Two weeks ago, I wrote a post called "Your explanation of the Monty Hall problem is wrong" about the "Monty Hall Problem" and what I thought was missing from the standard explanation.
That post used Bayesian inference which involved using some math that some people find a little more advanced than necessary to explain this problem. This is an attempt to explain the same thing as before, but using simpler math. Probability aficionados will likely call this a "frequentist" explanation, but I think of it as explaining probability by using the multiverse concept that so many people are familiar with now from all the ways it shows up in SF and pop culture.
Also, I liberally illustrated this with poorly drawn cartoons composited out of one gigantic image in a CSS crime, so that's something.
this has inspired in me an extremely satisfying explanation of the monty hall problem that doesn't need probability theory or counting cases
consider this version of the game:
there are three doors, A B C. without loss of generality, you choose door A. (if this doesn't sit well with you, just assume we labeled the doors after you picked one.)
monty hall now offers you a choice: would you like to open door A, or open both doors B and C?
and this is exactly the same as the regular problem. after all, what does it matter who opens the doors?
the only reason monty opens one is so that the final choice — stay or switch — seems to be a choice between two equivalent options, each a single door. and he deliberately opens a goat door to keep things ambiguous. but that's all just theatrics and has no bearing on the question.
switching has a ⅔ chance to win because you get to open two doors instead of one. that's it, that's the whole problem.
this also neatly explains all variants.
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say there are five doors, and you choose two of them. monty opens two more, revealing goats. would you like to keep your two doors and open both, or open only the one that remains? switching gives you a ⅗ chance to win.
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say monty always opens the goat door that's closest to him (and you know this). switching still gives you a ⅔ chance to win! his choice function just reveals extra information that divides the game into more cases: ⅓ of the time you now know exactly where the car is, and ⅔ of the time it's a coinflip. obviously if you know where the car is then you should go for it, but that still means switching, so your chance to win by switching is still ⅔.
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say monty opens a door at random, and if he accidentally reveals the car, the whole game is restarted from the beginning. this does change your chances to win, but specifically because the game can only restart if you didn't pick the car, which is exactly when you should switch. in half of those games, you would have won by switching, but the game is abandoned before you even get offered a choice. so you end up with ⅓ win by staying, ⅓ win by switching, and ⅓ the whole game resets.
on the other hand, if your goal is to win a goat, then you should simply pick the door that monty opens