4 ms·

Please explain to me why it is different whether to switch doors after opening all but one door vs after opening all but one door? (note: when you do this in r

Please explain to me why it is different whether to switch doors after opening all but one door vs after opening all but one door? (note: when you do this in real life, you stop talking now. let them explain it.) When you have N>2 doors, the host will with 100% certainty open N-2 doors with goats behind them. It doesn't matter whether N is 3 or 1,000,000. After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2. The key phrase is "host's hidden knowledge". Once the host starts opening the wrong doors, the fundamental nature of the problem is changed from being an independent probability problem to a dependent probability one.

henrikschroder 7y ago
> After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2. No, your numbers are wrong. In your initial choice, the probability of picking the winning door is 1/N. When Monty opens doors so that there is a single door remaining, the probability of that door containing the prize is the complement of the probability that you picked the winning door, i.e. 1-(1/N). So switching doors increases your probability from 1/3 to 2/3 if N=3, not to 1/2. If N=1000, switching doors increases your probability from 1/1000 to 999/1000, and that is why increasing the number of doors makes the correct solution more intuitive for some people.
- CydeWeys 7y ago
  And the point I'm making is that it's not clear at all why opening 998 doors out of 1,000 is analogous to opening 1 door out of 3 -- why wouldn't you still be opening just the 1 door? If you pick one of 1,000 doors, and then the host reveals a goat behind a single door, there's a slight (but only very slight) advantage to switching. The advantage is most pronounced with the fewest possible number of doors, i.e. 3. That's why this approach to teaching it has never made sense to me -- it's not clear at all why opening 998 doors should be analogous to opening 1 door, which it's much more obvious that opening 1 door is analogous to 1 door.

henrikschroder 7y ago

> After the host's hidden knowledge has reduced the problem space from N options to 2 options, the probability of getting the car is increased from 1/N to 1/2. No, your numbers are wrong. In your initial choice, the probability of picking the winning door is 1/N. When Monty opens doors so that there is a single door remaining, the probability of that door containing the prize is the complement of the probability that you picked the winning door, i.e. 1-(1/N). So switching doors increases your probability from 1/3 to 2/3 if N=3, not to 1/2. If N=1000, switching doors increases your probability from 1/1000 to 999/1000, and that is why increasing the number of doors makes the correct solution more intuitive for some people.

CydeWeys 7y ago
And the point I'm making is that it's not clear at all why opening 998 doors out of 1,000 is analogous to opening 1 door out of 3 -- why wouldn't you still be opening just the 1 door? If you pick one of 1,000 doors, and then the host reveals a goat behind a single door, there's a slight (but only very slight) advantage to switching. The advantage is most pronounced with the fewest possible number of doors, i.e. 3. That's why this approach to teaching it has never made sense to me -- it's not clear at all why opening 998 doors should be analogous to opening 1 door, which it's much more obvious that opening 1 door is analogous to 1 door.