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