## Courtesy at stake

### 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).

Fig 2

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.

Fig 3