2a) Suppose you are playing “Let’s Make a Deal”, the “prize” is worth $10,000, and a...

2a) Suppose you are playing “Let’s Make a Deal”, the “prize” is worth $10,000, and a goat is worth nothing. Further suppose that the initial pick of a door is not part of the game (since it’s random anyway). You’ve always picked door A initially before the game has begun. The game starts as nature randomly determines where the prize is (1/3 chance for any of the three doors); then Monty Hall selects one of doors B or C that has a goat (also a move by nature); then you choose to Stay at door A or to Switch to the unopened door. Finally, the outcome is resolved – either you find a goat or the prize behind the door you finally selected. If both B and C have a goat, Monty flips a coin (50/50 chance) to determine which one to open. If either B or C have the car, there is only one available door Monty can open, so he opens that one.

The problem assumes you’ve chosen A, but in words without solving anything, explain why your initial choice of door doesn’t really matter?

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Expert Solution

MONTY HALL PROBLEM

Monty Hall, the famous game show host on TV called "let's make a deal". In this show you can win a prize worth $10,000.There are 3 doors A,B,C. He asks you to choose a door. one door has a prize of $10,000 and 2 doors have goats. Obviously you want to win prize not goat. You have no idea which one has prize. Suppose initially you pick the door A. probability of getting a prize for choosing door A is 1/3. Monty knows which one has prize. He has 2 choices to open door either B or C. Monty opens either B or C revealing the goat. Now you left with 2 choices either to switch or stay with your initial choice. Suppose Monty open the door B revealing the goat. now you may choose A or C. Most people think that if door B has goat then door A or C have 50/50 chances to have prize. But now we don't know if doors are equally likely or not since here it is conditionally dependent on the new information emerges. There are 3 doors and your initial choice gives you probability of winning 1/3. There are 2 doors left which have together of 2/3 chance of winning the prize. Monty opening one of those doors does not change the probability. those probability will still be 2/3. we explain it by a tree diagram.

We have 3 branches. One is what you choose, second is which door has prize , third is which door Monty opens.

You choose door A. Initially by assumption we have equally likely probability of 1/3 for prize door. We choose door A, car is behind A then Monty can open either door B or C with equal probability 1/2. Now if we picked door A and car is behind B then Monty has no choice but to open C with probability 1.

if we pick A , car is behind C then Monty has no choice but to open B with probability 1 to reveal goat.

suppose Monty opens B then we have 2 paths only. when you condition on something you delete from your space everything that is irrelevant with what you observe.

since we remove other paths so we have to renormalized it to get total probability equal to 1.

P(success if switch|Monty opens door B)= 2/3.

hence if you switch your probability of success is 2/3 and if you stick with your initial choice then probability of success is 1/3. so it's better to switch.

orchestra answered 1 year ago

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