Timeline for How to explain Monty Hall problem when they just don't get it
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2014 at 12:37 | comment | added | devMomentum | For me I convinced myself with n-door generalization, and with opening all doors, except one. For n = 1 million, I have a 1 in a million chance, and all the others together have 999,999 in a million chance. After opening 999,9998 doors, I am drawn to choose the other closed door. Whether we open 1 door or n-2 doors has the same result, you still have more chance changing your door with one of the non opened doors. The larger n is, the smaller the increase in probability. Opening n-2 doors makes it much more obvious the choice to change door. | |
| Jun 19, 2014 at 6:31 | comment | added | Aron | I prefer to use 52 doors. Represent each door with identical paper cards, with a different pattern on the front. Choose a single pattern to represent the car, and call it the "lady". | |
| Jun 18, 2014 at 4:38 | comment | added | Izkata | @DavidG The numbers are too big to hold in your head and sort through. I could not understand the 100-door version until I had a better explanation of the 3-door version, and even now it's difficult to visualize. | |
| Jun 17, 2014 at 4:08 | comment | added | Ryan Reich | Both of the previous two comments perfectly demonstrate my objection that to someone who doesn't know what's going on, a generalization of a tricky problem may not convincingly illustrate the original problem at all. You have to know how it generalizes, and which features are relevant. | |
| Jun 16, 2014 at 23:40 | comment | added | senshin | Here's my problem with the $n$-door version - it's not at all obvious that the correct generalization of the Monty Hall problem is that the host opens $n-2$ doors rather than still only opening $1$ door. Both of these reduce to the same thing in the $n=3$ case, but only the opens-$n-2$-doors version gives you good intuition about how the problem works. | |
| Jun 16, 2014 at 18:52 | comment | added | Ben Aaronson | @DavidG The reason I don't like the 100 door version is because the intuition it gives you- for me at least- is not based on Monty knowing which door the car is behind. If Monty picked randomly (with the possibility of him unknowingly opening the door with the car) then switching would bring no benefits. So any explanation that gives you the right intuition in the case where he knows which door it's behind should also gives you the right intuition in the case where he doesn't. Otherwise you're just replacing one misleading intuition with another. | |
| Jun 15, 2014 at 21:16 | comment | added | Peter LeFanu Lumsdaine | @JoeTaxpayer: I think the thing which makes this explanation so good is that you don’t have to say anything slippery like “the 2/3 is packed into the one remaining door” to make it rigorous. Monty always picking a non-car door is encapsulated in the step “If you picked the car, you’ll lose it by switching,” etc. — which is a deterministic step. It avoids using conditional probability (which for many people is unintuitive) by breaking the argument into a conditional-but-deterministic part, plus a non-conditional probability part. | |
| Jun 15, 2014 at 17:24 | comment | added | David G | I have always prefered the 100 door extension and don't understand how it would be difficult to explain in that way. | |
| Jun 15, 2014 at 13:48 | comment | added | JTP - Apologise to Monica | Very nice. The only thing to spell out is the fact that Monty knows to remove a non-winner, so in effect the '2/3' is packed into that one remaining door. But your explanation is great, and nicely avoids the 100 door extension. | |
| Jun 15, 2014 at 7:26 | history | answered | ChrisA | CC BY-SA 3.0 |