December 9, 2012

Del Potro`s choice. The Monty Hall problem.

Spanish versionDel Potro y las tres pelotas. El problema de Monty Hall

  Davis Cup semifinals between Argentina and the United States of America.

They're going to play the fifth and intranscedental match of the final, which was resolved in the previous game.

Juan Martín del Potro asks Martín Jaite, team captain, to play this last game, even though Jaite has already decided that Eduardo Schwank will play the match.

Martín Jaite is not convinced that Del Potro plays, but at the insistence of the player, he decides to propose him the following riddle: You will play the match if you discover at a glance, without touching them, which of the 3 balls bounces badly.

As soon as Del Potro says that de defective ball is on the left, Martín Jaite tells him that the defective ball is not in the center, and gives him the opportunity of changing the choice of the ball. But Del Potro decides to keep his initial decision, because he thinks Martín Jaite doesn't want him to play today.

Is Del Potro succeeding by maintaining his first choice? Will he play the match?

Although, apparently, after discarding the ball on the center, the ball on the left and the ball on the right  are equally likely to be the defective one, Del Potro has really only a probability of 1/3 of playing the match.

Suppose that, from the beginning, the captain had proposed to Del Potro choosing which ball was faulty from a total of 10 balls. The probability of the defective ball of being chosen by Juan Martín would be  1/10. And for the other 9 balls would be of 9/10. So, at first, all of the balls have a 10% chance of being the faulty one: an 1/10 for the initially chosen, and  9/10 for all the other balls:

1/10 for the chosen ball A + ( 1/10 for the ball B + 1/10 for the ball C + 1/10 for the ball D + 1/10 for the ball E + ...)

As we eliminate balls in this group of 9 balls, what we do is change the percentage that corresponds to each of the balls of this probability of 9/10. Thus, when there are only 3 balls, plus the one we have chosen, the probabilities would be:

1/10 for the chosen ball A + ( 3/10 for the ball B + 3/10 for the ball C + 3/10 for the ball D)

And when there are only 2 options we have:

1/10 for the chosen ball A + ( 9/10 for ball B)

Thus, when the captain removes the remaining balls, the ball that rests 'inherits’ the probabilities of all the removed balls in such way that it would have a 9/10 probabilitiy of being defective.

Applying this argue to our problem, there is a 1/3 chance that the faulty ball is on the left, and 2/3 of being one of the other 2 balls (center and right). If we do choose among these 2 options, what would we choose? Would we choose the ball on the left, with a 1/3 de probability, or the set of (center and right), qith a 2/3 chance? Obviously, if we had to choose between these 2 options, we would take the set (center and right).

When the captain says that the center ball is not the faulty, this doesn't change the probabilities among the ball on the left and the set of balls (center and right), but what it makes is changing the probabilities in this last set. (Center and right) still have a 2/3 chance. The ball on the center will have a 0% of these 2/3, and the right one will have the 100% of them, that is, ‘inherits’ the 2/3 of the initial probability of the group.

So it would have been better than Del Potro had changed his mind after the elimination of the central ball, and should have said that the defective ball was that on the right.

This problem is known as the Monty Hall problem. Its name refers to the conductor of the famous American game show 'Let's make a deal'. In this type of competitions, which offer a good reward  from 3 possible choices, it's always better to switch the first choice, once the conductor rules out one of the options, which contained a 'secondary' prize..

From a mathematical point of view, we have: 

A = the event in which the contestant chooses the awarded option at first instance
A'= the contestant initially chooses a failed option
B = the contestant is right to maintain their initial choice 
B'= the contestant succeeds changing their initial choice 
P(A) = probability of succeeding the event A
P(A') = probability of succeding the event A'
P(B/A) = probability of hitting maintaining the initial choice, if you hit at first
P(B'/A') = probability of hitting by changing the initial choice, if you failed it
And so we apply the theorem of total probability, we have: 

P(B) = P(B/A) x P(A) = 1 x 1/3 = 1/3 
P(B') = P(B'/A') x P(A') = 1 x 2/3 = 2/3 

Since, in this case: 

P(A) = 1/3 
P(A') = 1-1/3 = 2/3 
P(B/A) = 1, P(B/A') = 0 
P(B/A') = 0, P (B'/A') = 1 

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