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I'm having a bit of trouble understanding this particular version of the Monty Hall problem.

Say that there are 5 doors, and I pay \$100 to switch doors, with a potential reward of \$1,000. If I switch doors, what is my expected winning value? In class, my professor said that the probability of us switching and winning (net win = \$900) is (4/15). Further, the probability of us switching and not winning is (11/15).

Can anybody offer any insight into why these are our probabilities?

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    Shall we assume there is only one prize and that the host *always* picks a prizeless door ? – 2017-02-07
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    Yes! Sorry should have clarified that. – 2017-02-07

1 Answers 1

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Assuming there is only one prize the probability of being right on the first pick is: $1/5$.

Thus the expectation given staying is $\mathsf E(W\mid S^\complement) = \tfrac 15\cdot 1000= 200$

Assuming the host always reveals a no-prize, then the probability of picking a prize given switching is $4/15$, because you can be right first time and certainly loose if you switch, or be wrong first time and then the prize is equally likely to be behind one from the three remaining doors. $$\tfrac 15\cdot 0+\tfrac 45\cdot \tfrac 13=\tfrac 4 {15}$$

The probability of losing given you switched is thus $11/15$.

Thus the expectation given switching is: $\mathsf E(W\mid S) = \frac 4{15}\times 900 - \frac {11}{15}\times 100 = \frac {500}3 = 166.\dot 6$

In conclusion, you are more likely to win if you switch, however it costs more than it is worth.

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    Hmm. Okay. So saying that P(W) is my prob of winning given that I switch. Then P(W)=P(W|S)P(S)+P(W|~S)P(~S), where S refers to the probability of me switching. In the solution you have above, what would P(W|S) be? I guess I am looking for advice in setting up this problem in a more methodical way so I have a template to work with when trying to solve this and more difficult versions of the Monty Hall problem. – 2017-02-07
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    @agra94 What we've calculate is: $\mathsf P(W=900\mid S) = \tfrac 4{15}, \mathsf P(W=-100\mid S)=\tfrac {11}{15}$ . However $S$ is not strictly an event; it is a strategy. You either choose to do so or do not ; so how do you assign a probability measure. – 2017-02-07