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I had the following question as part of a Finance course I'm doing. I don't struggle too much with Monty Hall problems in which there are $3$ doors, but this one with $4$ doors ($2$ with Prizes and $2$ with Goats, NONE empty) I am stumped even after several attempts:

"Consider a version of the Monty Hall problem where there are $4$ doors. There are goats behind $2$ of the doors and valuable prizes behind the other $2$ doors. The game is played as follows:

  • You first choose a door.
  • Before your chosen door is opened Monty Hall opens one of the other $3$ doors and reveals a goat behind that door.
  • He now gives you the opportunity to change your mind and select another door.
  • Your chosen door is now opened and your prize is revealed. If you behave optimally, what is the probability that you will win one of the valuable prizes?"

Has anyone encountered this before? Is there a procedure/formula for working it out? And can we generalize the formula (say for $n$ doors, $r$ prizes etc.)?

Thanks in Advance!

Rohit Singh
  • 1,143
Daniel Hardej
  • 25

1 Answers1

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Suppose we have $n$ doors and $1\le k<n-1$ prizes. You're allowed to pick one door, and after another door has been opened (with no prize), you can switch to another door. In this case, the initial probability of winning is $p_0=k/n$. The probability of winning if you switch can be computed as follows: \begin{align} p_{1}&=\frac{k-1}{n-2}\times p_0+\frac{k}{n-2}\times (1-p_0) \\ &=\frac{k(k-1)}{n(n-2)}+\frac{k(n-k)}{n(n-2)} \\ &=\frac{k(n-1)}{n(n-2)}. \end{align} It is always profitable to switch because $$ \frac{k(n-1)}{n(n-2)}>\frac{k}{n}=p_0. $$