### Non-Transitivity

The table represents the cycle of dice which we considered: in the column, which corresponds to each one of the considered dice, are indicated the number of spots in the dice and over that column is given the probability of that dice beating the dice in the next column (the column next the last one is the first one).

In the situation mentioned above, if the first player had chosen dice \(A\), it would be enough for the second player to choose dice \(D\) in order to have a probability of 2 to 1 to win. But if the first player had chosen any other dice, the second player would have a probability of 2 to 1 to win by choosing the dice before that one chosen by the first player. This means that the second player to choose has an advantage because he can always choose a better dice than the first player (courtesy notion is reversed in this context: being polite here is "serve yourself first" and not "let the opponent serve himself"...).

Conclusion: the original question is poorly worded and induces a wrong idea; before asking which dice is the best, it would be wise to be sure that there is always a dice that is better than the others (at least in a broad sense). As shown by the given example, this is not always the case.

In the exhibition "Matemática
Viva" there was a table (see the photo in figure
1) with four dice like the previous considered dice \(A\), \(B\), \(C\) and \(D\). Four
players around the table took one of the dice in such a way that
the neighbour to the right of the player with dice \(A\) is the player with dice \(B\), the player
to the right of the player with dice \(B\) is the player with dice \(C\), and so on. Starting with
the player with dice \(A\), each played a reasonable number of times with the
neighbour on his right and took note of the game result. With high probability^{2},
each player won with the neighbour on his right, which ended up by "suggesting"
that no dice was better than all the others^{3}.

**Assume that, for each pair of players, the dice are thrown not once but an odd number of times \(n\). If we consider that a player wins when he obtains a larger number than his opponent more times, then the probability of \(A\) beating \(B\) increases in relation to the case where the dice are thrown only once (we are assuming that dice \(A\) is better than \(B\)). That is why in the exhibition "Matemática Viva" visitors were advised to throw the dice several times. The reader may verify that this probability is given by the following formula:**

^{2}\(\sum_{=k}^{n}\left(_{j}^{n}\right)p^{j}q^{n-j}\), in which \(n=2k-1\) and \(\left(_{j}^{n}\right)=\frac{n!}{(j!(n-j)!)}\).

The first twenty values (for \(n = 1,3,...,39\) of this probability are (approximately): \(0.667\), \(0.741\), \(0.790\), \(0.827\), \(0.855\), \(0.878\), \(0.896\), \(0.912\), \(0.925\), \(0.935\), \(0.944\), \(0.952\), \(0.958\), \(0.964\), \(0.969\), \(0.973\), \(0.977\), \(0.980\), \(0.982\), \(0.984\)).

**By curiosity, note that if player (with dice) \(B\) played against player (with dice) \(D\), the probability of \(B\) winning would be \(\frac{1}{2}\), and if player \(A\) played against \(C\), the probability of \(A\) winning would be \(\frac{4}{9}\).**

^{3}