beats B beats C beats D beats E beats A (first chain); A beats C beats E beats B beats D beats A (second chain). However, when the game is played with two such sets, then the first chain remains the same (with one exception discussed later) but the second chain is reversed (i.e. A beats D beats B beats E beats C beats A). Consequently, whatever dice the two opponents choose, the third player can always find one of the remaining dice that beats them both (as long as the player is then allowed to choose between the one-die option and the two-die option): {| class="wikitable" |- align="center" ! colspan="2" rowspan="2" | Sets chosenby opponents ! colspan="2" | Winning set of dice |- align="center" ! Type ! Number |- align="center" | A || B || E || 1 |- align="center" | A || C || E || 2 |-align="center" | A || D || C || 2 |- align="center" | A || E || D || 1 |- align="center" | B || C || A || 1 |- align="center" | B || D || A || 2 |- align="center" | B || E || D || 2 |- align="center" | C || D || B || 1 |- align="center" | C || E || B || 2 |- align="center" | D || E || C || 1 |} There are two major issues with this set, however. The first one is that in the two-die option of the game, the first chain should stay exactly the same in order to make the game intransitive. In practice, though, D actually beats C. The second problem is that the third player would have to be allowed to choose between the one-die option and the two-die option – which may be seen as unfair to other players. Corrected Grime dice The above issue of D defeating C arises because the dice have 6 faces rather than 5. By replacing the lowest (or highest) face of each die with "reroll" (R), all five dice will function exactly as Dr. James Grime intended: A: R, 2, 2, 7, 7, 7 B: R, 1, 6, 6, 6, 6 C: R, 5, 5, 5, 5, 5 D: R, 4, 4, 4, 4, 9 E: R, 3, 3, 3, 8, 8 Alternatively, these faces could be mapped to a set of pentagonal-trapezohedral (10-sided) dice, with each number