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?
 A: Expected payout is
$$1,000,000 \frac{{6\choose 6}{34\choose 0}}{40\choose 6} + 1000 \frac{{6\choose 5}{34\choose 1}}{40\choose 6}+100 \frac{{6\choose 4}{34\choose 2}}{40\choose 6},$$
which if your answer is correct should equal 0.533.
A: 
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: $$\mathrm{E}(R(X)) = \dfrac{{6\choose 6}{34\choose 0}}{40\choose 6} \times\$10^6 + \dfrac{{6\choose 5}{34\choose 1}}{40\choose 6} \times\$1000 + \dfrac{{6\choose 4}{34\choose 2}}{40\choose 6}\times\$100 - \$1 \\ = -\frac{89644}{191919} \\ \approx -0.4670928881455\ldots$$
