Monty Hall variation, where goats are followed by random doors

Question

3 doors, 2 of which have undesirables behind them, 1 of which has a prize behind it. The items are randomly distributed.

Monty's behaviour:

If you first choose the door with the prize, then Monty will always show you a door with an undesirable, and lets you decide whether or not to switch.

If you first choose a door with an undesirable, 1/2 the time Monty just lets you lose, and 1/2 the time Monty shows you the other door with an undesirable, and lets you decide whether or not to switch.

You pick Door 1; Monty opens Door 2 to reveal an undesirable.

What are the chances there is a prize behind Door 1? Do you switch?

Answer 1

conditional probabilities.

Three things can happen.

You pick the right door and he shows you another. That happens $\frac 13$ of the time.

You pick a wrong door and he shows you another. That happens $\frac 23\frac 12 =\frac 13$ of the time.

You pick a wrong door and he doesn't show you another. That happens $\frac 23\frac 12 = \frac 13$ of the time.

The last one just didn't happen. Of the two that could have happened, they were each equally likely to happen.

So the probability you picked a good door or a bad door are equal. $\frac 12$. Switch or stay, the probability of switching leading to the prize (you picked a bad door) and the probability of switching leading to the bad door (you picked the good door) are equal.

Formally

$A =$ good door picked

$B = $ bad door picked

$C = $ door shown.

$P(A|C) = \frac {P(A)P(C|A)}{P(C)}=\frac {\frac 13\times 1}{P(C)}$

But $P(C) = P(C\&A) + P(C\&B)$ as $A$ and $B$ are mutually exclusive $=\frac 13*1 + \frac 23*\frac 12 = \frac 23$

So $P(A|C) = \frac {P(A)P(C|A)}{P(C)}= \frac {\frac 13}{\frac 23} =\frac 12$.

Answer 2

Label the door we first pick, "door $1$", without loss of generality.

$$\small\begin{array}{|c|c|c|c|} \hline \text{Prize behind door 1 } (33\%)&\text{Prize behind door 2 } (33\%)&\text{Prize behind door 3 } (33\%) \\ \hline \begin{array}{c|c}\text{M opens door 2}&\text{M opens door 3} \\50\%\text{ chance}&50\%\text{ chance} \\ \\\text{Switch to 3 loses}&\text{Switch to 2 loses} \end{array} &\begin{array}{c|c}\text{Game ends}&\text{M opens door 3} \\50\%\text{ chance}&50\%\text{ chance} \\ \\&\text{Switch to 2 wins} \end{array}& \begin{array}{c|c|c}\text{Game ends}&\text{M opens door 2} \\50\%\text{ chance}&50\%\text{ chance} \\ \\&\text{Switch to 3 wins} \end{array} \\ \hline \end{array} $$

Therefore, the probability that we win after switching is $2\cdot50\%\cdot33\%=33\%$, which is also the probability of disadvantageous switching. So, in this variation, it makes no difference whether or not we switch. The counterintuitive edge that switching would have given us in the normal Monty Hall variation is offset by the chance of the game ending.