I am trying to analyze a Monty Hall question with four doors (3 goats, 1 car) just so I can then apply the problem with n doors.
I applied Bayes' theorem, calculated the probabilities and am trying to do an expected value analysis. In order to do so, I am looking at every option a player of the game has but when I add up the probabilities, it gets a result greater than 1.
What I have so far (assuming only one goat is revealed):
You have a 1/4 chance of winning without switching, 3/8 if you win and switch, 3/8 if you lose after switching. I confirmed this answer with @Just_a_fool response here. However, shouldn't Just_a_fool also take in consideration the change of losing without switching (3/4 chance) in order to get every outcome? The problem with that is the probability adds up to 1.75. Am I missing something? Any help would be greatly appreciated!
Best Answer
First of all, the chance of not getting the price when switching is $\frac{5}{8}$, rather than $\frac{3}{8}$, since there is a $\frac{1}{4}$ chance the price is behind the door you chose initially, and there is a $\frac{3}{8}$ chance that it is the other of the two doors that you pick from when switching, so $P(Lose|Switch)=\frac{1}{4}+\frac{3}{8}=\frac{5}{8}$
As such, you have that $P(Win|Switch)=\frac{3}{8}$, and $P(Lose|Switch)=\frac{5}{8}$, which adds up to $1$ ... as it should: when you switch, you either win or lose!
You also have that $P(Lose|Stay)=\frac{3}{4}$ and $P(Win|Stay)=\frac{1}{4}$, which also adds up to $1$ .. as it should: when you stay, you either win or lose!
But you are adding up $P(Win|Switch)+P(Lose|Switch)+P(Lose|Stay)+P(Win|Stay)$ ... and are surprised that this is greater than $1$ ....
Well, you shouldn't, because that should add up to $2$, not $1$. And, as we saw already, you got $1.75$ because you didn't get $P(Lose|Switch)$ correct.