**Spanish version:**

**Del 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.

**asks**

**Juan Martín del Potro****, team**

**Martín Jaite****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

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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