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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?

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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.

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    $\begingroup$ Can you explain how you came up with this combinations? $\endgroup$ – ASKASK May 20 '14 at 4:29
  • $\begingroup$ and why should it equal 0.5333? $\endgroup$ – ASKASK May 20 '14 at 4:32
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    $\begingroup$ @Askask, because that is the expected winnings, from which you have to deduct the cost of the ticket to get the expected return: $0.5333-1 = - 0.467$. $\endgroup$ – Graham Kemp May 20 '14 at 6:11
  • $\begingroup$ You're picking six numbers out of forty of which six out of six have to match and none out of thirty four do not have to match in order to win the million. $\endgroup$ – JPi May 20 '14 at 10:02
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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: $$\mathrm{E}(R(X)) = \dfrac{{6\choose 6}{34\choose 0}}{40\choose 6} \times\$10^6 + \dfrac{{6\choose 5}{34\choose 1}}{40\choose 6} \times\$1000 + \dfrac{{6\choose 4}{34\choose 2}}{40\choose 6}\times\$100 - \$1 \\ = -\frac{89644}{191919} \\ \approx -0.4670928881455\ldots$$

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  • $\begingroup$ can you explain how you got that formula at the bottom? $\endgroup$ – ASKASK May 20 '14 at 4:28
  • $\begingroup$ @Askask Count the ways to pick $x$ "good" numbers and $6-x$ "bad" numbers. Then divide by the ways to select any $6$ numbers. $\endgroup$ – Graham Kemp May 20 '14 at 4:35
  • $\begingroup$ Why is it 36 losing numbers? $\endgroup$ – ASKASK May 20 '14 at 4:44
  • $\begingroup$ That I'm going to put down to ... lack of sleep? $\endgroup$ – Graham Kemp May 20 '14 at 4:51

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