You throw a perfect die (digits 1 to 6), adding up the results until the total score is over N.

What’s your estimate of the most likely final sum?

Please write down your guess prior to analytical evaluation or simulation.

This looks much like Die Toss.  Based on that problem, the most likely sum is N+1.  The expected sum for large N is (6*(N+1) + 5*(N+2) + 4*(N+3) + 3*(N+4) + 2*(N+5) + 1*(N+6))/21 = N+8/3.

Posted by Brian Smith on 2016-04-18 10:02:16