I find the easiest way to understand Monty Hall is to think of it in a meta way: Situation A - A person picks one of three doors, 1 n 3 chance of success. Situation B - A person picks one of two doors, 1 in 2 chance of success. If you were an observer of these two situations (not the person choosing doors) and you were gonna bet on which situation will more often succeed, clearly the second choice.
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zenharbinger@lemmy.world 1 year agoI found the easiest way to think about it as if there are 10 doors, you choose 1, then 8 other doors are opened. Do you stay with your first choice, or the other remaining door? Or scale up to 100. Then you really see the advantage of swapping doors. You have a higher probability of choosing the last remaining door than of having correctly choosen the correct door the first time.
_g_be@lemmy.world 1 year ago
zenharbinger@lemmy.world 1 year ago
But the issue is that by switching doors, you have a 66% chance of winning, it doesn’t drop to 50% just because there are 2 doors, it’s still 33% on the first door, 66% on the other doors (as a whole), for which we know one is not correct and won’t choose.
_g_be@lemmy.world 1 year ago
on the whole
is the key words here
individually the door has that 1:2 chance, but the scenario has more context and information and thus different odds. Choosing scenario B over scenario A is a better wager
zenharbinger@lemmy.world 1 year ago
you aren’t talking about the Monty Hall problem then
KarmaTrainCaboose@lemmy.world 1 year ago
I don’t understand how this relates to the problem. Yes 50 percent is greater than 33 percent, but that’s not what the Monty hall problem is about. The point of the exercise is to show that when the game show host knowingly (and it is important to state that the host knows where the prize is) opens a door, he is giving the contestant 33 percent extra odds.
dirtbiker509@lemm.ee 1 year ago
Holy shit this finally got it to click in my head.