### Which is the best dice?

Assume now that dice \(A\) has two sides without spots and four sides with 4 spots each, which we express by writing \(A = \left\{0, 0, 4, 4, 4, 4\right\}\) and, using a similar notation, suppose that \(B = \left\{3, 3, 3, 3, 3, 3\right\}\). Each time the two dice \(A\) and \(B\) are thrown (\(A\) followed by \(B\)) the game never ends in a tie. There are 36 pairs of equally likely results, 12 of them corresponding to the result \((0, 3)\), when 0 occurs in dice \(A\) and 3 occurs in dice \(B\), and 24 corresponding to the result \((3, 4)\), when 3 occurs in dice \(A\) and also in dice \(B\). Therefore the probability of \(B\) winning is \(\frac{12}{36}=\frac{1}{3}\) and the probability of \(A\) winning is \(\frac{24}{36}=\frac{2}{3}\). If the dice \(B\) and \(C = \left\{2, 2, 2, 2, 6, 6\right\}\) are thrown, they never tie because \(B\) beats \(C\) in 24 of the 36 equally likely results, and \(C\) beats \(B\) in the remaining 12 results. In this case we say that \(B\) is two times better than \(C\). The reader can confirm that \(C\) is two times better than dice \(D = \left\{1, 1, 1, 5, 5, 5\right\}\), (In order to check or avoid calculations, the reader may use an odds calculator).

Taking into account the previous calculations, it seems natural to conjecture that, among the considered dice \(A\), \(B\), \(C\) and \(D\), the dice \(A\) is the best of the four because it is better than \(B\), \(B\) is better than \(C\), and finally, \(C\) is better than \(D\). In particular, if two players decide to play against each other and, before starting the game, they have the opportunity to choose the dice they prefer playing with (between \(A\), \(B\), \(C\) and \(D\)), it appears that the first to choose has an advantage if he chooses dice \(A\).

The term "conjecture" used above instead of "conclude" may seem excessive prudence. But if the reader directly compares dice \(A\) and \(D\), he can verify that \(A\) wins when the result is \((4,1)\), corresponding to 12 equally likely results, and looses when the result is \((0,1)\), \((0,5)\) or \((4,5)\), corresponding to the remaining \(24 (= 6 + 6 + 12)\) equally likely results. This means that dice \(D\), which should be much (? 8 times) worse than \(A\), is in fact two times better than \(A\)... Returning to the conjecture, we see that underlying it is the wrong idea that the relation "better than" between dice is transitive.