# The drunken king | Science

Our drunken king last week it can only reach the edge of the board in three moves if three faces come out (if it is the white king, or three crosses if it is Baltasar), then the probability is 1/2 x 1/2 x 1/2 = 1/8.

The probability that the king returns to the starting point in three moves is 0, since this requires an even number of moves, since he has to retrace every step that moves away from his starting space.

As the number of rolls increases, the probability of 50% faces and 50% crosses increases, so the probability that the king returns to his starting square tends to 1.

Curiously, the final question, which seemed the simplest, is the one that has provoked the most debate among readers (see comments from last week). There are three possibilities: Ana wins, Berta wins or draw, and this explains why some give solutions close to 33%; but the three possibilities are not equiprobable. As it is easy to see through a table of possibilities, the equiprobables are the tie and the no-tie, which means that the probability of Ana winning (or that Berta wins) is 25%. This is one of those riddles that are more interesting from the point of view of cognitive psychology than as mathematical problems.

### The two-dimensional king

Like Marcuse's one-dimensional man, our drunken king could only move along a line (the first row of the chessboard). Let's place it now in the center of the board and allow it to move not only right and left but also forward and backward (but not diagonally). If all his steps are random and all possibilities equally likely, what is the probability that the king will reach the edge of the board in three moves? And in four moves or less? And in five moves or less? And the probability that he will return to the starting square in four moves or less? And the probability that, in an unlimited board, it returns to the starting point after endless moves?

As it is easier to have a coin than a tetrahedral die, we can determine the random plays by tossing a coin twice in a row: if two faces come out, the king moves one box to the right; if two crosses come out, one box moves to the left; if it comes out face and then cross, move one square forward; and if a cross comes out and then a face, a box moves backwards.

Obviously, the situation can be extended to three (or more) dimensions. If the drunken king was not in a grid but in a "cubicle" (three-dimensional network of cubic cells) and could also move up and down, what would be his probability of returning to the starting point after infinite plays?

Carlo Frabetti He is a writer and mathematician, a member of the New York Academy of Sciences. He has published more than 50 scientific dissemination works for adults, children and young people, among them Damn physics,Damn mathematics or The big game. He was a screenwriter The Cristal ball