4 ms·

When you choose the first door, you are partitioning the "board" into two parts: the door you chose, and "everything else". By opening a door, Monty lets you c

When you choose the first door, you are partitioning the "board" into two parts: the door you chose, and "everything else". By opening a door, Monty lets you cover 100% of the "everything else" partition using only one guess. So now you get to choose between partition 1, which covers 1/3 of the board, and partition 2, which covers 2/3 of the board.

spenczar5 5y ago
I have never heard this formulation, and I like it quite a lot. Thanks!
- selimthegrim 5y ago
  It is Bayes’ Rule essentially written out (sample spaces)
someguydave 5y ago
I think the confusing part for many people is the unstated fact that the host takes all prizes into account when revealing a door, and that the host is forbidden from revealing the car. If the host chooses randomly then his pick adds no information (other than the extra revealed prize)
- jdashg 5y ago
  I don't think people are confused by this in their initial solutions, but rather this after-the-fact problem modification happens to match the incorrect intuition people have. The instinct is strong that "I am right" so "I must have just solved the wrong problem", thus "the problem was confusing". Rather, I think you will find that most people don't expect the host will ever pick a door with a car until after you tell them that switching is better, and rationalization begins. Additionally, it's on the solver to realize that a problem is under-constrained, and to ask for more constraints. That means that even if you were earnestly confused by the question, you presented a solution to an under-constrained problem. This is like answering "what is the square-root of four" with "it's negative two".
  - someguydave 5y ago
    yeah, it’s reasonable to attribute those defects to the solver and not their situation and circumstances
- haberman 5y ago
  I don't think this actually matters. Suppose Monty doesn't actually know which door has the prize, and just picks randomly between the two. Now the only difference is that Monty might accidentally pick a prize. Suppose he does. What are the options now? 1. The game goes on, and you can just switch your guess to the door he picked. (WIN) 2. The game resets due to Monty's error and you play again. (REDO) 3. Monty just decides that you lose since he picked the prize. (LOSE, NO CHOICE) Options 1 and 2 don't deprive you of the car. But option 3 doesn't give you a choice, so there is no dilemma. So conditional on having a game where no prize has been revealed yet and you get a choice to switch, I think it still pays to always switch. Another way of putting it: you get to choose between partition 1 covering 1/3 of the board where you get 1 guess, or partition 2 covering 2/3 of the board where you get two guesses. It works even if all of the guesses are random.
  - 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 ago
      TL;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.

spenczar5 5y ago

I have never heard this formulation, and I like it quite a lot. Thanks!

selimthegrim 5y ago
It is Bayes’ Rule essentially written out (sample spaces)

someguydave 5y ago

I think the confusing part for many people is the unstated fact that the host takes all prizes into account when revealing a door, and that the host is forbidden from revealing the car. If the host chooses randomly then his pick adds no information (other than the extra revealed prize)

jdashg 5y ago
I don't think people are confused by this in their initial solutions, but rather this after-the-fact problem modification happens to match the incorrect intuition people have. The instinct is strong that "I am right" so "I must have just solved the wrong problem", thus "the problem was confusing". Rather, I think you will find that most people don't expect the host will ever pick a door with a car until after you tell them that switching is better, and rationalization begins. Additionally, it's on the solver to realize that a problem is under-constrained, and to ask for more constraints. That means that even if you were earnestly confused by the question, you presented a solution to an under-constrained problem. This is like answering "what is the square-root of four" with "it's negative two".
- someguydave 5y ago
  yeah, it’s reasonable to attribute those defects to the solver and not their situation and circumstances
haberman 5y ago
I don't think this actually matters. Suppose Monty doesn't actually know which door has the prize, and just picks randomly between the two. Now the only difference is that Monty might accidentally pick a prize. Suppose he does. What are the options now? 1. The game goes on, and you can just switch your guess to the door he picked. (WIN) 2. The game resets due to Monty's error and you play again. (REDO) 3. Monty just decides that you lose since he picked the prize. (LOSE, NO CHOICE) Options 1 and 2 don't deprive you of the car. But option 3 doesn't give you a choice, so there is no dilemma. So conditional on having a game where no prize has been revealed yet and you get a choice to switch, I think it still pays to always switch. Another way of putting it: you get to choose between partition 1 covering 1/3 of the board where you get 1 guess, or partition 2 covering 2/3 of the board where you get two guesses. It works even if all of the guesses are random.
- 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 ago
    TL;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.