Let's Make a Deal

Doing research and then trying to explain it to others isn't something new to me. I've always been of the opinion that if you're going to wonder about something, you might as well find out the answer, so I end up doing that a lot off of discussions I have with friends. Never in my life have I encountered such a storm of fury as when I tried to explain the Monty Hall Selection Theory.

And it's not just me, I've been reading on other people going ballistic over this, so I'm going to warn you up front: the solution to this problem is nonintuitive. That means, it doesn't work like you feel it "should." Keep an open mind and we'll get there.

The Monty Hall problem is named after its resemblance to the old TV game show Let's Make a Deal. Here are the rules to your game:

1. There are 3 doors to select, you may select any one of them.
2. Behind exactly one door is a prize.
3. After that selection, one door that does not have the prize behind it will be opened and eliminated as an option.
4. You then have the option of keeping your initial selection, or switching to the other unopened door.

The puzzle then says: what should you do to try to win? Keep your pick, switch to the other door, or does it not matter? Not to beat around the bush, the answer is: switch. Your odds of winning double if you change your selection.

This is where I'm going to remind you of what I said above about it being nonintuitive. Here's where I try to explain 4 ways. If one doesn't get you there, maybe the others will.

I. Logical

Your odds of getting the correct choice on the first pick are 1 in 3, obviously. Also obvious, if you don't change your selection, your odds don't change, so staying with your initial selection keeps your odds at 1 in 3.
However, if you didn't select correctly initially, as you won't 2 times in 3, switching will win.
Odds while staying: 1 in 3. Odds while switching: 2 in 3.

II. Mathematical/Rigorous
There are 3 options for where the prize is, we'll call them doors A, B, and C, and similarly 3 selections you can make, so there are 9 scenarios to consider.

If you keep:
Select A, prize A: Either eliminated, you win.
Select A, prize B: C eliminated, you lose.
Select A, prize C: B eliminated, you lose.
Select B, prize A: C eliminated, you lose.
Select B, prize B: Either eliminated, you win.
Select B, prize C: A eliminated, you lose.
Select C, prize A: B eliminated, you lose.
Select C, prize B: A eliminated, you lose.
Select C, prize C: Either eliminated, you win.
Odds of winning: 3 in 9 (1/3)

If you switch:
Select A, prize A: Either eliminated, switch to the other and lose.
Select A, prize B: C eliminated, switch to B and win.
Select A, prize C: B eliminated, switch to C and win.
Select B, prize A: C eliminated, switch to B and win.
Select B, prize B: Either eliminated, switch to the other and lose.
Select B, prize C: A eliminated, switch to C and win.
Select C, prize A: B eliminated, switch to A and win.
Select C, prize B: A eliminated, switch to B and win.
Select C, prize C: Either eliminated, switch to the other and lose.
Odds of winning: 6 in 9 (2/3)

III. Intuitive

Imagine the same game, but with a million doors instead of three. After 999,998 eliminations, you can keep your door or switch to the only one of the other 999,999 not eliminated. Does your first one-in-a-million pick still seem so good?

Does it help if you think of your choice not as "Door A. vs. Door C" but "Door A vs. everything else?"

IV. Experimental

As a last result, play the game. Keep track. After 50 or so trials, you'll see, you win twice as often by switching. Seeing it happen should also reinforce the first explanation: the keeping strategy only wins (1/3) when you were right the first time, and every other time (2/3) the switching strategy wins.

If you're really clever, perhaps you can get a friend to play with you. One with deep pockets is even better.