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The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious. Imagine a universe with many simultaneous Monty Hall clones playing…
The most intuitive and simple explanation that worked for me is: * if on the 1st try you choose the correct box (33% chance), then the one you can switch to will be wrong * if on the 1st try you choose the wrong box (66% chance), then the one you can switch to will be correct one therefore your goal is to pick the wrong box on the 1st try and then switch, and you have 66% chance to do it
Or to just imagine a 1000 boxes with the same problem formulation
(I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
1000 doors, choose 1.
Host opens 998 doors.
Do you switch?
The most intuitive and simple explanation that worked for me is: * if on the 1st try you choose the correct box (33% chance), then the one you can switch to will be wrong * if on the 1st try you choose the wrong box (66% chance), then the one you can switch to will be correct one therefore your goal is to pick the wrong box on the 1st try and then switch, and you have 66% chance to do it
"When we pick the original box, we know that the probability that the keys will be in there is 1/3. The probability that the keys will not be in the box you originally chose is 1 - 1/3 = 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3."
The article hides this away by lumping those two under "Host opens B or C" without further justification, but it's important to notice that this only works because those twelve outcomes have different probabilities.
Edit: In table form,
Car Guest Monty Swap? Probability
A A B No 1/3 * 1/3 * 1/2 = 1/18
A A C No 1/3 * 1/3 * 1/2 = 1/18
A B C Yes 1/3 * 1/3 = 1/9
A C B Yes 1/3 * 1/3 = 1/9
B A C Yes 1/3 * 1/3 = 1/9
B B A No 1/3 * 1/3 * 1/2 = 1/18
B B C No 1/3 * 1/3 * 1/2 = 1/18
B C A Yes 1/3 * 1/3 = 1/9
C A B Yes 1/3 * 1/3 = 1/9
C B A Yes 1/3 * 1/3 = 1/9
C C A No 1/3 * 1/3 * 1/2 = 1/18
C C B No 1/3 * 1/3 * 1/2 = 1/18
Total Yes = 6 * 1/9 = 6/9 = 2/3
Total No = 6 * 1/18 = 6/18 = 1/3Imagine that after you pick your box, Monte Hall invites an audience member up on stage and instructs them to choose one of the remaining two doors to open. This audience member doesn't know anything at all and just randomly picks one of the two doors. When their door is opened we see that it's empty. You're now given the option of switching just like in the standard game. Should you?
Cosmetically everything is identical with the standard game, but if you analyze the game carefully this time you're left with a 50/50 shot so there's no benefit of switching.
I think most of the arguments in this article would appear to work for this modified version of the game which means that they're not actually getting to the heart of the problem.
For completeness, the reason this now reduces to 50/50 is that there's also now a chance that the spectator opens the door with the car behind it, something that couldn't happen in the original Monte Hall problem. Put another way, there's actually a little bit of information that's conveyed to you when you see that the spectator happens to not open the door with the car and this extra information exactly cancels the usual benefit you get from eliminating the other empty door. In the example of "scaling up" in the article, if you did this with 20 doors and the spectator randomly picks 18 of the 19 unopened ones to open and then happen to not stumble upon the car, you might actually think that you could have been lucky all along. Ultimately you're left with a 50/50 chance.
The thing that is often de-emphasised in the presentation of the problem, in order to make it seem more mysteriously paradoxical, is that the presenter knows where the car is and this knowledge is always used perfectly. If the question always ended with "remember: Monty knows where the car is and will use this information", it would be more obvious. Imagine a universe with many simultaneous Monty Hall clones playing…
wow... what a grim approach!
You pick a box, and another empty box is revealed. Then, the two remaining boxes are shuffled so that you no longer can tell which is which (but the game host still knows). You then choose one of four options:
- take box A,
- take box B,
- ask the host to give you the box you picked initially, or
- ask the host to give you the box you didn't pick initially.