3 ms·
Nope. If the host shows the goat _for any reason_, then the 2/3 chance that the doors you didn't choose contain the car now applies to the one unchosen door rem
by KALSTR 5y ago
Nope. If the host shows the goat _for any reason_, then the 2/3 chance that the doors you didn't choose contain the car now applies to the one unchosen door remaining -- and you should switch.
- dougiejones 5y agoStaying is an equally good option as switching in this version if the host revealed a goat, since the host randomly revealing a goat now makes it more likely that your initial choice was correct. P(initial choice is correct) = 1/3 P(host shows goat) = 2/3 P(host shows goat | initial choice is correct) = 1 and vice versa P(host shows goat | initial choice is wrong) = 1/2 Applying Bayes: P(initial choice is correct | host shows goat) = P(host shows goat | initial choice is correct) * P(initial choice is correct) / P(host shows goat) = 1 * 1/3 / (2/3) = 1/2 So, initial choice now has 1/2 odds for being right. The host revealing a goat gave us information on whether he was playing the P=1/2 or the P=1 game. Intuitively applying this with the 100 door version: the host getting lucky and revealing 98 goats in a row makes it fairly likely that the initial door was correct all along and they didn't get super lucky avoiding the car 98 times, as there's a good chance the contestant helped them by hiding the car.
- 6gvONxR4sf7o 5y agoPerfect case study right here. It’s trivial to simulate and y’all aren’t budging. In this case, simulating, I see 2/3 for switching after the host shows a goat on purpose, and 1/2 for switching when the host happens to have shown a goat on accident.