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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 wi

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

qumpis 4y ago
Or to just imagine a 1000 boxes with the same problem formulation
- yesseri 4y ago
  And the key thing here is that all boxes except two gets removed, not only one.
- oleganza 4y ago
  That was a new spin to the explanation that I didn't think of before.
- jstx1 4y ago
  This doesn't do anything for me. (I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
  - coldtea 4y ago
    Because a 99/100 chance is much better than 2/3 to drive the point home...
  - curiousgal 4y ago
    Imagine there are 999 boxes with nothing in them and one box with the keys. After picking a box, the hosts opens 998 empty boxes. Would you still stick with your initial choice?
    - jstx1 4y ago
      I would change my choice because I understand the problem. But I would also change my choice in the scenario with 3 boxes. I'm not arguing with the conclusion, what I don't understand is the people who have their mind changed by the argument. Extending it to 1000 boxes/doors still doesn't explain why the remaining unopened box is different from the box your picked originally.
      - curiousgal 4y ago
        > the remaining unopened box is different from the box your picked originally. Because you now know that every other box is empty. So by process of elimination you know that your box and the remaining one are different.
  - brigandish 4y ago
    It's because it makes the initial choice so increasingly unlikely (increasing with the number of doors) to be correct that when the doors are taken away and you're left with only two, one of which must be right, it means that the other door is incredibly likely to be the right one.
    - jstx1 4y ago
      > It's because it makes the initial choice so increasingly unlikely But you still need to conivnce people that the one remaining unopened door is more likely than the door you originally selected. They were both unlikely to begin with, ramping up the number of doors doesn't explaing why one of them should be preferred.
      - brigandish 4y ago
        There are 1000 doors. You choose 1. Anyone knows it's incredibly unlikely that the correct one is chosen first time. Now 998 doors are removed. There is 1 door from the others and the door you choose. Given that your choice is almost certainly wrong, and that your opponent couldn't remove the correct door from amongst the 998, that means the other door is the correct one. Is that convincing enough?
  - deleted 4y ago
    [deleted]
  - __s 4y ago
    I've often found it easier to understand things intuitively by putting an idea to the slippery slope test. If such & such were true, imagine changing some parameters to an extreme, how absurd does it become? For monotonic functions it's useful
- wanderingstan 4y ago
  This was the one that worked when explaining it to my friends. It gives a mental image of the host opening 998 boxes, leaving only your selected box and one other. From here it’s easier to see that there must be something special about that one box the host left un-opened! (Though even then there were people who clung to the “2 boxes means 1-in-2 chance” fallacy, failing to see that the host has revealed information.) Edit: an other version was to change the hosts proposal: what if he let you choose one box, and then said he would let you switch to having whatever was in the other 999 boxes? Of course you would switch! The crux is understanding that this offer is actually the same as in the first proposal, since the host is not opening the boxes at random.
  - charlieflowers 4y ago
    To me, THAT is the most powerful intuitive description.
kasperni 4y ago
I think their explanation is a lot easier to understand "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."
- jasode 4y ago
  >= 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. Your sentence the particular way you worded it is not the correct mathematical model. The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply. Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
  - kasperni 4y ago
    > Your sentence the particular way you worded it is not the correct mathematical model. It wasn't really my sentence I just quoted the article. Nonetheless I disagree with you. The probability that the other boxes have the key is 2/3 and that is all that really matters. Opening a door doesn't change anything.
  - basch 4y ago
    You do get to switch to both other doors. One of the two remaining doors is a goat, and it is opened for you. Another way to phrase that is getting to pick both doors and the goat doesnt count against you. The Monty Hall problem distills to simply "would you like one or two doors, (if the prize is behind any door in the set you pick you win.)"
justatdotin 4y ago
yes, that's how I got it

qumpis 4y ago

Or to just imagine a 1000 boxes with the same problem formulation

yesseri 4y ago
And the key thing here is that all boxes except two gets removed, not only one.
oleganza 4y ago
That was a new spin to the explanation that I didn't think of before.
jstx1 4y ago
This doesn't do anything for me. (I understand the Monty Hall problem, I just don't see how changing the number of doors makes a difference to anyone's intuition.)
- coldtea 4y ago
  Because a 99/100 chance is much better than 2/3 to drive the point home...
- curiousgal 4y ago
  Imagine there are 999 boxes with nothing in them and one box with the keys. After picking a box, the hosts opens 998 empty boxes. Would you still stick with your initial choice?
  - jstx1 4y ago
    I would change my choice because I understand the problem. But I would also change my choice in the scenario with 3 boxes. I'm not arguing with the conclusion, what I don't understand is the people who have their mind changed by the argument. Extending it to 1000 boxes/doors still doesn't explain why the remaining unopened box is different from the box your picked originally.
    - curiousgal 4y ago
      > the remaining unopened box is different from the box your picked originally. Because you now know that every other box is empty. So by process of elimination you know that your box and the remaining one are different.
- brigandish 4y ago
  It's because it makes the initial choice so increasingly unlikely (increasing with the number of doors) to be correct that when the doors are taken away and you're left with only two, one of which must be right, it means that the other door is incredibly likely to be the right one.
  - jstx1 4y ago
    > It's because it makes the initial choice so increasingly unlikely But you still need to conivnce people that the one remaining unopened door is more likely than the door you originally selected. They were both unlikely to begin with, ramping up the number of doors doesn't explaing why one of them should be preferred.
    - brigandish 4y ago
      There are 1000 doors. You choose 1. Anyone knows it's incredibly unlikely that the correct one is chosen first time. Now 998 doors are removed. There is 1 door from the others and the door you choose. Given that your choice is almost certainly wrong, and that your opponent couldn't remove the correct door from amongst the 998, that means the other door is the correct one. Is that convincing enough?
- deleted 4y ago
  [deleted]
- __s 4y ago
  I've often found it easier to understand things intuitively by putting an idea to the slippery slope test. If such & such were true, imagine changing some parameters to an extreme, how absurd does it become? For monotonic functions it's useful
wanderingstan 4y ago
This was the one that worked when explaining it to my friends. It gives a mental image of the host opening 998 boxes, leaving only your selected box and one other. From here it’s easier to see that there must be something special about that one box the host left un-opened! (Though even then there were people who clung to the “2 boxes means 1-in-2 chance” fallacy, failing to see that the host has revealed information.) Edit: an other version was to change the hosts proposal: what if he let you choose one box, and then said he would let you switch to having whatever was in the other 999 boxes? Of course you would switch! The crux is understanding that this offer is actually the same as in the first proposal, since the host is not opening the boxes at random.
- charlieflowers 4y ago
  To me, THAT is the most powerful intuitive description.

kasperni 4y ago

I think their explanation is a lot easier to understand "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."

jasode 4y ago
>= 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. Your sentence the particular way you worded it is not the correct mathematical model. The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply. Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
- kasperni 4y ago
  > Your sentence the particular way you worded it is not the correct mathematical model. It wasn't really my sentence I just quoted the article. Nonetheless I disagree with you. The probability that the other boxes have the key is 2/3 and that is all that really matters. Opening a door doesn't change anything.
- basch 4y ago
  You do get to switch to both other doors. One of the two remaining doors is a goat, and it is opened for you. Another way to phrase that is getting to pick both doors and the goat doesnt count against you. The Monty Hall problem distills to simply "would you like one or two doors, (if the prize is behind any door in the set you pick you win.)"

justatdotin 4y ago

yes, that's how I got it