like on paper the odds on your original door was 1/3 and the option door is 1/2, but in reality with the original information both doors were 1/3 and now with the new information both doors are 1/2.
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Ethalis@jlai.lu 1 year agoI know it to be true, I’ve heard it dozens of times, but my dumb brain still refuses to accept the solution everytime. It’s kind of crazy really
Num10ck@lemmy.world 1 year ago
Marvin42@feddit.nl 1 year ago
Yes, you don’t actually have to switch. You could also throw a coin to decide to stay at the current door or to switch. By throwing a coin, you actually improved your chances of winning the price.
Aosih@lemm.ee 1 year ago
This is incorrect. The way the Monty Hall problem is formulated means staying at the current door has 1/3 chance of winning, and switching gives you 2/3 chance. Flipping a coin doesn’t change anything. I’m not going to give a long explanation on why this is true since there are plenty other explanations in other comments already.
This is a common misconception that switching is better because it improves your chances from 1/3 to 1/2, whereas it actually increases to 2/3.
emokidforever@lemmy.world 1 year ago
This explanation really helped me make sense of it: Monty Hall Problem (best explanation) - Numberphile
Elderos@lemmings.world 1 year ago
To me, it makes sense because there was initially 2 chances out of 3 for the prize to be in the doors you did not pick. Revealing a door does not reset the odds of the whole problem, it is still more likely that the prize is in one of the door you did not pick, and a door was removed from that pool. Imo, the key element here is that your own door cannot be revealed early, so it is never “tested”, and this ultimately make the other door more “vouched” for, statistically, and since you know that the door was more likely to be in the other set to begin with, well, might as well switch!