**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.
**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