OT: The Monty Hall Problem

[Dacks]

A stricter mathematical proof for the problem.

Events

C1 = car is behind door one
C2 = car is behind door two
C3 = car is behind door three

MH1 = Monty Hall reveals what is behind door one
MH2 = Monty Hall reveals what is behind door one
MH3 = Monty Hall reveals what is behind door three

We will assume that the car is placed randomly, so P(C1) = P(C2) = P(C3). The probabilities of MH1, MH2 and MH3 will depend on other factors, so we'll get into those as they come.

Original Problem

Assume the player selects door number one. (This is a safe assumption since we can just renumber the doors after he chooses.) We can eliminate the event MH1 since Monty can't show what's behind the door the player selects.

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P(winning car if he stays) = P("C1 and MH2" or "C1 and MH3")
= P(C1)*P(MH2|C1) + P(C1)*P(MH3|C1)
= 1/3 * 1/2 + 1/3 * 1/2
= 1/3

(To win if he stays, C1 must be true. If C1 is true, then Monty has the choice of either MH2 or MH3. Assuming Monty chooses randomly, P(MH2|C1) = P(MH3|C1) = 1/2.)

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P(winning car if he switches) = P("C2 and MH3" or "C3 and MH2")
= P(C2)*P(MH3|C2) + P(C3)*P(MH2|C3)
= 1/3 * 1 + 1/3 * 1
= 2/3

(To win if he switches, either C2 or C3 must be true. If C2 is true, then Monty must reveal door three, hence P(MH3|C2) = 1. Similarly, P(MH2|C3) = 1.)

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I'll see if I can put together a similar solution for the 3 player problem.

 

[Dacks]

The 3 Player Problem

Let's assume player 1 selects door 1, player 2 selects door 2, and player 3 selects door 3. Again, a fair assumption.

This time, however, Monty can reveal any door without a car behind it. Since each door has the same likelihood of not concealing the car, P(MH1) = P(MH2) = P(MH3) = 1/3.

Now, let's look at player 1's chances of winning given that he is not eliminated by Monty.

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P("winning if he stays" | "MH2 or MH3") = P("MH2 or MH3" and "winning if he stays") / P(MH2 or MH3)

(This is just a standard re-arrangement of the "A given B" formula)

Let's look at P("MH2 or MH3" and "winning if he stays") on it's own. This is the same as P(winning if he stays) while removing the possibility of MH1. As in the original problem, we get

P(winning car if he stays) = P("C1 and MH2" or "C1 and MH3")
= P(C1)*P(MH2|C1) + P(C1)*P(MH3|C1)
= 1/3 * 1/2 + 1/3 * 1/2
= 1/3

P(MH2 or MH3) = P(MH2) + P(MH3) = 1/3 + 1/3 = 2/3.

Then simple division gives you P("winning if he stays" | "MH2 or MH3") = 1/2.

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P("winning if he switches" | "MH2 or MH3") = P("MH2 or MH3" and "winning if he switches") / P(MH2 or MH3)

Let's look at P("MH2 or MH3" and "winning if he switches"). This is the same as P(winning if he switches) while removing the possibility of MH1:

P(winning if he switches) = P("C2 and MH3" or "C3 and MH2")
= P(C2)*P(MH3|C2) + P(C3)*P(MH2|C3)
= 1/3 * 1/2 + 1/3 * 1/2
= 1/3

(In this game, Monty has no restrictions, so P(MH3|C2) and P(MH2|C3) are only 1/2, not guarantees.)

P(MH2 or MH3) = P(MH2) + P(MH3) = 1/3 + 1/3 = 2/3.

Then simple division gives you P("winning if he switches" | "MH2 or MH3") = 1/2.

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[dash]

I'd give rep to both posts if I could.

I took care of Dacks, dare...I love the "simple division" gets us to at the end of his post.

 

[juliansteed]

Fuck Monty Hall...Chuck Barris was were it's at....

What about Monty Hall's current successor Wayne Brady?

YouTube - Wayne Brady Terrorizes Dave Chappelle

 

[ritari330]

Don't worry, mine does too! I'm applying for a job at statscan and their entrance competition is at the end of October, figured this would be a good way to start reviewing.

hope you get it Dacks!