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Lets explore an extension of the monty hall problem. Assume the usual scenario with two goats and one car. Also assume that there are two types of hosts, and you do not know which type your host is.

Host type A is the standard host that gives you a choice to switch. Host type B only gives you a choice if you choose the right door at the beginning, otherwise he reveals that you have lost immediately. You know that host A appears with probability P(A), and host B with probability 1-P(A)

what does P(A) have to be so there is no dominant strategy if you have been given the choice?

My first attempt is to state that the probability of winning P(W) has to be 1/2 if you always switch, so there is no dominant strategy.

Thus

P(W)=P(W|A)P(A) + P(W|B)(1-P(A))=1/2 Note that P(W|B)=0 if you always switch.

Hence

P(A)= P(W)/P(W|A)=1/2/2/3=3/4

Is this correct? I cannot take away the feeling that the fact that you have a choice gives you new information that should change the problem, but I may be simply over thinking everything.

1 Answers 1

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You choose your strategy to switch or to not switch regardless of the host.

Let $W_s$ be your probability of winning if you employ the switch strategy and $W_n$ be your probability of winning if you employ the no-switch strategy

If $B$ is your host, you only get the opportunity to engage the strategy $\frac 13$ of the time.

If $A$ is your host you get to opportunity to engage your strategy every time.

$P(A)P(W_s|A) + \frac 13 P(B)P(W_s|B) = P(A)P(W_n|A) + \frac 13P(B)P(W_n|B)\\ P(W_s|A) = \frac 2{3}\\ P(W_n|A) = \frac 1{3}\\ P(W_s|B) = 0\\ P(W_n|B) = 1 \\ P(A) \frac 23 = P(A) \frac 13 + \frac 13 P(B)\\ P(A) \frac 23 = P(A) \frac 13 + \frac 13 (1 - P(A))\\ P(A) \frac 23 = \frac 13 \\ P(A) = \frac 12$

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    I don't think the first sentence is quite right. Of course if you've decided *in advance* that you won't switch, your chances at that time are $\frac 1 3$, but if you have chosen a door *and been given the option to switch,* and don't, the odds *now* of winning are bound up in $P(A)$ and $P(B)$. Right? (Put another way, it's not necessarily true that $\frac 1 3$ of the times you are *given the option of switching* and don't switch, you will win.) – 2017-01-12
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    I have added more detail to answer your concern, but it doesn't change the result. – 2017-01-12