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The Monty Hall Problem: A Statistical Illusion
- herodotus 7y agoThe 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.
- justinsaccount 7y agoI 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.
- indigodaddy 7y agoThis was the key for me as well in having a real understanding of it.
- resonantjacket5 7y agoAgreed, 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.
- dang 7y agoMany previous threads: https://hn.algolia.com/?dateRange=all&page=0&prefix=true&query=%22monty%20hall%22%20comments%3E10&sort=byDate&type=story https://hn.algolia.com/?dateRange=all&page=0&prefix=true&que... (Sharing for curiosity purposes.)
- bhk 7y agoThis 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.
- indigodaddy 7y agoThat’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.
- tromp 7y agoFor 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.
- BoiledCabbage 7y agoI'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.