The Monty Hall Problem

In September 1991 a reader of Marilyn Vos Savant's Parade column asked the following question:

Marilyn's conclusion was that one should switch.

The confusion arises thus...

The game is really two games!

The first game is a three-door problem. Your odds of choosing correctly are 1 in 3.

The second game is a two-door problem. Your odds of choosing correctly are 1 in 2.

Because your odds are worse in the first game, any choice in the first game is more likely to be incorrect than any choice in the second. Marilyn and friends therefore say that you should switch. But they're comparing apples and oranges.

The probability for a door in the first game cannot be applied to the same door in the second game; it's a new game! The odds for all remaining doors increase when a door is removed from play. The door selected in the first game, which had a 1/3 chance, has a 1/2 chance in the second game.

Marilyn said...

No. Not unless he said that he was going to open all the doors except the one that I chose and the one with the prize. Think about that... If he opened all but mine, then I'd know I chose correctly. If he opened all but mine and one other, then I'd know I chose incorrectly.

What he actually did was open all the doors except mine and one other. Then we know which doors don't have a prize, but we still don't know which of the two remaining does. 50/50!

Okay, let's look at it this way. Let's say that the car is behind door A (with A being 1, 2 or 3, pick any one). Goats are behind doors B and C (the remaining doors). Here are the possible outcomes...

So the odds are 50/50, whether or not you switch!

Hmm... Something doesn't seem quite right...

Let's expand the table, and show where the car is, with all the cases for all the doors...

Giving...

• 6 occurrences of Switch & Lose
• 6 occurrences of Stay & Lose
• 6 occurrences of Stay & Win
• 6 occurrences of Switch & Win

50/50! Hmm... Something still doesn't seem quite right... Hey! The first choice was correct in half the cases. That can't be right. Aha! The probabilities of the cases themselves are unequal. The table must be weighted...

Giving...

• 6 occurrences of Switch & Lose
• 12 occurrences of Stay & Lose
• 6 occurrences of Stay & Win
• 12 occurrences of Switch & Win

Or, returning to the more concise form...

Here's a short and sweet explanation: The odds that you made the wrong first choice are 2 in 3. So if you get a chance to change your mind, take it! Assuming that the host isn't trying to trick you... like giving you the second chance only when your first choice was correct...

I think.

• Monty Hall problem - at Wikipedia
• Ask Dr. Math
• 1990 A.D.
• Let's Make A Deal
• Marilyn is tricked by a game show host
• Monty Hall Problem - At the Official Let's Make A Deal Website
• Probability and Nonsense

*
* *

Some time later, while going through some old files, I came across a clipping from the Mensa Bulletin of March 1996. It was a letter to the editor from John Dwyer, and appears to describe a related Conundrum, a simplified version of the same problem...

Here's an old problem I've never been able to solve:

Suppose there are two sealed envelopes, one of which contains twice as much money as the other — that's all you know. You may choose one, and the rule is you may switch whenever you want until you open an envelope (at which time the game is over).

You pick an envelope, and common sense tells you there's no point in switching. But using probability theory, you reason as follows: This envelope contains $x and the other contains $y. Either y=2x or y=x/2. So if I switch I stand to gain $x or lose $x/2, with equal probability. I'll switch.

Having switched, you think: either x=2y or x=y/2, so... I'd better switch back!

Is probability theory nonsense?...