3 ms·
I think it's actually about 1^-12. Given that you have N doors, you (N-1)/N chance of picking a door with a goat. If the host then shows you another door with a
by asdfologist 13y ago
I think it's actually about 1^-12. Given that you have N doors, you (N-1)/N chance of picking a door with a goat. If the host then shows you another door with a goat, and you decide to switch, then you have 1/(N-2) odds of picking the door with the car. So your overall probability of picking the door with the car if you switch is (N-1)/N * 1/(N-2) = (N-1)/(N(N-2)). For N=3, this is 2/3. For N=10^6, this equals roughly 10^-6 + 10^-12. So you have only a 10^-12 margin over the original 10^-6 shot of picking the car.
- StavrosK 13y agoI don't think that's right. Why is the probability of winning after switching the product of the initial probability times the one after the switch (which I think is also wrong)? You pick one door, the probability of winning is 1/N. The probability of losing is (N-1)/N, which stays fixed no matter how many doors the host opens. If you switch, the probability of winning is (1/(N-2))-(1/(1/N)). If he opens all doors except two, and you switch, the probability of either of the two having the car is 50% minus the original 10^-6/2. Each door decreases the probability by one in a million, not one in a trillion. If that were the case, it would be highly unlikely that you won even if you switched after all but one of the doors were opened.
- asdfologist 13y agoYou're wrong and I wrote a simulation to prove it. To repeat my original assertion, the probability of winning if you switch is (N-1)/(N(N-2)). For N=4, this equals 3/8. Run the following simulation and you'll get roughly 0.375, as my formula predicts. http://pastebin.com/i89M4GZU http://pastebin.com/i89M4GZU