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C G G
G C G
G G C
You choose a door, say Door 1. Monty opens another door with a goat behind it.
C x G
G C x
G x C
Now look at the grid. Staying in column 1 gives you the probability space C G G, a 1/3 chance of getting a car. Switching gives you C C G, a 2/3 chance of getting a car.
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 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…
In the scenario where they are shot that you described, it is still better to switch!
Earlier quoted context omitted.
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.)
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?
Extending it to 1000 boxes/doors still doesn't explain why the remaining unopened box is different from the box your picked originally.
In the table, why does the host picking one of the other boxes get combined into a single probability row? If the host could pick either box, and both of those choices result in a loss, should we count that as additional possibilities?
Consider the Honty Mall problem: it’s like the original problem, except after you pick a box, Honty offers you both of the other boxes. It’s much easier to see swapping is better in this problem, and it’s also easier to see that the chance is 2/3 if you swap. Then you just have to show that the Honty Mall problem is equivalent to the Monty Hall problem, by stipulating that Monty will always open a box that’s empty.
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…
And perhaps also, “remember: Monty will always open a door, and the contestant knows it”.
Makes me wonder if there were similar shows where the host can choose not to open a door.
There's another version of the Monte Hall problem that highlights why this is such a counterintuitive problem. Imagine 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…
- a car is revealed, I lose immediately (there is no option to switch anymore)
- no car is revealed, which means I again have 2/3 chances when switching, not a 50% chance as you've stated
In your example, the spectator opens an empty door, so there is no difference to the host opening an empty door in regards to the probability. Again, if the spectator opens a car, I just lose.
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 associated line of reasoning resolved the paradox for me. If I stick with my original choice, it is as if I ignored the new information. If I switch the choice, I react to the new information.
The extreme of this is to pick among countably infinite doors, having the presenter open countably infinite doors and leaving just yours and another closed. Who could reasonable suggest that the chance is still 50/50, assuming you don't flip a coin and base your choice on that?
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
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."
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.