There are three closed doors. One has a new car, one has the keys to the car and one has a goat. These prizes are randomly assigned.
There are two players: the first player has to find the car, the second player has to find the keys to the car. If both players succeed they win the car.
The first player enters the room and may open any two of the three doors, one after the other. If successful, the doors are closed again and the second player enters the room. The second player may also open two of the three doors, but cannot communicate with the first player.
Surprisingly there is a strategy where the probability of winning is better than (2/3)2. Find it.
The big problem arising from the lack of communication between the two players is that the second player can't be sure that he is not choosing as one of his doors the same door behind which the first player found the car (if he didn't find the car, then it doesn't matter anyway).
So the idea is to minimize the risk that the second player will make one of his door choices the same as the car found by player one. Well, since there's no restriction on planning a strategy ahead of time, have player 1 choose doors 1 and 2 while player 2 will choose doors 2 and 3. At least in the 50% case that the car was in door 1, this in fact guarantees finding the keys.
In order for this strategy to fail, either the car is behind door 3 or the keys are behind door 1. What's the probability of a win, where neither of these takes place? The probability the car is not behind door 3 is 2/3; that is the probability it's behind either door 1 or door 2, each with a 50% probability under the condition given. If it was door 1, the pair is sure to win as the keys are behind 2 or 3. If the car was behind door 2, they still have a 50% chance of finding the keys.
So with this strategy, the probability of winning the car is (2/3)*(1/2) + (2/3)*(1/2)*(1/2) = 1/3 + 1/6 = 1/2, rather than the (2/3)^2 = 4/9.
Posted by Charlie
on 2018-09-30 09:34:57