5 ms·

If both contestants choose the same door, and the host shows a goat, then yes, they both would increase their odds by switching.

kgwgk 6y ago
And if they could have chosen the same but didn’t? The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)
- deleted 6y ago
  [deleted]
- the_af 6y ago
  If participants are allowed to pick the same door, even if they don't in practice, then yes: after Monty shows a goat behind the third goat then both participants should switch. This is because they are essentially playing two simultaneous but independent games, with no interference between each other. As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.
  - kgwgk 6y ago
    After they are shown a goat behind the unchosen door they are less sure that they got it wrong in the first try. It becomes a 50/50 bet.
    - the_af 6y ago
      No, your intuition is misleading you: it's not a choice between two doors. If the probabilities don't convince you, you can easily simulate this by using three cards and considering one is the car and the other two goats. Pick one at random, then pick the remaining "goat" as if you were Monty; lastly, switch your initial choice as a player. Repeat the experiment 10 times and count how many times you win; remember you must always switch. You will win more often than lose! I tried this, that's how I know this will convince you. Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable. The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)
      - kgwgk 6y ago
        So when they don’t switch each one has probability 1/3 of getting the car? Where does the car go if no one gets it? And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing? Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again. Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
        the_af 6y ago
        > So when they don’t switch each one has probability 1/3 of getting the car? Yes. > Where does the car go if no one gets it? I don't understand your question. What does this have to do with probabilities? > And if they switch both have probability 2/3 of getting the car? Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously. They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's. > Is this some form of car sharing? That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch. > Maybe you should try to play with three cards and one ace. There are only 3 doors, so you should only use 3 cards. > Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2? I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".
        kgwgk 6y ago
        If following has happened already: Player #1 has picked door A Player #2 has picked door B The host has opened door C and there was a goat At this point: What are the chances of winning of player #1 if he keeps his pick of door A? And if he switches to door B? What are the chances of winning of player #2?
        6y ago

kgwgk 6y ago

And if they could have chosen the same but didn’t? The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)

deleted 6y ago
[deleted]
the_af 6y ago
If participants are allowed to pick the same door, even if they don't in practice, then yes: after Monty shows a goat behind the third goat then both participants should switch. This is because they are essentially playing two simultaneous but independent games, with no interference between each other. As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.
- kgwgk 6y ago
  After they are shown a goat behind the unchosen door they are less sure that they got it wrong in the first try. It becomes a 50/50 bet.
  - the_af 6y ago
    No, your intuition is misleading you: it's not a choice between two doors. If the probabilities don't convince you, you can easily simulate this by using three cards and considering one is the car and the other two goats. Pick one at random, then pick the remaining "goat" as if you were Monty; lastly, switch your initial choice as a player. Repeat the experiment 10 times and count how many times you win; remember you must always switch. You will win more often than lose! I tried this, that's how I know this will convince you. Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable. The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)
    - kgwgk 6y ago
      So when they don’t switch each one has probability 1/3 of getting the car? Where does the car go if no one gets it? And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing? Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again. Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?
      - the_af 6y ago
        > So when they don’t switch each one has probability 1/3 of getting the car? Yes. > Where does the car go if no one gets it? I don't understand your question. What does this have to do with probabilities? > And if they switch both have probability 2/3 of getting the car? Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously. They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's. > Is this some form of car sharing? That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch. > Maybe you should try to play with three cards and one ace. There are only 3 doors, so you should only use 3 cards. > Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2? I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".
        kgwgk 6y ago
        If following has happened already: Player #1 has picked door A Player #2 has picked door B The host has opened door C and there was a goat At this point: What are the chances of winning of player #1 if he keeps his pick of door A? And if he switches to door B? What are the chances of winning of player #2?
        6y ago