# Expected value of prime lottery ticket

Below is a problem I think that I have solved correctly, but cannot seem to get the correct answer. Any help would be greatly appreciated.

You pay $\$13.00$for a ticket. When you buy a ticket, seven numbers are drawn at random without replacement from the set$\{1,2,3,4,\dots, 46,47\}$. If$x$is the number of prime numbers selected, then you win nothing if$x = 0$,$x = 1$,$x = 4$, or$x= 6$. You get back$\$1.00$ if $x = 2$, $\$10$if$x = 3$,$\$100$ if $x = 5$, and $\$10,000$if$x = 7$. Your expected loss from a single ticket is: My proposed solution: I can only win a sum of money when I have$2$,$3$,$5$,$7$primes in my list. Thus, the chances of getting these would be: $$P(2\text{ primes}) = \frac{{33 \choose 5}{14 \choose 2}}{{47 \choose 7}} = 0.34341$$ $$P(3\text{ primes}) = \frac{{33 \choose 4}{14 \choose 3}}{{47 \choose 7}} = 0.23683$$ $$P(5\text{ primes}) = \frac{{33 \choose 2}{14 \choose 5}}{{47 \choose 7}} = 0.01680$$ $$P(7\text{ primes}) = \frac{{33 \choose 0}{14 \choose 7}}{{47 \choose 7}} = 0.00005457$$ With$2$primes, we win$\$1$ leading to a profit of $\$1\times0.34341 = \$0.34341$

With $3$ primes, we win $\$10$leading to a profit of$\$10\times0.23683 = \$2.3683$With$5$primes, we win$\$100$ leading to a profit of $\$100\times0.01680 = \$1.68076$

With $7$ primes, we win $\$10000$leading to a profit of$\$10000\times0.00005457 = \$0.5457$Thus our expected value to win should be$-\$13$ (the cost of the ticket) $+$ the above amounts, which yields $-\$8.06$. I'm sure there is some sort of error in my reasoning, but I cannot figure out where it is. • Your expected loss is$8.06 Jan 22, 2015 at 22:10
• There are $15$ primes in the set. They are $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$$ Thus, you should be making selections from $32$ non-primes and $15$ primes. Jan 22, 2015 at 22:36
• @N.F.Taussig You are completely right! Somehow stupid me skipped the prime $7$. I think you should post this as an answer. Jan 22, 2015 at 22:43