# monty hall multiple prizes

Really surprised I didn't see this already posted on math SE...(as far as I can tell)

I understand the N door, k-revealed Monty Hall extension: Monty hall problem extended.

The solution is to switch if: (n-1)/n * (1 / (n-k-1)) > 1/n which occurs for any k>=1

Now let's extend this to say: of the N doors, P are prize doors and N-P are non-prize doors. Monty Hall reveals k doors. In this case, k will have to be small enough that Monty Hall can reveal a nonprize door, meaning you cannot have k so big that Monty is forced to revealed one or more of the prize doors. So as usual, he can only reveal nonprize doors. And also does not open your original door.

In this case, what is the inequality determing if you should switch or not, similarly to the inequality in the accepted answer given above for the Ndoor, k revealed version? This time it will also be a function of P the #prize doors.

If there are $N$ doors of which $P$ contain a prize, the probability that you win if you don't switch is $P/N$. Suppose Monty then reveals $k$ randomly selected[*] non-prize doors (with $k\leq N-P-1$, which means he always has $k$ available doors to reveal).

Then there are two possibilities:

• Your original door contains a prize. This happens with probability $P/N$. Of the $N-k-1$ remaining doors, there will be $P-1$ which contain a prize, so the probability that this happens and then you win the prize after switching to a random unopened door is $$\frac{P}{N}\frac{P-1}{N-k-1}$$
• Your original door does not contain a prize. This happens with probability $\frac{N-P}{N}$ and means that $P$ of the remaining doors contain prizes, so the probability that this happens and you win after switching is $$\frac{N-P}{N}\frac{P}{N-k-1}$$

That is, your total probability of winning after switching is $$\frac{P}{N}\frac{P-1}{N-k-1}+\frac{N-P}{N}\frac{P}{N-k-1}=\frac{P^2-P+NP-P^2}{N(N-k-1)}=\frac{P(N-1)}{N(N-k-1)}$$

So the inequality you're looking for is: you should switch when $$\frac{P(N-1)}{N(N-k-1)} \geq \frac{P}{N}$$

That is, switching multiplies your probability of winning by a factor of $\frac{N-1}{N-k-1}$, which is always greater than $1$ when $k$ is positive. So you should always switch, a result which is not surprising as this problem is qualitatively the same as the original Monty Hall problem. In both cases, after Monty opens some doors, you have more information about the location of the prizes than you did before the doors were opened, so you have a better chance of making the right decision.

[*] If the doors Monty opens are not randomly selected, it is possible that he could leak some extra information to you about what to do based on them. For example, if you know that he's lazy and always opens the leftmost doors that he's allowed to, then you clearly want to switch to one of the doors that he skips over. I think that's beyond the intent of this question — but it might make a good follow-up!

• yes so the inequality is actually exactly the same as single prize case (multiplied by P on both sides cancels out) so actually number of prizes doesn't matter, except that it constrains how many doors can be opened – dektorpan Sep 13 '17 at 3:48
• @dektorpan: Indeed. You could think of this as a consequence of linearity of expectation. – Micah Sep 13 '17 at 4:26

You pick a door hiding a prise with probability of $P/N$, and $1-P/N$ is the probability for picking a door hiding a no-prise.

Monty is guaranteed to reveal $K$ of the no-prise doors, where $0\leq K\leq (N-P-1)$.

The probability that you will pick a prize if you switch is then:$${\frac PN\cdot\frac{P-1}{N-K-1}+\frac{N-P}{N}\cdot\frac{P}{N-K-1} \\~\\~\\ = \frac{(N-1)P}{N(N-K-1)}}$$

So when is it better to switch? $\phantom{K>0}$