##### 2021-05-21   |   by CusiGO

In last week’s four probability problems, the first one has attracted a lot of comments and different solutions, because to solve this problem, you have to start from some meaningful assumptions. It is taken from Jos é P é rez villaplana’s book the problem of probability computation (paraninfo, 1965). According to the author, the solution is 8 / 13, which leads to a meta problem, that is, finding part of the hypothesis.

The second problem can and must be solved without calculation, because as long as we realize that black and white balls are interchangeable, the required probability is 1 / 2 considering the symmetry of the situation.

The third question also comes from Perez villapana’s book (highly recommended, by the way). Its solution is that, considering the separability standard of 11, 59 must be divided into two parts, and the difference is a multiple of 0, 11 or 11; Therefore, the seven digit number must make the odd four digit number add 35 and the even three digit number add 24. The only possibility of adding 35 to four digit number is three 9s and one 8, and the only possibility of adding 24 to three digit number is three 8s; 9, 8 and 7; Or two nines and a six. By looking at different combinations of three ternary and quaternions and putting them together, you get a probability of 4 / 11.

The fourth, though not obvious, is related to Monty Hall’s dilemma, which we have dealt with more than once, which in turn is a variant of the Bertrand box paradox (I wrote an article a few years ago). It is worth noting that the coin is more likely (specifically, twice as many on both sides) to be issued face up on the first issue. In other words, the probability of a positive coin is 2 / 3. In this case, it will definitely appear positive in the second toss, and the probability of a positive coin is 1 / 3. In this case, the probability of a positive coin appearing in the second toss is 1 / 2, so the required probability is 2 / 3 x 1 + 1 / 3 x 1 / 2 = 5 / 6.

Let’s look at another pair of contradictory probabilities that often lead to false estimates:

Many people think that in a family with four children, the most likely gender distribution is half and half, that is, every two children; However, it’s easy to see that there are likely to be three offspring of different genders. What is the most likely gender distribution for five children? If it’s six? Is there any rule as the number of children increases?

On the bridge, things get complicated because there are no two possibilities (gender), only four possibilities (stick). For bridge players, the most unlikely distribution is that 13 cards come from the same stick (1 out of 1587533899); But what is the most likely distribution bar? Players usually think it’s 4-3-3-3, but they’re wrong. What is it and what are its possibilities?

Carlo frebetty is a writer and mathematician at the New York Academy of Sciences. He has published more than 50 popular science books for adults, children and teenagers, including damned physics, damned mathematics or great games. He’s a screenwriter for crystal ball.