Sunday, November 4, 2007

The Monty Hall Problem

Difficulty: 3 of 10

Here's a puzzle that is truly a classic. It is based on an ancient game show with a host named Monty Hall.

In part of the show, a contestant chooses between three doors, only one of which has a fabulous prize. After choosing a single door, but before opening it, the game show host asks, "Would you change your mind if you saw this?" One of the other doors is opened, to reveal a goat (in case it isn't obvious, the goat is not the fabulous prize). Note that the game show host does this every single time, and always reveals a goat. The contestant is given the opportunity to either stick with the original decision or switch to the last unopened door.

So which would you choose? What is the probability of winning for each decision?

At first glance, it seems you have two equally likely choices, so you have a 1/2 chance of winning either way. But that just begs the question: are they in fact equally likely? Before any door is opened, you have a 1/3 probability of being correct. Does that probability suddenly jump up to 1/2 when a door is opened? No it doesn't. Therefore, you have a 2/3 probability of winning if you switch.

However, this puzzle has tripped up quite a number of extremely intelligent people, so allow me to argue the point. First consider three equally likely possibilities. Either you've picked goat 1, goat 2, or the fabulous prize. If you've picked goat 1, Monty Hall shows you goat 2. If you've picked goat 2, Monty Hall shows you goat 1. If you've picked the fabulous prize, Monty Hall has a 1/2 probability of showing you goat 1, and a 1/2 probability of picking goat 2. Here is a table of all the possibilities and their respective probabilities


You pick goat 1, Hall picks goat 2


You pick goat 2, Hall picks goat 1


You pick prize, Hall picks goat 1


You pick prize, Hall picks goat 2

Note that there are now four possibilities, but they are not equally likely. That is because the last two possibilities are merely the "splitting" of the last of the original three equally likely possibilities. For the first two possibilities, it would be better to switch. For the last two, it would be better to stay. So adding up the probabilities, you have a 2/3 chance of winning if you switch, and a 1/3 chance of winning if you stay.

So why is it that the door you picked stays at 1/3 probability, but the other door jumps from 1/3 to 2/3 as soon as a door is opened? It is because when Hall opens a door, you are given new information about the other two doors. Because Hall has zero chance of opening the door you picked, no new information is given about that door.


Anonymous said...

I disagree. In each case you are shown a goat (100% probability). Your chances of having chosen the door with the great prize remains exactly what it always was: 1 out ot 3.

miller said...

That's correct, unless you switch to the other unopened door, in which case you have 2 out of 3.

Or did you mean to say that it's 1 out of 3 even if you switch? If so, that implies that if you could open both of the leftover doors, you would only have a 2/3 chance of winning. That wouldn't make sense, since one of the doors has a prize.

Anonymous said...

OK, OK, I was sleepy when I said that. I was hoping I could get back and correct myself before you did.

Anonymous said...

The problem I have with this is that after the host eliminates 1 of the 3 doors as the winner, the game essentially resets with only 2 doors remaining -- 1 a winner and 1 a goat. That's a 1 in 2 chance, not 2 and 3 as the third door is no longer a possible choice in the game.

miller said...

Anonymous, it does not "essentially reset". There are two doors left, but those two doors do not have the same history. That is, one of the doors was chosen by me (and therefore could not be revealed by Monty Hall), while the other one could have been revealed by Monty Hall. The situation is not symmetrical between the two doors, so we can't assume that they have identical probabilities.