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Not quite true. If the host gives you the option to switch based on a coin flip, that flip will be independent of the state of the game and just average the ov
by tel 17y ago
Not quite true.
If the host gives you the option to switch based on a coin flip, that flip will be independent of the state of the game and just average the overall probabilities between a regular game and a switch-style game.
Now, allowing the host to act totally randomly will remove the advantage. In particular, the host needs to randomly choose which door to open from all three (including your own). If he finds the goat, you reset and play again perhaps, if he doesn't find the goat then you've got a 50/50 chance between the remaining doors. The probabilities are simple because each door was treated exchangeably and there could be no possible information flow that is mutual to, entangled with, the location of the goat.
- code0 17y agoAs you said, "average the overall probabilities between a regular game and a switch-style game", which is 1/3 for a game with no-switching and 2/3 for a game with always-switching. Averaging gives you a probability of 1/2. So if you switch, you win 50 % of the time. But so do you if you do not switch. So there is no benefit in switching, which was basically my point. EDIT: In both the versions of the game, I am assuming an omniscience host who always opens the door with a goat behind, if he does open a door at all.
- tel 17y agoYou're absolutely right and my argument fails, but it doesn't invalidate the point I wanted to get at. In the long run, your coin flipping method does cause the game to behave as though there were only two doors, but in any given game you'll know whether or not the host is going to choose a door. That means that, conditional on a good flip, you may still be able to act to your advantage since there's information transfer from the host to you.