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Comment by superb-owl | Hacker News Reader

The trick that finally got me to accept the conclusion was to imagine there were 100 doors. You pick one, then Monty opens 98 other doors. Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right.

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What helped me is the fact that the only way you WON'T win by switching, is if you correctly choose it the first time. And that only happens 1/3 of the time

The problem I always had with this (and still have) is that you're not asked about the probability (i.e. what trend you expect to see in a string of repeated experiments), but rather you're asked:

"Is it to your advantage to switch your choice?"

Well that depends on where the prize is. If the prize is behind the other door, then yes it's to your advantage to switch. Otherwise, it isn't. How can there be any more to it than that? You're not going to be given the opportunity to play the game 100 times. This is your one shot and the location of the prize is already fixed.

Well if you don't switch then your odds of winning are a simple 1 out of 3. Switching inverts your initial guess, so instead of losing 2 out of 3 times because your initial pick was wrong, switching makes you win 2 out of 3 times when your initial pick is wrong.

Imagine there are 100 doors. You pick one at random. You have a 1 in 100 chance of getting the right answer.

If the host opens all the doors that contain the booby prizes bar one (in this case 98 goats) and leaves one door remaining, plus your original guess, it's pretty obvious you are unlikely to have guessed correctly on your first try.

Scale it down to 3 doors and the host always revealing a goat and you have the same situation except there's a 1 in 3 chance you'll win by sticking with your first guess and a 2 in 3 chance if you swap.

Right, it's your one shot. So what gives you better odds (that is, is to your advantage). If you don't switch, you're selecting the lower odds chance of winning, putting yourself at a disadvantage. Since you don't know where it is, but do have a chance to double your chance of winning, then switching is advantageous.

Expected value is the thing to consider.

> Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right.

It's not obvious to me that this is true. Why do we update the chance of the "new" door but not the one we already selected? Intuitively, I would expect the odds of both doors to increase to 50% as the other 98 doors are opened.

I know that you're correct, I just don't find this to be an intuitive explanation.

The odds of your original door being right are 1%—that never changes. It’s determined by the options when the game was set up and when you picked it.

If your original door was wrong—99% chance—then it’s in one of the 99 other doors. But the host opens 98 of them, so if it was one of the 99 other doors—again, 99% chance—then the only option left is the last door.

The problem I have with this is that the original Monty Hall problem has him opening one door out of 2. With the 100 doors analogy, wouldn’t Monty be opening just one of the remaining doors? Or half? Why all 98?

The rule is "after the player has guessed a door, open ALL the other doors except the door that was guessed plus one other door".

I think your confusion is because if there are 100 doors then the rule is self-evident. But when there are only 3 doors, the rule of "open all doors except the door that was guessed plus one other" is indistinguishable from "open just one door".

The key is knowing that the rule is the "open all..." rule.

That’s not the rule from the question in the link nor in the original game (the game’s rule isn’t quite the same as the classic problem, it’s even messier). Your proposed rule doesn’t uniquely extend the original, it’s just one of the possible generalizations.

Sure, but why is the rule “open all”? In Monty Hall, the rule could also be “open half” or “open one” and it would be the same game!

i think the idea is to demonstrate more easily the information leak by exaggerating the amount of information being leaked.

with 3 doors, there are only so many the host can chose to open. with a 100, there is up to 98. sure, it can be 1 or 2... or 98.

I agree, I had the same thought last year: https://twitter.com/colmanhumphrey/status/139338571650012365.... It still helps but now only a small amount: (99/100)*(1/98) ≈ 0.0101, vs 1/100 without switching

Then the question of "switching" wouldn't make sense. Which of the remaining 98 would you be asked to switch to? The whole point is that you end up with a choice of sticking with the door you picked, or switching to the one remaining door.

Seems fine to me. Do you want to stick with your original or switch to one of the remaining 98? It’s still worth switching too, just a smaller increase in probability.

> is that the original Monty Hall problem has him opening one door out of 2

The better way of thinking about it is that he is leaving one door closed

I’ve found pointing this out is an “aha” moment for people with even little understanding of statistics.

Regardless of the number of doors, the space collapses to two options when the host reveals the rest. That new option is the set of all doors the player never picked.

The player’s initial option, 1/3 (33%) for 3 doors or 1% for 100 doors… versus the other option that is the rest of the set. 2 doors or 66% for 3 doors up to 99% of the doors for the 100 case.

That other option being the rest of the set must have better probability for any door count greater than two.

The point of opening the 98 doors is to make the math - and the fact that switching doors is safer - more obvious.

I've hear that before many times, but caution. You're just leaning on your intuition, probably not understanding why you got to the wrong answer in the first place.

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