To those who insist on the 50/50, you’re working on the same logical fallacy that you have a 50/50 chance of winning the lottery, either you’ll win or you won’t. It’s just that this fallacy is cleverly disguised in the statement of the problem.

## Kocher says << You select | Monty reveals

Good Prize | Dud 1

Good Prize | Dud 2

Dud 1 | Dud 2

Dud 2 | Dud 1 >>

Yes. But just because there are four outcomes doesn’t mean they have equal probability. That’s the fallacy. The odds of Good Prize is 1/3, the odds of Dud1 is 1/2, so the probability of your first alternative is 1/3*1/2 = 1/6.

Similar for Good Prize/Dud2, chance is 1/6.

The odds of Dud1/Dud2 are 1/3, and so are the odds of Dud2/Dud1.

You win if you switch in the latter two cases, sum 1/3 + 1/3 = 2/3. Manduck has spelled this out in more detail using Nanny and Billy instead of Dud1 and Dud2, but same idea.

Kocher goes on: << Perhaps it would clarify the situation if the rules allowed you to not select a door in the first part and Monty just opened one of the Dud doors at the start. Your probability of winning is the same before and after the door is opened (50%). You do not gain any new information when the door is opened. >>

This is dead wrong. Your probablility of choosing the right door at start is 1/3. After one door is opened, your odds are improved to 1/2. How can you possibly think that opening the door doesn’t give new information?

And, as noted, if you had already picked a door at 1/3 odds, you will improve your chances if you change your selection when the odds are changed to 1/2. The initial door still had 1/3 odds, the new door has had it’s status improved to 1/2.

This is why people give the example, suppose there are 100 doors. You select one (odds: 1/100). Monte opens 98 of the others to show duds. Do you switch? You bet, because the odds backing your door are still 1/100, the other door has odds of 99/100 (NOT of 1/2)