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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?

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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.

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