This page consists of different ways of approaching the classic Monty Hall problem.
You have a 1 in 3 chance of picking the right door on the first try. So if you stay with that door, you have a 1 in 3 chance of winning. That means if you switch, you have a 2 in 3 chance of winning, doesn't it?
This is the noted logician Raymond Smullyan's argument.
Imagine this slightly different game. It begins the same with the host hiding a prize, then letting you select a door. Once you've picked your door, the host allows you to take either what's behind your door or everything behind all the other doors. Which is better?
As you might expect, it's better to take what's behind more doors. You have a 2 in 3 chance of getting the prize this way, but only 1 in 3 if you stay with your original choice.
This scenario might seem different, but is functionally identical to the original game. The only difference is that in the first case you're shown that there's a booby prize behind one door. But that doesn't matter! You already know that there's a booby prize behind one of the doors you didn't choose. So how could seeing it change the probabilities?
One way to verify this is to look at every possible case, and to calculate the probabilities. Here's part of a table that shows this.
PRIZE LOCATION |
FIRST CHOICE |
DOOR REVEALED |
ACTION | RESULT | PROBABILITY |
---|---|---|---|---|---|
Door 1 1/3 |
Choose 1 1/3 |
Shows 2 1/2 |
Stay 1/2 |
WIN | 1/36 |
Switch 1/2 |
lose | 1/36 | |||
Shows 3 1/2 |
Stay 1/2 |
WIN | 1/36 | ||
Switch 1/2 |
lose | 1/36 | |||
Choose 2 1/3 |
Shows 3 1 |
Stay 1/2 |
lose | 1/18 | |
Switch 1/2 |
WIN | 1/18 | |||
Choose 3 1/3 |
Shows 2 1 |
Stay 1/2 |
lose | 1/18 | |
Switch 1/2 |
WIN | 1/18 |
This table holds all the cases when the prize is behind door 1. To calculate the probability of a particular sequence of (prize, choose, reveal, action), just multiply the numbers in each column. For example, let's say I choose door 2 (one in three probability), then the host reveals what's behind door 3 (probability one, since he can't reveal door 1), then I stay with door 2 (one in two probability). Multiplying the probabilities, we have 1/3 (for door 1 holding the prize) × 1/3 × 1 × 1/2 = 1/18. The odds of this happening are one in 18. This is given in the last column.
Now let's look at the table. Note the results of switching (light green boxes): the sum of their winning probabilities is 1/18 + 1/18 = 1/9. The sum of the losing probabilities is 1/36 + 1/36 = 1/18, which is 1/2 of the winning probabilities.
Now look at the cases where we stay with our original door. Adding up, we find that we lose 1/9 of the time, while winning only 1/18 of the time. This is exactly the reverse of if we switched. So it's better to switch.
Though this table holds only the results for door 1, the results are the same for the other doors. If you're a real masochist you can work it out for yourself with the full table. Just add up all the probabilities. (They're color-coded to make it easy.) Compare:
Lose | Win | |
---|---|---|
Stay | 1/3 | 1/6 |
Switch | 1/6 | 1/3 |
If you look at the first table, you'll realize that the difference in winning probability between switching and staying is due to the fact that if the door you first choose is the one with the prize behind it, the host can reveal what's behind either of the other doors. If the prize isn't behind the door you choose, the host doesn't have a choice about which door to reveal. That lack of choice, which signifies the prize is behind the other door, occurs in two out of three cases.
Consider this gedankenexperiment: what would happen if instead of three doors, there were a hundred? After you've chosen one, the host opens every door except the one you've chosen and one other. Now you're facing two closed doors and 98 open doors. Does it still seem likely that there is a 50% chance of finding the prize behind the original door? That would mean you had a 50% chance of choosing the right door the first time. How did you do that when there were a hundred closed doors?
Now, how about a million doors? Let's say you choose #1. The host opens a door, and there's nothing behind it. The host continues opening doors - dozens, hundreds of doors - nothing. He's running like mad, opening door after door after door - thousands, tens of thousands, hundreds of thousands of doors, every door he passes - nothing! Suddenly and deliberately he skips door #443,291, then continues opening other doors. Finally he stops. The only doors left unopened are your choice, #1, and the one he so carefully avoided opening - door #443,291.
Now would you switch?
If not, combine this with the "switcher takes (almost) all" version of the game. Imagine being shown a million doors, choosing one, then being told you could have either what's behind that door or what's behind every other door.
Would you switch now?
Imagine this version. There are a thousand doors. You get to choose a door, which has a 1 in 1000 chance of being right. The other doors together have a 999/1000 chance of having the prize. You can either stay with your first choice, or take everything behind all the other 999 doors. Which would you do?
You'd probably take the 999 doors, figuring there was a better chance of them having the prize than your measly one door. Makes sense. Let's say that's what you did. So now you have a 999/1000 chance of winning the prize. That's important: you have a 999 in 1000 chance of winning the prize. Of course, you'll also have at least 998 booby prizes, but that's okay.
Before the host opens the 999 doors, he does one last thing. He opens all of the 999 doors but one, revealing 998 booby prizes. Now you're left with only two doors closed: your original door, and the one out of 999 he didn't open. He offers once again to let you choose either of the two doors. Which would you take?
If chose your original door, think about this. You selected the 999 doors because there was a 999/1000 chance it had the prize. Of course, it also had at least 998 booby prizes. Now you know where those booby prizes are. The host offered you the choice of your original door (1/1000, you've already said) or 998 booby prizes and the other door (which, all together, had a 999/1000 chance of having the prize). You already knew the booby prizes were there... why would seeing exactly where they are change the probabilities?
If you're still not convinced, try it yourself.
The "One last thought experiment" section was inspired by a conversation with my friend Dave.
Last updated 7 August 2003
http://www.rdrop.com/~half/Creations/Puzzles/LetsMakeADeal/MoreApproaches.html
All contents ©1998-2003 Mark L. Irons