Summary: We have 3 doors with a car behind one (the prize) and goats behind the other two (non-prizes). You pick one of the doors; Monty Hall then opens one of the remaining doors with a goat behind it (he never opens the door with the prize). The issue is: do you stick with your original choice or switch.

Marilyn says you should switch, as do many of the sites I've visited, claiming a 2/3 probability of win if you switch vs. 1/3 if you don't. The case analysis that Marilyn did is shown in Hoffman's book, and I believe it is flawed. She combined the two different action options of my Case A into a single option.

My decomposition of cases and their outcomes below shows that the probability of winning is the same (50%) whether you switch or not. The critical case is Case A in which the host (Monty) has two different doors he can open.

Here's my analysis:

(Note: I apologize for the clumsy table format, but having it in tabular form makes it much clearer. I tried html but couldn't get it to work correctly, and extra spaces are automatically deleted.)

I select Door #1, but I consider all cases of prize arrangement, so the same analysis applies for any initially selected door.

Option 1: Choose Door & Always Switch:
_____|______ Door ______| ___________Actions__________ | _______
Case |__1__|___2__|__3 _|_ 1stPick | HostOpens | 2ndPick _| Outcome
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 2 ___|___ 3 ___|__ Lose
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 3 ___|___ 2 ___|__ Lose
B ___| Goat |_ Car_| Goat | ___ 1 __|_____ 3 ___|___ 2 ___|__ Win
C ___| Goat | Goat_| Car _| ___ 1 __|_____ 2 ___|___ 3 ___|__ Win

Option 2: Choose Door & Always Stick with It:
_____|______ Door ______| ___________Actions__________ | _______
Case |__1__|___2__|__3 _|_ 1stPick | HostOpens | 2ndPick _| Outcome
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 2 ___|___ 1 ___|__ Win
A ___| Car _| Goat_| Goat | ___ 1 __|_____ 3 ___|___ 1 ___|__ Win
B ___| Goat |_ Car_| Goat | ___ 1 __|_____ 3 ___|___ 1 ___|__ Lose
C ___| Goat | Goat_| Car _| ___ 1 __|_____ 2 ___|___ 1 ___|__ Lose

For both options, the probability of win is 1/2.

How is my analysis flawed?

Thanks,

Don Link
Columbia, MD

How did you manage to get 4 cases after the pick?

If you decided to distinguish goats at thi stage, why not to distinguish them before the first pick? You would have B and C repeated twice.

- If you selected the winning door to begin with you are guaranteed to lose.
- If you selected a losing door to begin with you are guaranteed to win.

Then it is easy to see that the you will win with this strategy 2/3 of the time.

Your analysis here uses the strategy of exploring all possible outcomes. The flaw in your analysis is that you do not explore all possible outcomes (for probability's sake). Basically there are a total of 6 possible outcomes (you pick 1 of 3, host "picks" 1 of remaining 2).

In your case A, the probability of either of the 2 outcomes is actually 1/6 yet for cases B and C the probability is 1/3 (each since the host does not really have a choice).

From there you should see how the correct outcome comes to light.