Two players alternate throwing a six-sided die. The first player who fails to roll a higher number than the preceding roll loses. What is the probability that the first player wins?
What if the die is n-sided?
If you are using an n-sided die, and the last roll was not a
losing roll, and there are k numbers available that are
higher than the last number rolled, then the person who is not about to
roll has a probability of winning of p(k) = ((n-1)/n)^k and other
person has a probability of winning of 1-((n-1)/n)^k.
In particular, at the start of the game, the second player has
probability ((n-1)/n)^n of winning. This was Fredericko's theory.
I unfortunately was no longer logged in when I tried and failed to post
a proof of this, lost the whole thing, and I don't feel like retyping
it. I did a proof by induction on k.
General Result
