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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: W

by 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?

the_af 6y ago
Player #1 had a 1/3 probability of picking the right door. Therefore, he has a 2/3 probability of winning if he switches. Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches. In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).
- kgwgk 6y ago
  So you think that at that point (after #1 selected door A, #2 selected door B, and a goat walked out of door C): The probability of player #1 getting the car if he keeps his door, which is equal to the probability that the car is behind door A, is 1/3 The probability of player #1 getting the car if he switches, which is equal to the probability that the car is behind door B, is 2/3 The probability of player #2 if he keeps his door, which is equal to the probability that the car is behind door B, is 1/3 The probability of player #2 getting the car if he switches, which is equal to the probability that the car is behind door A, is 2/3 In summary, the car is: Behind door C with probability 0 Behind door A with probability 1/3 Behind door B with probability 2/3 Behind door B with probability 1/3 Behind door A with probability 2/3
  - the_af 6y ago
    No. Because the experiments are completely independent (i.e. player #1 & player #2 don't interact at all; their choices don't interfere), the events are effectively "choosing door A and switching to door B" and "choosing door A' and switching to door B'". You have doors A, B, A' & B'. This of course assumes all doors are revealed simultaenously; i.e. no player gets to see the result of the other's choice beforehand. You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent. But don't trust me: try it!
    - kgwgk 6y ago
      Why don’t you try to give an answer to the following questions? After player #1 has picked door A, player #2 has picked door B and door C has been opened revealing a goat: a) if player #1 stays with door A his probability of winning is ___ b) if player #1 switches to door B his probability of winning is ___ c) if player #2 stays with door B his probability of winning is ___ d) if player #2 switches to door A his probability of winning is ___ e) the probabilty that the car is behind door A is ___ f) the probabilty that the car is behind door B is ___ g) the probabilty that the car is behind door C is ___
      - the_af 6y ago
        These are two independent games, so for every player: a) The probability of winning if staying: 1/3 b) The probability of winning if switching: 2/3 c) The probability of the car being behind door the player chose is the same as a): 1/3 d) The probability of the car being behind the door Monty didn't open is 2/3 e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat! Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.
        kgwgk 6y ago
        There is one single game. It's my version and I make the rules! Rules2: "One player picks a door, the second picks a door (maybe the same), the host opens a door that has not been picked (if there are two he picks at random). Then the players can either keep their choice or switch to the other closed door." Scenario2: "Player #1 picked one door, player #2 picked another door, the third door was opened and there was a goat behind it." In that scenario, what are the probabilities? I agree it's different from the original problem, that's the point! Rules0: "The player picks a door, the host opens a door that has not been picked with full knowledge that there will be a goat behind it. Then the player can either keep his choice or switch to the other closed door." Scenario0: "The player picked one door, the host opened another door and there was a goat behind it." But the game I proposed is equivalent to the version of the problem that some people, including you apparently [0], insist in this thread that is equivalent to the original problem. Rules1: "The player picks a door, the host opens another door selected at random. Then the player can either keep his choice or switch to the other closed door." Scenario1: "The player picked one door, the host opened another door and there was a goat behind it." The latter is not equivalent to the original problem. The situation is similar but the rules are different. The solution is different. However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors. They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat. [0] The problem I proposed at https://news.ycombinator.com/item?id=24707305 https://news.ycombinator.com/item?id=24707305 is also equivalent: Rules1': "The player picks a door, and opens another door selected at random. Then he can either keep his choice or switch to the other closed door." Scenario1': "The player picked one door, then opened another door and there was a goat behind it."

by kgwgk 6y ago

the_af 6y ago

Player #1 had a 1/3 probability of picking the right door. Therefore, he has a 2/3 probability of winning if he switches. Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches. In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).

kgwgk 6y ago
So you think that at that point (after #1 selected door A, #2 selected door B, and a goat walked out of door C): The probability of player #1 getting the car if he keeps his door, which is equal to the probability that the car is behind door A, is 1/3 The probability of player #1 getting the car if he switches, which is equal to the probability that the car is behind door B, is 2/3 The probability of player #2 if he keeps his door, which is equal to the probability that the car is behind door B, is 1/3 The probability of player #2 getting the car if he switches, which is equal to the probability that the car is behind door A, is 2/3 In summary, the car is: Behind door C with probability 0 Behind door A with probability 1/3 Behind door B with probability 2/3 Behind door B with probability 1/3 Behind door A with probability 2/3
- the_af 6y ago
  No. Because the experiments are completely independent (i.e. player #1 & player #2 don't interact at all; their choices don't interfere), the events are effectively "choosing door A and switching to door B" and "choosing door A' and switching to door B'". You have doors A, B, A' & B'. This of course assumes all doors are revealed simultaenously; i.e. no player gets to see the result of the other's choice beforehand. You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent. But don't trust me: try it!
  - kgwgk 6y ago
    Why don’t you try to give an answer to the following questions? After player #1 has picked door A, player #2 has picked door B and door C has been opened revealing a goat: a) if player #1 stays with door A his probability of winning is ___ b) if player #1 switches to door B his probability of winning is ___ c) if player #2 stays with door B his probability of winning is ___ d) if player #2 switches to door A his probability of winning is ___ e) the probabilty that the car is behind door A is ___ f) the probabilty that the car is behind door B is ___ g) the probabilty that the car is behind door C is ___
    - the_af 6y ago
      These are two independent games, so for every player: a) The probability of winning if staying: 1/3 b) The probability of winning if switching: 2/3 c) The probability of the car being behind door the player chose is the same as a): 1/3 d) The probability of the car being behind the door Monty didn't open is 2/3 e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat! Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.
      - kgwgk 6y ago
        There is one single game. It's my version and I make the rules! Rules2: "One player picks a door, the second picks a door (maybe the same), the host opens a door that has not been picked (if there are two he picks at random). Then the players can either keep their choice or switch to the other closed door." Scenario2: "Player #1 picked one door, player #2 picked another door, the third door was opened and there was a goat behind it." In that scenario, what are the probabilities? I agree it's different from the original problem, that's the point! Rules0: "The player picks a door, the host opens a door that has not been picked with full knowledge that there will be a goat behind it. Then the player can either keep his choice or switch to the other closed door." Scenario0: "The player picked one door, the host opened another door and there was a goat behind it." But the game I proposed is equivalent to the version of the problem that some people, including you apparently [0], insist in this thread that is equivalent to the original problem. Rules1: "The player picks a door, the host opens another door selected at random. Then the player can either keep his choice or switch to the other closed door." Scenario1: "The player picked one door, the host opened another door and there was a goat behind it." The latter is not equivalent to the original problem. The situation is similar but the rules are different. The solution is different. However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors. They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat. [0] The problem I proposed at https://news.ycombinator.com/item?id=24707305 https://news.ycombinator.com/item?id=24707305 is also equivalent: Rules1': "The player picks a door, and opens another door selected at random. Then he can either keep his choice or switch to the other closed door." Scenario1': "The player picked one door, then opened another door and there was a goat behind it."