Answer:
In the long run, ou expect to lose $4 per game
Step-by-step explanation:
Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.
Assuming X be the toss on which the first head appears.
then the geometric distribution of X is:
X
geom(p = 1/2)
the probability function P can be computed as:

where
n = 1,2,3 ...
If I agree to pay you $n^2 if heads comes up first on the nth toss.
this implies that , you need to be paid 

![\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7Bn%7D_%7Bi%3D1%7D%20n%5E2%20P%28X%3Dn%29%20%3DVar%20%28X%29%20%2B%20%5BE%28X%29%5D%5E2)
∵ X
geom(p = 1/2)








Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6
= $4
∴
In the long run, you expect to lose $4 per game