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Does someone know a solution to the following generalization of the Monty Hall Problem:

The Problem: Assume you are on Let's Make a Deal and are presented with the regular dilemma of the Monty Hall Problem. Unlike normally, however, the car isn't placed randomly behind a door, but according to a certain probability. So the probability that the game show host puts the car behind door $1$ is $A_1$, behind door $2$ is $A_2$ and behind door $3$ is $A_3$. (With $A_1+A_2+A_3=1$, of course)

Additionally, you have a preference for choosing some of the doors over others, meaning that the probability of you choosing a random one of the three doors on your first try is not necessarily $1/3$. So the probability of you choosing door $1$ is some $B_1$, the probability for choosing door $2$ is $B_2$ and the door $3$ has probability $B_3$ of being chosen by you. (Once again with $B_1+B_2+B_3=1$)

Now what is the probability of you winning the car?

As the normal Monty Hall Problem is already tough enough for me, I thought I might ask you all for help. If my notations are bad please do not hesitate to change them. Nonetheless, my progress so far:

My Progress: Obviously the probability of choosing the $1st$ door and the car being behind it is $A_1\cdot B_1$ if you do not switch. The probability of you choosing the $2$nd door and finding the car behind it if you do not switch is $A_2\cdot B_2$ and the accordingly the probability of finding the car behind door $3$ without switching is $A_3 \cdot B_3$.

And this is right about where my head starts spinning, because I don't know where to go from here on

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    $\begingroup$ Has the participant any knowledge about $A_1,A_2,A_3$? If so then he must drop his original preferences and choose at first for the door $i$ where $A_i$ is minimal. Then if the host has opened a door and invites him to switch he must do that. His chance to win is then $1-A_i$. $\endgroup$
    – drhab
    Commented Mar 23, 2021 at 15:43
  • $\begingroup$ @drhab The participatent has no knowledge about $A_1,A_2,A_3$. $\endgroup$
    – emily.harris
    Commented Mar 23, 2021 at 15:47
  • $\begingroup$ Then nothing changes because there is no way to improve his strategy. $\endgroup$
    – drhab
    Commented Mar 23, 2021 at 15:50
  • $\begingroup$ @drhab As I read the query, the participant doesn't get to choose initially, but is assigned door $k$ as his initial guess, with probability $B_k$. $\endgroup$
    – user2661923
    Commented Mar 23, 2021 at 15:54
  • $\begingroup$ Suppose that the $A_k$ probabilities for doors 1,2,3 are $(1/4), (5/8),$ and $(1/8)$ respectively. Further suppose that the participant has chosen Door 1. $(5/8)$ of the time, door 3 will be shown and $(1/8)$ of the time, door 2 will be shown, by Monty. However, in the $(1/4)$ of the time that door 1 is correct, within that possibility, what percentage of the time will Door 2, be shown, rather than Door 3? Are we to assume that it will be $(1/2)$ of the time that Door 2 will be shown? $\endgroup$
    – user2661923
    Commented Mar 23, 2021 at 16:03

1 Answer 1

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Since you don't know the values of $A_1, A_2, A_3$ then the safe play is to always switch. Suppose that if you were assigned door $k$, that your win factor was $p(k)$.

Then, your overall expectation will be $\sum_{i=1}^3 B_k \times p(k).$

Therefore, the problem is reduced to calculating $p(k)$ based on the OP's query and response to my comment/question.

I will calculate $p(1)$, with the idea that the calculation of $p(2)$ and $p(3)$ should result in a symmetrical formula.

Assuming that the participant was assigned door 1:

$A_2$ of the time, door 2 is correct and door 3 is shown.

$A_3$ of the time, door 3 is correct and door 2 is shown.

$\frac{A_1}{2}$ of the time, door 1 is correct and door 2 is shown.

$\frac{A_1}{2}$ of the time, door 1 is correct and door 3 is shown.

Therefore, the chance that door 2 will be shown is $A_3 + \frac{A_1}{2}.$ When that happens, the participant's chances of winning will be $\frac{A_3}{A_3 + (A_1/2)}.$

Similarly, the chance that door 3 will be shown is $A_2 + \frac{A_1}{2}.$ When that happens, the participant's chances of winning will be $\frac{A_2}{A_2 + (A_1/2)}.$

Therefore,

$$p(1) = \left\{[A_3 + (A_1/2)] \times \frac{A_3}{A_3 + (A_1/2)}\right\}$$ $$+ \left\{[A_2 + (A_1/2)] \times \frac{A_2}{A_2 + (A_1/2)}\right\}$$ $$ = A_3 + A_2 = (1 - A_1).$$

Addendum
Examine the above computation.
Suppose that you are assigned door 1, you are computing $p(1)$, and you are focusing on what happens when Door 1 is correct.

It is actually totally irrelevant what percentage of the time Monty Hall will show Door 2, rather than Door 3. This is demonstrated by the fact that in each of the two terms used to compute $p(1)$, the denominator is always exactly cancelled by the applied scalar.

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