Variation on the Monty Hall Problem

Suppose the prize is in A, for argument's sake. There are 6 equally likely options a priori:

• you pick A and the host picks B (no switch)
• you pick A and the host picks C (no switch)
• you pick B and the host picks A (this did not happen, as we then wouldn't see the goat)
• you pick B and the host picks C (switch)
• you pick C and the host picks A (ruled out, as before)
• you pick C and the host picks B. (switch)

By knowing that we saw a goat when the host picked (without information) we have that we have 4 situations we could be in (all same probability) and in 2 of them we need to switch. So we now have a 1/2 chance.

I had prepared a great argument to this, typed it all up, and realized I was wrong at the last moment. This question has already been answered, but I thought I'd add my explanation in case it helps anyone later.

Bayes Theorem: $$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$$ For this case, I'm using this:

• Let Winning be the probability of winning if you switch
• Let Goat be the probability of Monty Hall revealing a goat

$$P(Winning|Goat)=\frac{P(Goat|Winning) \cdot P(Winning)}{P(Goat)} $$

For the original Monty Hall problem, because Monty knows where the goat is, there's a 100% chance he'll reveal a goat whether or not switching will make you win (P(Goat|Winning)). There's also a 100% chance he'll just pick a goat (P(Goat)). There are 4 out of 6 possible winning outcomes if you adopt a switching strategy. The formula works out like this: $$P(Winning|Goat)=\frac{1 \cdot .66}{1}=.66$$

If Monty Hall does not know where the goats are, the probabilities are a little different and is a little complicated (at least it is for me).

The probability of Monty Hall picking a goat ($P(Goat)$) is pretty easy; that's just 2 out of 3 or about 66%.

The other two variables depend on the rules of the game. For example, if Monty Hall will let you switch to the car if he reveals the car, then the probability of winning the game if you adopt a switching strategy ($P(Winning)$) is 66% because there are 4 winning outcomes out of 6; however, if you're not allowed to switch to an open door, then the two outcomes where Monty Hall picks the car at random become losing scenarios and $P(Winning) = \frac{2}{6} = 33\%$.

Finally, P(Goat|Winning) means "the probability of Monty Hall picking a goat given that you will win if you switch." If you are guaranteed to win if you switch doors, that means that you must have selected one of the two goats.

For the first rule there are four outcomes:

• You picked goat B, Monty reveals car A, you switch, you win
• You picked goat C, Monty reveals car A, you switch, you win
• You picked goat B, Monty reveals goat C, you switch, you win
• You picked goat C, Monty reveals goat B, you switch, you win

So, of the winning switching outcomes, Monty picks a goat half the time. Thus, $P(Goat|Winning)=50\%$ for the first rule.

In the second rule, there are two winning outcomes:

• You picked goat B, Monty reveals goat C, you switch, you win
• You picked goat C, Monty reveals goat B, you switch, you win

Remember, if Monty reveals a car for the second rule, you lose automatically. As a result, $P(Goat|Winning)=100\%$ for the second rule.

If you are allowed to switch to the door Monty reveals, then: $$P(Winning|Goat)=\frac{.5 \cdot .66}{.66} = .5$$

On the other hand, if Monty revealing a car means you lose, then: $$P(Winning|Goat)=\frac{1 \cdot .33}{.66} = .5$$