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>Now the only difference is that Monty might accidentally pick a prize. This a mistaken conclusion, as actually there is a difference if a goat is revealed by
by jml7c5 5y ago
>Now the only difference is that Monty might accidentally pick a prize.
This a mistaken conclusion, as actually there is a difference if a goat is revealed by devious Monty (who makes sure he doesn't reveal the car) and dumb Monty (who picks at random), because the rules of the game really are different. The apparent outcome (you pick door, Monty reveals a goat) is the same. But the process is different. In the latter version dumb Monty has all the same information as you, in the former version devious Monty gets more information that you do and changes his behavior based on that information.
In programming terms, devious Monty has an "if" statement in there. (Something like, `if (random_choice == prize_door) { swap_choice(); open_door(); } else { open_door(); }`.)
- haberman 5y agoTL;DR: I was wrong (but I did understand the distinction you are drawing) I'm aware that there is a difference between devious Monty and dumb Monty. The question I was trying to answer is whether this difference affects the actual outcome of the game, or the choice you should make. I ran a Monte Carlo simulation to test my analysis, and this proved that my previous analysis was in error. If Monty doesn't know to avoid the door with the car, then it makes no difference statistically speaking whether the player switches or not after Monty reveals a goat. If the rules are such that you win when (dumb) Monty picks the car, then your overall chances are now 2/3 instead of 1/3, since you effectively get two guesses (your guess and Monty's guess). But it makes no difference whether you switch after Monty picks a goat. On the other hand, if (dumb) Monty choosing the car results in either (1) you losing or (2) a reshuffle and redo, then your overall chances are only 1/3. In this case, it also makes no difference whether you switch after Monty's guess.