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That simulates a different scenario than what’s being described here; it simulates a situation where we count the times when Monty picks the car. But those situ

That simulates a different scenario than what’s being described here; it simulates a situation where we count the times when Monty picks the car. But those situations are irrelevant because they have no bearing on the fundamental question of whether or not to switch doors when Monty shows you a goat. Effectively, we discard all outcomes when Monty shows you the car, making it an identical “game” as when Monty simply does not ever choose the car. As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.

Arnavion 6y ago
>it simulates a situation where we count the times when Monty picks the car. No, it does not. >> if (montys_choice === car_door) { i -= 1; continue; } This discards the game where Monty chose a car. >> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); This makes Monty never choose a door with a car. Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.
- IggleSniggle 6y ago
  Thank you (and everyone else in this thread) for sticking with it with me. It was a struggle to read that single line of code on a cell-phone screen, and that was compounded by the fact that it turns out that I don't understand the Monty Hall problem when I thought that I did. I'm going to sit with this one for awhile.
dllthomas 6y ago
So let's say you and I sit down, and do the following: 1) I roll a 3-sided die (or a 6 sided, wrapping) and keep it covered. 2) You pick a number, 1-3 3) I flip a coin. If it's heads, I pick the lower available number; if it's tails I pick the higher. 4) I peak at the die. If it's my number, we reveal and start over. 5) I (always, at this point) offer you a wager: if the die shows your number (so you would have lost if you switch), you pay me $7; if the die doesn't show your number, I pay you $5. Assuming you believe the die and coin are fair, etc, would you agree to play that game 1000 times? In those games, is there a reason you would turn down the wager? Is it the same game if I flip the coin secretly, peek at the die, announce my number (picked algorithmically in the obvious way), and then pay out according to whether switching would win (as above)? Because (assuming I've explained these games as I intend... it's getting late) I would play the former with you not the latter (but I would play the latter if we switched the payments around - I chose 5 and 7 because 7/12 is halfway between 1/2 and 2/3).
rcxdude 6y ago
It will be different. The question comes down to P(winning by switching | monty reveals a goat). The key difference is whether P(mony reveals a goat | you chose a goat door the first time) is 50% or 100% (if you choose the car, there's a 100% chance he reveals a goat in either case). Since 'winning by switching' is the same as 'choosing a goat door first', you can apply bayes theorem to see how the results change. To put it another way, if Monty is choosing randomly, half the time where you would win by switching, instead the game just ends/isn't counted, but the same is not true of the case where you win by not switching. From a bayesian point of view, Monty randomly revealing a goat should increase your belief that you picked the car the first time. That said, switching is not worse than not switching unless Monty is biased towards revealing the car instead.

Arnavion 6y ago

>it simulates a situation where we count the times when Monty picks the car. No, it does not. >> if (montys_choice === car_door) { i -= 1; continue; } This discards the game where Monty chose a car. >> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); This makes Monty never choose a door with a car. Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.

IggleSniggle 6y ago
Thank you (and everyone else in this thread) for sticking with it with me. It was a struggle to read that single line of code on a cell-phone screen, and that was compounded by the fact that it turns out that I don't understand the Monty Hall problem when I thought that I did. I'm going to sit with this one for awhile.

dllthomas 6y ago

So let's say you and I sit down, and do the following: 1) I roll a 3-sided die (or a 6 sided, wrapping) and keep it covered. 2) You pick a number, 1-3 3) I flip a coin. If it's heads, I pick the lower available number; if it's tails I pick the higher. 4) I peak at the die. If it's my number, we reveal and start over. 5) I (always, at this point) offer you a wager: if the die shows your number (so you would have lost if you switch), you pay me $7; if the die doesn't show your number, I pay you $5. Assuming you believe the die and coin are fair, etc, would you agree to play that game 1000 times? In those games, is there a reason you would turn down the wager? Is it the same game if I flip the coin secretly, peek at the die, announce my number (picked algorithmically in the obvious way), and then pay out according to whether switching would win (as above)? Because (assuming I've explained these games as I intend... it's getting late) I would play the former with you not the latter (but I would play the latter if we switched the payments around - I chose 5 and 7 because 7/12 is halfway between 1/2 and 2/3).

rcxdude 6y ago

It will be different. The question comes down to P(winning by switching | monty reveals a goat). The key difference is whether P(mony reveals a goat | you chose a goat door the first time) is 50% or 100% (if you choose the car, there's a 100% chance he reveals a goat in either case). Since 'winning by switching' is the same as 'choosing a goat door first', you can apply bayes theorem to see how the results change. To put it another way, if Monty is choosing randomly, half the time where you would win by switching, instead the game just ends/isn't counted, but the same is not true of the case where you win by not switching. From a bayesian point of view, Monty randomly revealing a goat should increase your belief that you picked the car the first time. That said, switching is not worse than not switching unless Monty is biased towards revealing the car instead.