# Expected win of a carnival sharpshooter game

A carnival sharpshooter game charges $\$25$for$25$shots at a target. • If the shooter hits the bullseye fewer than$5$times he gets no prize. • If the shooter hits the bullseye$5$times he gets back$\$10$.

• For each additional bullseye over $5$ he gets back an additional $\$5$. The shooter estimates that he has a$0.2$probability of hitting the bullseye on any given shot. What is the shooter's expected gain if he plays the game? I know I can calculate this by brute force, but is there a faster way to solve this than multiplying 0-25 by their respective rewards? ## 1 Answer I'd begin with just awarding$\$5$ per bullseye, less $\$15$; this yields total expected winnings of$25 \times 0.2 \times 5 - 15 = \$10$.

However, this miscounts those situations where he hits fewer than $5$ bullseyes. The discrepancy is

$$\sum_{k=0}^4 \binom{25}{k}5(k-3)(0.2)^k(0.8)^{25-k}$$

which evaluates to about $\$0.29$. Subtract that from the$10$to get$\$9.71$ actual expected earnings.

Of course, that's also not accounting for the $\$25\$ the shooter pays to play the game in the first place...

ETA: The exact expected winnings is apparently

$$\frac{578973231197800522}{59604644775390625}$$

I suspect that's not pleasant to arrive at with pencil and paper.