# Contrasted Probabilities | Science game

Of the four probabilistic problems that occurred last week, the former generated a large number of comments and various solutions, since to solve it it is necessary to start from some opinionable assumptions. It is taken from Jose Perez Villaplana’s book *Probability calculation problems* (Paraninfo, 1965), and according to the author the solution is 8/13, giving rise to the basic problem of detecting any hypothetical part.

The second can – and should – be solved without calculations, because it suffices to realize that the black and white balls are interchangeable, so that, given the symmetry of the situation, the required probability is 1/2.

The third problem is also taken from Perez Villaplana’s book (highly recommended, by the way). It is solved by seeing that, taking into account the criterion of divisibility by 11, 59 must be divided into two parts with a difference of 0 or 11 or a multiple of 11. The only possibility is 59 = 35 + 24; Therefore, the seven digits of the number must be this way so that the sum of the four numbers in an odd place reaches 35, while the sum of the three numbers in an even place reaches 24. They are three 9s and one 8, and for three numbers to add up to 24 it should be : Three 8s; A 9, 8 and 7; Or 9 and 9 one 6. Looking at the different combinations of triples and three and quartets and adding them together, the probability is 4/11.

The fourth problem is related, although not clear, to the Monty Hall dilemma, which we have dealt with on more than one occasion, which in turn is a form of the Bertrand Fund paradox (to which I devoted an article for two years). Keep in mind that if the first lot appears two-sided, it is more likely (double specifically) that it will be a two-sided coin. That is, the probability that the coin is face to face is 2/3, and in this case it is certain that on the second toss it will appear face to face as well, and the probability that the coin will be face to face is 1/3. / 2, so the desired probability is 2/3 x 1 + 1/3 x 1/2 = 5/6.

Let’s look at another pair of contrasting possibilities that often lead to false estimates:

### Babies and sticks

Many people believe that in a family with four children, the most likely distribution by gender is one and a half, that is, two of each; However, it is easy to see that there are likely three offspring of one gender and the other. And if there are five children, what is the most likely gender distribution? And in the case of the six? Is there a pattern as the number of children increases?

Things get complicated at the bridge, because there are not two (genders) but four (allowances). Obviously, the most likely distribution, for the bridge hand, is that all 13 cards are of the same type (probability is 1 in 158,753,389,899); But what is the most likely distribution for the club? Often players think it is 4-3-3-3, but they are wrong. What is it actually, and what is its likelihood?

**Carlo Frappetti ***Writer, mathematician and member of the New York Academy of Sciences. He has published more than 50 popular scholarly works for adults, children and teens, including “Damn Physics,” “Damn math,” or “The Great Game”. He was a screenwriter for “La bola de cristal”.*

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