Little math problems October 2019

[Thomas Okken]

I don't know how to prove it rigorously, but it seems to me like you can't do better than four dice rolls. Because of the way the probabilities are distributed, whatever scheme you come up with must have a multiple of 16 possible outcomes, and since a die roll has 6 = 2 * 3 possible outcomes, getting the required factor of 2^4 requires 4 dice rolls.

That is, unless you allow re-rolls; then you could do something like this: roll two dice and add the numbers, and then say 3 -> 0; 2, 4, 5 -> 1; 6, 7, 12 -> 2; 8, 10 -> 3; 11 -> 4; 9 -> try again. That means you end up needing, on average, 2 1/4 dice rolls, but of course there is no upper limit to the number of rolls you might end up needing on any given turn.