Question

Dont worry about the amount of words, it's a short question.

A prisoner has to play a variation of the Monty Hall game with the jailer every day, not knowing which of the three doors the car is hidden behind. After the jailer's first choice, the prisoner therefore chooses one of the two remaining doors at random and opens it. In the event that he accidentally opens the door with the car, the jailer wins. If the jailer loses, the game must be played again the next day, with the car again hiding behind a random door. The prisoner may leave the cell as soon as the jailer has won the car.

Assume that the jailer plays with the "do not switch doors" strategy. What is the chance that the prisoner will be released after ten days? And how big is the chance that he will have to spend at least 10 days in jail?

Answer 1

Assume that you always start by picking door A, and the host then always shows you some other door which does not contain car

If the car is inside door A, then after you pick door A, the host will open another door (either B or C), and you will not switch to the remaining door (either B or C), thus WINNING.

If the car is inside door B, then after you pick door A, the host will be forced to open door C, and you will not switch to door B, thus LOOSING.

If the car is behind door C, then after you pick door A, the host will be forced to open door B, and you will NOT switch to box C, thus LOOSING.

Hence, in 1 of the 3 (equally-likely) possibilities, you will win. the probability of winning WITHOUT switching is 1/3.

in order to find the required probability we use geometric distribution

k number of failures until the first success:

for k = 0, 1, 2, 3, ....

here p=1/3

chance that the prisoner will be released after ten days =p(x>=10){prisoner will be opening the door which has the car inside it in 11th attempt or 12th attempt or 13th attempt and so on}

chance that he will have to spend at least 10 days in jail means chance jailer has been failing at least 10 days.

nothing but