I don't think you can and get the desired probability distribution. Consider the case of rolling six consecutive snake-eyes, which is a 1 : 36^6 = 2,176,782,336 occurrence. If you map just one base 68024448 digit to this, it gets a probability of 1 : 68,024,448, which is 32 times too large. And omitting it entirely gives it a probability of 0, which is infinitely too small. On May 7, 2016 21:25, "Bill Gosper" <billgosper@gmail.com> wrote:

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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun