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The 9/11 collapses -- an unlikely sequence

Tests for divisibility by 9 and 11

**The Monty Hall problem**

*A player is shown three closed doors. One hides a nice prize and the other two hide booby prizes.
The player picks a door. Monty then opens another door, to reveal a booby prize. The player is asked whether he or she would like to stand pat or switch his choice to the remaining closed door. Should he or she switch?*

The reason this problem is so popular is that, for most
of us, the answer is counterintuitive. Of course, in the original TV game show, *Let's Make a Deal*, the player was not given the option to switch.

My mathematician son, Jim Conant, almost instinctively knew the right answer: switch! He saw immediately that in 1/3 of cases standing pat fails so that in 2/3 of cases, switching must succeed.

But to the duller-witted of us, that reasoning doesn't seem to account for the apparent point that once the set is whittled down to two choices, switching gives a 50-50 chance of success.

However, probabilities are about information, and the information the player obtains by Monty's opening of the door affects the probabilities.

I confess I found this very difficult to grasp (and retain!) until I was able to come up with the neat proof below.

I arrived at the proof by thinking: Suppose we divide the doors into subsets {A} and {B,C} and Monty doesn't open a door once a player selects A. There is a 1/3 probability of success if he chooses set A and 2/3 for {B,C}. Now if someone tells him he can't choose, say, element B, that doesn't affect the 2/3 probability for {B,C}. So if he switches to {B,C} he must take C, which then has a 2/3 probability of success.

However, here's the proof over easy:

A B C case 1. 0 (0 1) case 2. 0 (1 0) case 3. 1 (0 0) case 1: Monty opens B, switch succeeds case 2: Monty opens C, switch succeeds case 3: Monty opens B or C, switch failsIn 2/3 of cases, switch succeeds.

I suggest that a source of confusion is the "or" in case 3. Despite there being two options, Monty opens only one door, as in cases 1 and 2.

If you're still unhappy, Jim Conant suggests a light bulb might go on if you consider this scenario:

We have 100 doors, with a prize behind only one. You choose a door. Monty now opens 98 doors, none of which hides a prize. Should you switch? Of course, since the probability that the other door hides the prize is influenced by the knowledge that 98 other doors hid nothing. What is your chance of having chosen the winner? Still, 1 percent. What is the chance that the prize is behind one of the other 99 doors? Still 99 percent. So the knowledge you are given squeezes that probability onto the other closed door.

Let's have a bit of fun with information theory.

The base-2 information content, or value, of the three closed doors is simply log_{2}3 = 1.58 bit. Each door's information value is of course 1/3 log 3 =
0.528 bit. When one of the doors is opened, the remaining two doors still have information values of 0.528 bit each.

On the other hand, in the case of two closed doors, the information content of each door is 1/2 log 2 = 0.5, which corresponds to maximum uncertainty. Hence the 0.028 difference in information corresponds with your awareness of an asymmetric change in probabilities that occurs once a door is opened. Because one door is opened, the information content of that door, once you get past your surprise, becomes 0. In this case, the information content of the unopened door then becomes 1.06 bit.

Now if we have 100 doors, the scenario opens with a total information content of log 100 = 2 log 10 = 6.64 bit. A single door has an information content of 0.07 bit.

The information content of the 98 doors Monty opens is 6.51, which is far greater than 0.07. Indeed, I suspect you feel much more confident that the prize is behind the other remaining closed door in the 100-door scenario.

We also have that the information content of the set of 99 doors is 6.57. Once 98 of those doors are opened, the information content of those doors becomes 0 and hence the information content of the remaining door in the set is 6.57 versus the measly 0.07 content of the door you chose.

Paul Conant's Erdos number is not an element of Z