I bought a Dice Lab d120 http://www.newyorker.com/tech/elements/the-dice-you-never-knew-you-needed and asked young Rohan how many rolls, on the average would I need to simulate a roll of two ordinary dice. I expected the answer 1.1: Divide the range into 36+36+36+12, and if you land in the 12, roll again for a trit. But Rohan responded with a rapid-fire sequence of constructions that converged toward an answer of .748+ Indeed, log(36)/log(120) can be approached by on-line radix-converting a real fraction base 120 to base 36. But lg 36 ~5.17 bits is the information assuming the dice are distinguishable. It seems to me that for indistinguishable dice, the ensemble is 1+1,2+2,...,6+6, 1+2|2+1, ... , 5+6|6+5 , for which I get 2 lg 3 + 7/6 ~ 4.3366 bits. For six rolls, this gives lg(3^12*2^7 = 68024448) bits. Given a "digit" base 68024448, can anyone think of a way to convert it to six rolls of a pair of indistinguishable dice? Or am I completely confused? --rwg