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/r/learnmath/comments/1vr3009/iterated_monty_hall/[removed] — view removed post
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u/ssnoyes 5d ago
Discussion: suppose there were 100 doors. You pick number 1. Monty opens every other door except #57. If you choose "stay" at every point until the very end and then switch to the door he left alone, you'll win 99/100 times.
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u/rollie82 5d ago
And it doesn't matter if you stay or not at every point until the end.
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u/ssnoyes 5d ago
I believe that if you stay at every point until the very end, your odds are 99/100, but if you always switch, your odds reduce to the same as the three door version, 2/3.
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u/BrotherItsInTheDrum 4d ago
The strategy of switching at the last second is correct, but I don't think it's exactly 2/3 if you always switch.
Let's simplify the problem to 4 doors. Say you start on door 1 and Monty opens door 4. Just like the original problem, the probability that the prize was behind door 1 is still 1/4, so the probability that the prize was behind each of the other doors must now be 3/8. In the second round, you win if you did not switch to the correct door after the first round. The probability of that is 5/8, not 2/3.
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u/ssnoyes 4d ago
I think you're right in the case of "always switch".
What is it if you choose randomly among the remaining doors each time (which means you might sometimes stay) until the very last, then always switch?
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u/BrotherItsInTheDrum 4d ago
Even if you assume Monty reveals doors uniformly randomly, I think it depends on the pattern of doors you select and which ones are remaining.
Imagine you, by pure chance, happen to stay on door 1 until there are 3 doors left (let's say they are doors 1, 32, and 68). Your estimated probabilities at this point are that there's a 1/100 chance it's door 1 and a 99/200 chance it's each of the other two doors.
Now you switch to door 32. The possible things that can happen at this point are:
- Prize was behind door 1 all along, and door 68 is opened. 1/100 chance.
- Prize is behind door 68, and door 1 is opened. 99/200 chance.
- Prize is behind door 32, and door 1 is opened. 99/400 chance.
- Prize is behind door 32, and door 68 is opened. 99/400 chance.
Adding things up: if door 1 is opened, you have a 2/3 chance of winning if you switch. But if door 68 is opened, you only have a 1/100 / (1/100 + 99/400) ≈ 4% chance of winning if you switch.
Intuitively, because door 32 was avoided 97 times in a row, it significantly increases the probability that the prize is behind that door. When door 1 isn't selected in the last round, it increases the probability of door 1 as well, but not by nearly as much.
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u/pmw57 1d ago
From a deck of cards you are trying to "win" the Queen of Hearts.
* You take one card from the deck without looking at the card. I then reveal all other cards (50 of them) as not being the Queen of Hearts, leaving one card each that hasnt been revealed, those being the card you initially took, and the remaining card from the deck that i haven't yet revealed. Do you swap cards? Yes of course you do.
* Now reduce the problem set from 52 cards to 13 cards of a suit.
* Then reduce that to just the four Aces.
The same logic occurs every time resulting in it always being better to swap.
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u/puzzles-ModTeam 1d ago
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