The Monty Hall
Three Door Dilemma:
Explanation

Did you play the game long enough to determine what the best strategy was? If not, go back and play a few more rounds.

Well, how was your intuition? If your initial strategy was not correct you are not alone, as most people do not get this right on the first pass. This is a very counter-intuitive problem.
How confident were you in your selection of a strategy? Most people are quite confident in their intuition.

Do you believe that the best strategy is to always switch? It is better by a factor of 2 than not switching.
Does your intuition tell you that it can't be so?

If you don't believe that the best strategy is to ALWAYS switch, play 100 rounds of the game with some other strategy. Then play 100 rounds where you always switch. When you always switch, you will win approximately 67% of the time. You can play 100 rounds in a few minutes, so it doesn't take long to test out strategies.

Here are three ways I know for explaining this problem. If you have a another way, please drop me a note.

Explanation # 1. Consider these 3 possibilities for the location of the winning door ($)...
Case 1: $XX
Case 2: X$X
Case 3: XX$
Let's assume that you pick door number 1 initially. In the first case ($XX) you are better off to stick with your first choice, as that is where the $ is. In the second case (X$X), you initially pick door # 1, and Monty will show you that door # 3 is empty. Here, you would be better off to switch to door # 2. In the 3rd case (XX$) you choose door number 1 first, Monty shows you that door #2 is empty, so again you would be better off to switch. So, in 2 out of 3 cases you are better off to switch to the other door. If you switch every game you will win 66% of the time.

Explanation # 2. When you start out there are three doors. Your chance of guessing right on the first pass is 1 in 3. We all probably agree on this. The probability that you were wrong is 2 in 3. Most likely the money was behind one of the 2 doors that you didn't choose, as a matter of fact twice as likely. Before Monty shows you the empty door, would you be willing to switch your one door for the other two? I assume you would (that's actually what you get to do if you switch: you get both of the other doors; at least one just happens to be empty.) You know that at least one of the two remaining doors must be empty. They both can't be hiding the money. So when Monty shows you that one of those two doors is empty, it shouldn't be a surprise. He is just telling you something you already know - that one of the two remaining doors is empty. The odds have not changed. Most likely you still picked the wrong door initially; most likely the money is behind one of the two other doors; Monty does give you clue about which one NOT to switch to (since it is empty), and therefore you should switch to the other closed door.

Explanation # 3. Imagine that there are one thousand doors. You pick one. Your chances are 1 in 1000 of guessing right. In 1000 trials you will be wrong (statistically) 999 times. If Monty shows you what is behind 998 of the remaining 999 doors, and they were all empty, that should not surprise you either. After all he isn't going to show you the million dollars. Again you are left with two doors: your original pick and the one Monty chose not to show you. Because there are only two doors left, your luck can not miraculously improve from 1 out of 1000 to 1 out of 2 every time you play this game. If you stay with your first pick your chances are still 1 in 1000. With 1000 doors you are 999 times better off by switching.

Did your intuition fight with these explanations?

By the way, this is an honest game (this computer doesn't cheat).

If you have any comments, I'd like to hear from you ---- gary (at) dcity.org
Gary