Monty Hall problem
Monty Hall problem

Three Prisoners problem
Three Prisoners problem

Monty Hall problem

The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall.
The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990. (Wikipedia: Monty Hall problem)


Monty Hall problem.


Suppose you're on a game show, and you're given the choice of three doors.
Behind one door is a car, behind the others, goats.
You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat.
He says to you, "Do you want to pick door #2?"
Is it to your advantage to switch your choice of doors?


     

The answer is "yes". The chances of winning are doubled by switching.
The odds go up from the original 1/3 to 2/3 if you switch.

This may be the simplest way to explain what is going on here.

You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat.
The probability that the car is behind door #1 is 1/3.
That means the probability that it is behind other two doors (#2 and #3) is 2/3.
The host now opens one of the doors, #2 or #3, which has a goat. Let's suppose he opens door #3.
Notice that the host knows where the car is located. (This is the crucial piece of information and is the key to solve the problem.)

You now have 3 relevant pieces of information:

1. The probability that the car is behind door #1 is 1/3.
2. The probability that the car is behind door #2 or #3 (i.e., not behind door #1) is 2/3.
3. The car is not behind door #3.

With these 3 pieces of information you can conclude that the probability that the car is behind door #2 is 2/3.
Therefore it is to your advantage to switch from the original choice of door #1 to door #2.
The reason is simple. your original guess is right with probability 1/3, whereas one of the other two choices is correct with probability 2/3.



Three Prisoners problem.


Out of three prisoners scheduled to be executed, A, B, and C, one of them will be pardoned.
A asks the warden to tell him the name of one of the others who will be executed.
As the question is not directly about A's fate, the warden obliges and truthfully says:" B will be executed".
A thinks he is now one of two remaining prisoners, his chances of being pardoned have risen from 1/3 to 1/2. Is he right?
To make the analogy to the Monty Hall problem more explicit:
if the warden says A could switch fates with C, should he?


     

The answer is he didn't gain information about his own fate, but he should switch with C if he can.
Before the warden's revelation, A estimates his own chances of being pardoned are 1/3, the same as both B and C, if the pardon is a "random" choice.
The confirmation of the execution of either one of the other two prisoners does not changes that.
It's irrelevant to the fate of A, whose probability of survival remains 1/3.

However, things are looking brighter for C who now has a probability of 2/3 of being pardoned.

Concerning switching fates, if the warden says "B will be executed" and A wants to live, he's better off switching fates with C.