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The one which made it clear for me was: imagine you have 4 doors instead. You choose one at random. The host then chooses two of the remaining doors he knows ha
by ComplexSystems 3y ago
The one which made it clear for me was: imagine you have 4 doors instead. You choose one at random. The host then chooses two of the remaining doors he knows has goats. Is it in your best interest to switch to the one he didn't open?
Suppose you have 100 doors. You choose one at random. The host then chooses 98 of the remaining doors he knows has goats. Is it in your best interest to switch?
In all of these situations, the basic question is: what is more likely - that you chose the one with the car, or that you didn't, and the door remaining is only remaining because the host knew it was the one with the car?
- TulliusCicero 3y agoYeah, simply imagining a much larger number of doors makes the logic obvious.
- ksaj 3y agoI used an entire deck of cards (52 of them) to demonstrate this to a friend, and he still wouldn't believe it. I gave up explaining it after that. It's really easy to get people arguing about this one, even though the correct answer is so well documented.
- bombcar 3y agoI remember reading someone who couldn’t understand it - so he build a computer program (in python I seem to recall) to simulate it over millions of goes. And the numbers worked out as expected.
- ksaj 3y agoIt's an easy python program to write, because there's nothing complex about the process. Pick a language here: https://rosettacode.org/wiki/Monty_Hall_problem https://rosettacode.org/wiki/Monty_Hall_problem Every one of these prove the effect beyond a doubt.
- bombcar 3y agoTo be precise they demonstrate it; a proof (mathematically) comes from axioms.