Monty Hall Problem with Five Doors

Solution 1:

Think about it this way:

You have five doors, and you choose one. You already know you had $\frac{1}{5}$ chance of being right. Now the presenter must open one of the other doors that he/she knows is empty. That means the probability that the one of the three doors left is a winner is $\frac{4}{5}$, while the probability that the door you've chosen is correct is still $\frac{1}{5}$.

So if you stick with your first choice, you have a $\frac{1}{5}$ chance of winning. Now, if you decide to change to one of the other three doors, you know that you'll have a $\frac{4}{5}$ chance of winning (meaning if you were allowed to say "I choose this group of three doors" but not specify any single one, you would win $4$ out of $5$ times).

But you still only get to pick ONE door, and you have $3$ to choose from, so if you are going to choose to pick from the remaining $3$ doors, which together have $\frac{4}{5}$ chance of winning, you'll have a $\frac{1}{3}$ chance of being right from the selection of these three (since it is equally likely that you'll choose any of these three).

But put all together, that means you have a $\frac{4}{5} \cdot \frac{1}{3} = \frac{4}{15}$ chance of winning by switching to ONE door from the remaining $3$.

Here is a picture:

Solution 2:

The computation is straightforward, though it is easy to get tangled up.

The chance of having chosen the correct door was one fifth when you chose it. The host's action would have been possible whether or not you have chosen the correct one. The probablity remains one fifth if you stick.

The probability that the prize is behind one of the other doors is therefore four fifths. There are three indistinguishable possibilities, each of which is correct with probability four fifteenths. The probability if you change is four fifteenths.

Solution 3:

I assume there is only one prize behind the doors. The probability that your originally picked door ($A$, say) is the winner is $\frac15$ and this does not change because the host really does not give you any informationm about that door. He does give you information about the other doors, though. If he open $B$, clearly the probability of $B$ being the winner door is $0$. And (unless there is some known bias in the host's choices) the remaining doors $C,D,E$ have equal winning probabilities $p$. From $\frac15+0+p+p+p=1$, you can compute $p=\frac4{15}>\frac15$.

Solution 4:

You mentioned,

All the extended Monty Hall problems I found online had the host open all but one, so they weren't really helpful with this particular problem my class is working on.

The reason all the online Monty Hall problems have the host opening all but one is to convince you that, indeed, you get new information about where the prize is based on the host's door. If you have a hundred doors to choose from, you pick one, and then the host opens 98 of the others, do you now have a 50% chance? No, you have a 1% chance, and 99% the prize is behind the other door. Always change the door.

What if the host only opens 97 of the others? Do you now have a 1/3 chance? No, you still have a 1% chance. 99% of the time, the prize is still behind one of the other two doors, so now each of the other doors has something like a 49% chance of having the prize. You switch.

The fewer doors the host opens, the lesser the chances of getting the prize when you switch, but they're always higher than the chance of getting the prize when you don't switch. Even if the host only opens one door, that still elevates the chances of the other doors. You should always switch.