this post was submitted on 07 Dec 2023
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I was stubborn about this for so long, and I'm still not entirely sure I understand it, but here is a perspective that made me doubt my belief.
Imagine the Monty Hall Problem, but with 100 doors and only one grand prize. You pick one; it obviously has a 1/100 chance of being a grand prize. Then Monty reveals 98 doors without grand prizes in them such that the only doors left are the one you chose and one that Monty left unopened. Monty obviously arranged for one of those two doors to have the grand prize behind it. The "choice to switch" is really just a second round of the game, ~~but with a 1/2 chance of winning~~ (wrong, your odds change only if you "participate" in round two).
If you stick with your door, you are relying on your initial 1/100 chance of winning. If you switch, you are getting the ~~1/2~~ odds of the "second round".
Apparently with three doors, switching gives you a 2/3 chance of winning, but I don't understand the math of how to get that answer and I wouldn't be able to calculate the odds of the 100 door version. I just know intuitivey that switching is better.
The "second round" of the game is always just, "flip your odds of winning if you swap". That's all it is.
Monty will always open the proper doors to ensure this happens every time. Did you pick the winning door in the first round? Monty will eliminate all other doors but leave one of the losers. Did you pick a losing door in the first round? Monty will eliminate all the other losers and only leave the winner. It's always the opposite of what you picked. Therefore, if you swap, you will simply get the opposite odds of the first round.
100 doors to pick from, only 1 winner? 1/100 chance to win if you just picked at random and ended it there. Now Monty offers a swap. Without the swap, you have 99 different ways to lose this. But with the swap, all 99 of those ways become winners, because Monty will always swap the opposite with you.