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?

(In reply to Answer by K Sengupta)

Let P(f) = Prob.(First player eventually wins).

Then, after some computations, we observe that for n-sided die:

P(f)     =      1/n + (n-1)^2/n^3 + ......+ (n-1)^(n-1)/n^n .....(i)
P(f)*(n-1)/n =         (n-1)^2/n^3 + ......+ (n-1)^(n-1)/n^n + (n-1)^n/n^(n+1)  ......(ii)

Subtracting (ii) from (i), we have:

P(f) = 1 - (1- 1/n)^n

Thus, for n rolls the required probability is:

1 - (1- 1/n)^n

Substituting n=6, it trivially follows thatfor 6-sided die, the required probability is:

1 - (5/6)^6 ~ 0.6651(correct to 4 places)

For very large n, we observe that:

As n -> infinity, the expression (1-1/n)^n -> 1/e

Consequently, for very large n, the required probability would be:

1 - 1/e ~ .63212 (correct to 5 places)

Edited on December 3, 2008, 12:18 pm

Posted by K Sengupta on 2008-12-03 11:57:58