4 ms·
Can’t that be expressed even more simply as: 1/3: you picked car, switch fails 1/3: you picked goat, monty ruins. Not included 1/3: you picked goat, monty op
by graeme 6y ago
Can’t that be expressed even more simply as:
1/3: you picked car, switch fails
1/3: you picked goat, monty ruins. Not included
1/3: you picked goat, monty open goat. Switch wins.
Of the two non ruined scenarios, half win
We can also demonstrate by using the 100 door example. Monty opens 98 of the 99 doors you didn’t choose. 98 times out of 99, monty ruins the game.
So random choice dramatically changes things.
- tasty_freeze 6y agoI think where I differ (also expressed on the parent comment) is that when Monty ruins the game, you are tossing out many cases where the "I'm sticking with my original choice" strategy failed, while I'm counting them as a failure in the strategy.
- irishsultan 6y agoYou are free to count them as a failure of the "I'm sticking with my original choice" strategy, but only if you also count them as a failure of the "I'm switching" strategy, because when Monty ruins the game there is no way to win the car by switching to the remaining closed door. If Monty chooses randomly both strategies fail 2/3 of the time.
- graeme 6y agoPrecisely. Either way switch or no switch have the same success rate if Monty ruins the game. You can count it as 1/2 or 2/3 but either way both choices are the same. So it’s a very different scenario from when Monty knows the door is a goat.