> So our odds of winning with this strategy are still 2/3. That’s true before he opens a door. Then either A) he shows a car and the odds of winning with this

> So our odds of winning with this strategy are still 2/3. That’s true before he opens a door. Then either A) he shows a car and the odds of winning with this strategy are 100% or B) he shows a goat and the odds of winning with this strategy are between 1/2 and 2/3 depending on how the choice of door was made If the choice is random, he shows a car with probability 1/3. The probability of winning is 1/3 x 1 + 2/3 x 1/2 = 2/3 But that analysis is for the unconditional problem and what we’re asked is what to do in case B, after a goat has been unveiled. If he shows always a goat, the probability of winning is 0 x 1 + 1 x 2/3 = 2/3. Here there is no difference between the unconditional and the conditional problems because A never happens.