Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the Monty Hall Problem, there are three doors. Let's take two scenarios. In both the host reveals that there is a goat behind door number two. In the first scenario, I choose the first door. According to the "correct" solution, after the host opens door 2, door 1 has a 1/3 chance of having the car, and door 3 2/3. But say I choose door 3. Then door 3 has a 1/3 chance, and door 1 2/3. How can I get two different answers, differing solely based on my initial choice?

share|improve this question
    
Haven't read the question, but here's a way of seeing the intuition behind the whole Monty Hall problem in general: if you switch the doors in the last step, you will win if and only if your door has a goat, and lose if and only if your door has a car. Having the goat behind the door is more likely, so you're more likely to win if you switch the doors. –  mathh May 6 at 14:58

1 Answer 1

The two scenarios you describe both exclude the possibility, which was present at the beginning, that the car is behind the middle door. This cuts out one third of the sample space and leaves the two other possibilities equal.

If the car had been behind the middle door, one of the other doors would have been opened, and you would have needed to swap to win it.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.