Question

X, Y and Z in that order throws a tetrahedral die the first throws a 4 wins. The game continues indefinitely until someone wins. Find the probability that X wins, Y wins and Z wins.

Answer 1

outcomes: the last letter is a winner, Prob of each is found:

X = 1/4 = (1/4)
XY = (3/4)(1/4) = (1/4)3/4)
XYZ =(3/4)(3/4)(1/4) = (1/4)(3/4)^2
XYZX = (3/4)(3/4)(3/4)(1/4) = (1/4)(3/4)^3
XYZXY = ...
XYZXYZ = ..
XYZXYZX = (1/4)(3/4)^6

Prob that X wins
= (1/4) + (1/4)(3/4)^3 + (1/4)(3/4)^6
this is a GP with a = 1/4 , r = (3/4)^3
sum to infinity = a/(1-r)
= (1/4) / (1- 27/64)
= (1/4) / (37/64)
= (1/4)(64/37)
= 16/37

prob that Y wins
.....
follow my steps