[AmericanChopper]

The trick to understanding the Monty Hall problem isn’t to run it 1000 times, it’s to imagine it with 1000 doors. Thinking about the problem with an arbitrarily large number of doors make the solution intuitively obvious. No code required.

[mcprwklzpq]

This trick did not help me. Intuition told me in the end i have to choose between 2 doors and the chance of winning is 1/2. And it is if i choose in the end at random. This is what made the question confusing for me.

[AmericanChopper]

The opening of doors prior to having you make your decision is a red herring. The person opening the doors knows which one the prize is behind, and they’re only going to open doors which have no prize. So if there’s 1000 doors, you have a 1/1000 chance of choosing the correct door, and there’s a 999/1000 chance that the prize is behind a different door. So if the host selectively opens 998 losing doors, leaving your original selection and one other door closed, there is still a 999/1000 chance that the prize is behind the door you didn’t choose. If the host instead opened the doors randomly, there would be a 998/1000 chance that they’d reveal the prize before narrowing it down to the two doors. But they don’t open them randomly, which is why the door opening doesn’t matter. You’re effectively being given a choice between the one random door you originally chose, or every other door.

[mcprwklzpq]

This was not what bothered me. I needed to see how this puzzle does not contradict my intuition that if i flip a coin to choose between two last doors i do get 1/2 chance of winning whatever are the chances for each door. So with 1000 doors: (1/1000)/2 + (999/1000)/2 = 1/2 chance of winning. And in general (A+B)/2 = 1/2 if A+B=1.

[AmericanChopper]

If you flip a coin at the decision stage, then you would have a 50/50 chance of winning. But that approach would mean you’re disregarding information that you can use to increase your chances of success above 50/50. When you make your initial choice you have a 1/1000 chance of guessing correctly, meaning there is a 999/1000 chance that the prize is behind another door. By eliminating 998 doors from contention, you’re not changing those odds. There is still a 999/1000 chance for the prize to be behind a door other than the one you initially picked. Only now, instead of having 999 doors to choose from, you only have one. So by switching your choice, you increase your chance of success from being whatever it was initially, to being (1 - whatever it was initially). Which will always be greater than 0.5.

This is materially the same as being given a choice between opening one door randomly, or being able to open every other door. Because the conditions of opening doors prior to the final choice is that the door you initially picked can’t be opened at this phase, and neither can the winning door.

[theelous3]

Yes! This was how I first figured out how to understand it, and how I explain it to anyone if it comes up. It's immediately obvious with a large number of doors.

[jonsen]

I pick 333 doors, and Monty opens 333 doors?

[kemotep]

No the variation with 1000 doors is that you open 1 door, Monty opens 998 doors and asks if you want to switch.

[AmericanChopper]

Yeah, the use of 3 doors, and the bit where one is opened just kinda hides what’s happening. Which is that you’re actually choosing between one random door, or ALL the other doors.

[scubbo]

The Eureka moment for me was similar, but slightly different. Consider a variant game with N doors. You pick a door, then Monty offers you two switch your payout to "the best single result from the remaining N-1 doors". Clearly, you would swap.

Now set N to 3, and consider that, by opening a door without the prize, that's exactly what Monty's offering you.