4 ms·

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/

by 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?
  - 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!

by kgwgk 6y ago

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?
- 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!