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We are given a total of 3,000,000 people buy lottery tickets for this game and each ticket costs $20. The layout for the prizes is as follows:

Prize | # Winners

20 | 522,000

40 | 261,000

50 | 195,000

500 | 4,000

10,000 | 300

1,000,000 | 3

  1. Calculate the the pmf of p = amount of prize for random ticket
  2. If I purchase one ticket what is my expected gain?
  3. If I pay an extra $10 my prize is multiplied by 2. What is my expected gain if buy a ticket and pay an extra 10?

Solution: This is what I have so far

For the pmf I have p(20) = 522,000/3,000,000 = .174 p(40) = 261,000/3,000,000 = .087 p(50) = .065 p(500) = .00133 p(10,000) = .0001 p(1,000,000)=.000001

To calculate expected gain I know i have to add up the probabilities of winning each of the prizes less the cost. So the expected gain of the $20 prize would be 0(.174) because the cost of the ticket is 20 and so on... thus the expected gain would be:

= 0(.174)+20(.087)+30(.065)+480(.00133)+9980(.0001)+999980(.000001) Is that correct?

The question continues to ask:

Suppose that I continue to buy tickets until I win a prize (of any amount). After I win a prize, I will not buy any more tickets. Let T be the number of tickets that I will buy.

  1. Find the pmf of T if the tickets are not independent of each other
  2. Under the assumption of independence, what is the distribution of T? Write the pmf. How many tickets would I expect to buy?

Any suggestions for these last two questions?

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    "Question continues to ask:..." should begin another post, with a link back to this one. – 2017-02-22

1 Answers 1

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Mostly.   You forgot to include "not winning any prize", which still costs the ticket price so has a very probable negative gain.

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    how to I include this loss in the equation? – 2017-02-22
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    $+(-20)(\tfrac{982,303}{3,000,000})$ or simply don't subtract the ticket price until after weighting all the prizes. – 2017-02-22
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    It's easier to calculate $\mathbb{E}[\text{Gain}]=\mathbb{E}[\text{Winnings}]-20$, first because you won't forget your 0, then because knowing $\mathbb{E}[\text{Winnings}]$ independent of price is always useful. – 2017-02-22
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    Precisely so. $\ddot\smile$ – 2017-02-22
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    so it would be E= 20(.174) + 40(.087) + 50 (.065) + 500(.00133)+ 10000(.0001) + 1,000,000(.000001) - 20 – 2017-02-22
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    Yes. Though I would also advise against doing the division until after the sum to reduce rounding errots. $$\frac{20(522000)+\overset{\text{etc.}}{\ldots}+1000000(3)}{3000000}-20$$ – 2017-02-22
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    is it possible to get a negative number? – 2017-02-22
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    Possible? It's how lotteries make a profit. – 2017-02-22