2 ms·
You can distill it all to this: If your initial guess is correct, then switching always loses. If your initial guess is wrong, then switching always wins. Ev
by nkrisc 3y ago
You can distill it all to this:
If your initial guess is correct, then switching always loses.
If your initial guess is wrong, then switching always wins.
Everyone who understands the game can agree on those facts without controversy. If they don’t then there’s no point proceeding because they do not understand the rules of the game.
Therefore, the odds of winning when you switch are the inverse of the odds of winning from your initial guess.
If your initial guess wins 1/3 of the time, that means always switching loses 1/3 of the time and conversely always switching wins 2/3 of the time.
- diputsmonro 3y agoA slightly different reframing of this idea makes it even more clear to me: When you make a choice, you divide the doors into two sets: the set containing your choice, and the set containing the other doors. Obviously your set has a 1/3 chance of containing the prize, and the other set has a 2/3 chance. When the host reveals a door in the second set, it has no effect on those initial probabilities. The set as a whole still has a 2/3 chance of containing the door, and now you have the choice of selecting the only item in that set which you know isn't wrong. This reasoning is just as obvious when you scale up the doors too. With 100 doors, your set has a 1% chance of containing the prize, and the set of unchosen doors a 99% chance. All but one door in that set are revealed to be wrong, then you get a "50/50" choice of swapping to that last remaining item... of course that's what you want to do!
- twiceaday 3y agoConsider an alternative game 1. Pick a door 2. Monty asks if you want to switch to both of the other two doors 3. If you switched, Monty will remove a losing door from one of your two doors I think it is easy to see that this game has the same odds as the original, and also that switching increases your odds from 1/3 to 2/3.