Monty has just shown you the goat behind door 3. Let Aj be the event that the prize is behind door j. Let B be the event that Monty shows you the goat behind door 3 AND tells you to stay. Then

P(A1|B) = P(B|A1)P(A1) / sum{P(B|Aj)P(Aj)}
= P(B|A1) / sum{P(B|Aj)}
= [(1/2)*(2/3)] / [(1/2)*(2/3) + (1)*(1/3) + (0)*(1/3)]
= [1/3] / [2/3]
= 1/2.

On the other hand, if we change B to be the event that Monty shows you the goat behind door 3 AND tells you nothing, then

P(A1|B) = P(B|A1)P(A1) / sum{P(B|Aj)P(Aj)}
= P(B|A1) / sum{P(B|Aj)}
= [(1/2)*(1/3)] / [(1/2)*(1/3) + (1)*(2/3) + (0)*(2/3)]
= [1/6] / [5/6]
= 1/5.

[jason1990]

Incidentally, I was assuming Monty will intentionally show you the door with the goat AND will behave randomly as you described. If he ONLY behaves as you describe, then the calculations are different. But curiously, the final answer remains the same. The difference of course is that, when you switch, you have two doors to switch to. By symmetry, the probabilities of finding the prize behind each of those doors is the same.

So if you pick door 1 and Monty tells you to stay, then the probability the prize is behind

door 1 is 50%
door 2 is 25%
door 3 is 25%

If you pick door 1 and Monty says nothing, then the probability the prize is behind

door 1 is 20%
door 2 is 40%
door 3 is 40%

[Dusseldorf]

That is basically what i was coming up with.. Is there any way monty can make the P() 50% always?

[set57hike]

P(Right|Monty says stay)=(2/9)/(4/9)=1/2
P(Wrong|Monty says stay)=(2/9)/(4/9)=1/2

P(Right|Monty says nothing)=(1/9)/(5/9)=1/5
P(Wrong|Monty says nothing)=(4/9)/(5/9)=4/5

Should we change our strategy of always switching depending on what Monty does?
Clearly, if Monty says nothing, we should always switch, as we will win with probability 80% when Monty says nothing and we switch.

If Monty says stay, we are free to choose any strategy, since we will win with probability 50% no matter what we do. Among these choices, we could choose to always switch.

Hence always switch remains optimal and we will win with probability 2/3. However, there are now an infinite number of alternative strategies which are also optimal.

Note that the probability distribution on Monty's comments were very carefully chosen to give this interesting result. If the information he gave was a little bit better than this, we would have a new optimal strategy.

To make this point, suppose Monty did the following...
If your original choice was correct, with 100% probability Monty will say "I think you should keep your first choice".

If you original choice was incorrect, with 0% probability he will say "I think you should keep your first choice".

Otherwise he will say nothing. How does this affect the probability distribution.

Now the optimal strategy is to alwatys switch when he says nothing and to always stay when he says stay. You will win with probability 1.

[ QUOTE ]
We are familiar with the monty hall problem or we can check wikipedia for an outline. We agree that the probability, after the host intentionally reveals the door with a goat, is 1/3 for your original choice and 2/3 for the door you can switch to.

My question is this. Suppose now that Monty will behave randomly as follows:

If your original choice was correct, with 2/3 probability Monty will say "I think you should keep your first choice".

If you original choice was incorrect, with 1/3 probability he will say "I think you should keep your first choice".

Otherwise he will say nothing. How does this affect the probability distribution.

Thanks any replies. Supposedly I am teaching a lesson on the monty hall problem.

[/ QUOTE ]

[jason1990]

This is probably not what you're looking for, but imagine this: Monty gets sick and his brother-in-law takes over the show. His brother-in-law has no idea how the game works, so when I pick door 1, he randomly picks one of the other two doors to show me (without regard for where the goats are). He happens to pick door 3 and there happens to be a goat behind door 3. Now the probabilities of switching vs. not are 50-50. And yet, from an objective standpoint, we have the same information as when Monty hosts the game: a goat is behind door 3.

It might be instructive to have students think about this variant. I think it emphasizes the fact that we don't condition on "information", we condition on "events"; and the event that Monty shows us the goat behind door 3 is different from the corresponding event involving his brother-in-law.