4 ms·

Imagine there are two contestants and each one selects a door. The host opens the remaining one and it happens to show goat. Do you think they will both increas

• basch 6y ago
  Can both select the same door? If the answer is no, you've changed the parameters of the game, because the person who chooses second can only pick from two doors. The subset of games where they close different doors is smaller than all possible games.
  • deleted 6y ago
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  • kgwgk 6y ago
    And if they could have selected the same?
    • diggernet 6y ago
      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
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        • 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?
• diggernet 6y ago
  In your example, the host didn't get to choose their door, which kept them from choosing goat. As I follow this discussion, the key is the knowledge that given a choice, the host will pick a non-winning door. If the host can't choose, you lose the benefit of that fact.
  • kgwgk 6y ago
    Of course. I was replying to someone who says that the host could pick randomly and it wouldn’t change anything.
• noxer 6y ago
  The host has no choice in this case so his action has no effect. The chance to win is 1/3 for all. And after the host gets a goat one of the contestant has to wins so its 1/2 for both whether they switch or not.
  • kgwgk 6y ago
    That’s indeed what happens when the host picks at random (the hypothesis in this subthread). He has a 1/3 chance of showing the car. If he doesn’t, the car is behind any of the other doors with probability 1/2.
• spydum 6y ago
  This breaks most people's brains because there WILL be games where the host can't show a goat in the third door. That has to be accounted for.
  • kgwgk 6y ago
    If he picks at random (the hypothesis in this subthread) there will be (one third of) games where he doesn’t show a goat.