# Lottery expected value probability question

The question is as follows:

Let's say that you play the lottery (when you are old enough). Six numbers without repetition are chosen from 1-40, If you pick all six numbers, you win \$1 million. If you pick five of the six, you win \$1000. If you pick four of the six, you win \$100. What is the expected value of a \$1 lottery ticket? Note: the way you play this lottery game if by receiving a card with an empty circle under each number from 1-40. You will fill in the circle underneath each of the six numbers you choose.

I have tried all sorts of work but they all result to either a ridiculously small answer or a ridiculously large answer, and the correct answer is -0.467. Can someone help me out here?

Expected payout is

$$1,000,000 \frac{{6\choose 6}{34\choose 0}}{40\choose 6} + 1000 \frac{{6\choose 5}{34\choose 1}}{40\choose 6}+100 \frac{{6\choose 4}{34\choose 2}}{40\choose 6},$$

• @Askask, because that is the expected winnings, from which you have to deduct the cost of the ticket to get the expected return: $0.5333-1 = - 0.467$. – Graham Kemp May 20 '14 at 6:11
Let's say that you play the lottery (when you are old enough). Six numbers without repetition are chosen from 1-40, If you pick all six numbers, you win \$1 million. If you pick five of the six, you win \$1000. If you pick four of the six, you win \$100. Expected value of return is the sum of probabilities times their values, minus the outlay (cost of ticket). $$\mathrm{E}(R(X)) = \sum\limits_{X=0}^6 R(x) \mathrm{P}(X=x) \\ = \mathrm{P}(X=6) \times\10^6 + \mathrm{P}(X=5) \times\1000 + \mathrm{P}(X=4)\times\100 - \1$$ Note: the actual returns are \$999999, \$999, and$99 on the three winning conditions, and -\$1 on each of the loosing conditions. Or simply subtract the whole cost from the expectation of prizes. Now, the probability of choosing$x$winners is calculated by: Count the ways to choose$x$of the$6$winning numbers, and choose any$6-x$other numbers from the remaining$34$'losing' numbers, then divide the product by the ways to choose any$6$of$40$numbers. $$P(X=x) =\dfrac{{6\choose x}{34\choose 6-x}}{40\choose 6}$$ So: $$\mathrm{E}(R(X)) = \dfrac{{6\choose 6}{34\choose 0}}{40\choose 6} \times\10^6 + \dfrac{{6\choose 5}{34\choose 1}}{40\choose 6} \times\1000 + \dfrac{{6\choose 4}{34\choose 2}}{40\choose 6}\times\100 - \1 \\ = -\frac{89644}{191919} \\ \approx -0.4670928881455\ldots$$ • can you explain how you got that formula at the bottom? – ASKASK May 20 '14 at 4:28 • @Askask Count the ways to pick$x$"good" numbers and$6-x$"bad" numbers. Then divide by the ways to select any$6\$ numbers. – Graham Kemp May 20 '14 at 4:35