Question 1178261
Let X = r.v. equal to the value of the die on the 1st throw.

Then for X = 1, 2, 3, 4, 5, 6, it can be shown combinatorially that {{{P(X = x) = 1/6^2 + 5/6^3 }}}+...+{{{5^(x-1)/6^(x+1) = (1/6)*(1-(5/6)^x)}}}.  

The probability of a '3' turning up on any throw is 1/6. Hence the expectation for the number of wins in this game is 


{{{E(X) = 1*(1/6^2) + 2*(1/6)*(1-(5/6)^2) + 3*(1/6)*(1-(5/6)^3) + 4* (1/6)*(1-(5/6)^4 ) + 5* (1/6)*(1-(5/6)^5) + 6* (1/6)*(1-(5/6)^6) = 1.85}}} to 2 d.p.