The Monty Hall Problem: A Statistical Illusion | Hacker News Reader

The Monty Hall Problem: A Statistical Illusion

statisticsbyjim.com

30 pointsgaryclarke2712 comments

The intuitive explanation I like is this: Suppose there are 10000 boxes. You pick one. Monty Hall knows where the prize is. He opens 9998 boxes, leaving just two closed: your original choice, and one other one. Should you switch? Before Monty Hall opened any boxes, the chances of the prize being in one of the boxes other than the one you chose was 9999 in 10000. This probability does not change when he opens 9998 of these. So you should choose the remaining box, not your original choice.

The same argument applies with 3 boxes, but intuition is much less reliable in this case.

I always found that one makes things more complicated. I like this explanation better:

You win by switching as long as you were wrong in the initial guess. You have a 2/3 chance of being wrong in the initial guess. Therefore you will win by switching 2 times out of 3.

Can also phrase it in reverse: You lose by switching if you were right in the initial guess. You have a 1/3 chance of being right in the initial guess. Therefore you lose by switching 1 time out of 3.

This was the key for me as well in having a real understanding of it.

Agreed, one main problem with the original monty hall problem description was that it wasn't clearly said that monty hall would only open a door that must have a goat.

This problem is confusing partly because the premises are often incorrectly stated... as in TFA.

If it is given that Monty always picks a door that does not have a prize behind it, then indeed the correct answer is to switch doors. But otherwise, there is no advantage in switching: 1/3 of the time he would expose the prize, and in the other 2/3 of cases you are just as likely to win without switching.

That’s the basis of the problem though, you are shown a door with no prize. I don’t understand the confusion. In that scenario you switch. No other scenario is offered.

For an article focussing on assumptions, it's rather sloppy to state the situation as

"Monty opens one of the other two doors, and there is no prize behind it."

rather than

"Monty opens one of the other two doors, one which he knows has no prize behind it."

which rules out the interpretation of Monty opening a door that by chance happens to have no prize behind it, leading to a different answer.

I'm not convinced him knowing where the prize is before opening the door matters. If he opens a door and knew there was no prize you now get the same info as if he opened the door and was surprised there was no prize. In both cases switching its better. And if he unwittingly opens the door and there is the prize, again switching is better.

His knowledge before hand doesn't matter, switching its always better.

That said, I feel like there is a limitation in our modeling of probability that this problem isn't easier. Yes anyone who knows stats can solve it, but somehow we're using the wrong "language"that the problem is still tricky. Geometry isn't tricky becsuse we have the right language for it. Geometry is "hard" when problems are hard/complex. Monty Hall the problem isn't complex, but solving it correctly is still difficult.

To take a programming term, probability applied to this problemfeels like it has accidental complexity.

This is not correct. In case Monty opens a random door which may or may not have the prize behind it, there is no reason to switch. Probability table is:

1/3 chance you pick the prize on first choice. (And Monty opens a random goat-door)

2/3 chance you don't, in which case 1/2 of the time, Monty opens the prize door and you lose. The other 1/2 of the time Monty opens a random goat)

So after Monty has opened a door there is now 3 possible scenarios:

1. 1/3 chance you picked the prize. Don't switch

2. 1/3 = 2/3 * 1/2 chance you didn't pick the prize and Monty opened the prize door. You lost.

3. 1/3 = 2/3 * 1/2 chance you didn't pick the prize and Monty opened a goat door. In this case, switch.

The probability of 1 and 3 are equal and switching or not switching doesn't matter.

Now if Monty knows where the prize is and always opens a goat door, things are very different.

1/3 chance you picked the prize. Monty opens a goat. You shouldn't switch.

2/3 change you picked a goat. Monty opens a goat. You should switch.

Now 2/3 of the time you can switch and get the prize. So switching is correct.

It all depends on the unstated assumption that Monty may actually have told you something, rather than taken a risk. And that's why this problem is mostly a good lesson in what can happen if you don't state your problem well.

What if Monty only opens a door when he knows you've chosen the prize door?

Extension: what if Monty tosses a coin to decide whether to open a door, conditioned on you having chosen the prize door?

Specifying Monty's behaviour matters, I think.

Monti's a priori knowledge absolutely matters and should have been stated in the article. If he had no knowledge beforehand about which door the prize was behind, then you would have to expand the probability table to include cases in which Monti opens one of the remaining two doors and accidentally reveals the prize, in which case the user just loses without getting the opportunity to switch or not switch. Monti knowing the door is hiding the prize and actively selecting the non-prize door versus just guessing and sometimes accidentally revealing the prize, are two different statistics problems.

Many previous threads: https://hn.algolia.com/?dateRange=all&page=0&prefix=true&que...

(Sharing for curiosity purposes.)

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