I must says other interesting thing with the mechanic of *sum-and-reroll* design-oriented. You can calculates the probability for *j* rerolls happen and create something like a "average probability of rerolls" at P(j) probability:

Where P(j) is the probability for an exactly number *j* of rerolls happen. By example: with a probability of reroll of 70% we want to know what is the probability to 10 rerolls, that is P(j)=(0.7^10)*0.3=0.84% !!!

Inversely the number of *j* for a P(j)=0.5 (average of rerolls per throw) is (ln0.5-ln0.3)/ln0.7=**-.43**!!!, is negative because NEVER CAN HAPPEN cause is bigger that the probability of one reroll that is 0.7*0.3.

This happen because we are calculating *exact* number of *j* in a mechanic with a high probability for reroll (70%). If we calculate the same for the probability of *at least* a *j* amount of rerolls the numbers grows.

Following this if you have a P(≥j)=50% when you p=70% then you have a probability of 50% for *at least* 2 rerolls per throw (j=ln0.5/ln0.7).

This maybe useful to predict the "annoyance level" of a game.

P.S.: still I dont know how to solve the problem of 1's, seems I fail in something related to dependency of events.

Sorry Masacroso, but I can only see the initial explanation. I tried both in Firefox and Chrome, and also as java and html5, but it is not working.