Monday, September 12, 2011

Monty Hall Problem

So someone (I won't mention who to prevent grievous bodily harm to this person) requested that I post about what I'm doing in my classes and such.  So that's what today's post is about. Consider yourself warned.

Picture this: Your the contestant on a game show with host Monty Hall. Monty asks you to choose one of 3 closed doors. Two doors hold goats and one holds $50,000. You choose a door. Monty then opens one of the doors you did not choose. If the opened door contains a goat, Monty asks if you would like to switch doors. What do you do?
Does it make a difference if Monty knowingly opens a door with a goat behind it?
Think about it...



Instead of telling you the answer straight up, I'm going to explain a few things first just to torture you.
To give you a little background, conditional probability of A given B means the probability of event A given that event B has occurred. Also, the conditional probability of A given B equals the probability of A and B divided by the probability of B. Notation-wise, P(A) denotes the probability of event A occurring. Also P(A|B) denotes the conditional probability of A given B. Also also, P(A|B) = P(A and B)/P(A).

So back to Monty Hall.  Let's assume:
W = the event that you win the $50,000
G = the event that a goat is revealed
M = the event that the money is revealed
Since there are 3 doors and only one holds $50,000; the probability of winning is 1/3, or using our handy notation, P(W) = 1/3.

If Monty randomly opens one of the un-chosen doors then either the money is revealed (denoted M) or a goat is revealed (denoted G).
So P(W) = P(W and G) + P(W and M).
Using the above information about conditional probabilities and doing a little algebra:
P(W) = P(W|G)*P(G) + P(W|M)*P(M)
We know that P(W) = 1/3, P(G) = 2/3, P(M) = 1/3 and that P(W|M) = 0 (because you cannot win if Monty opens a door that you did not chose and it reveals the money.) Filling in what we know, we now have
1/3 = P(W|G)*(2/3) + 0*(1/3)
Using simple algebra, we find that P(W|G) = 1/2.
This tells us that if we chose a door and Monty randomly opens one of the un-chosen doors and it reveals a goat, it doesn't matter if we switch doors or not.

Now what if Monty knowingly opens a door with a goat behind it? Should this affect our decision to switch doors? Let's find out.
So we still have P(W) = P(W|G)*P(G) + P(W|M)*P(M) just like in the last part, but now some of the probabilities will be different because Monty knows which doors have goats and will select one of those to open. P(W) = 1/3, P(W|M) = 0, but now P(G) = 1 because Monty always opens a door with a goat in this scenario and for the same reason, P(M) = 0. So we have:
1/3 = P(W|G)*1 + 0*0
Now we see that P(W|G) = 1/3.
This tells us that if we chose a door and Monty knowingly opens one of the un-chosen doors to reveal a goat, we have a 33.3% chance that our door contains the money and a 66.7% chance that the door that we didn't chose holds the money. In this scenario it would be in our best interest to switch doors.

So in on scenario it doesn't matter if we switch or not and in the other, it is in our best interest to switch. Bottom line: switch doors!

4 comments:

  1. This puzzle made me think of Princess Bride and the wine. Good movie.

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  2. Praise God He called you for this and not me!!!
    I'll just take the goat and save myself the trouble. :)
    -Amy

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  3. I LOVE THAT MOVIE! Lisa in your next example you should use something out of the Princess Bride! Your kidies would love you :)

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  4. That's hideous. There is no excuse.

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