# If I purchase one ticket for a lottery game, what is my expected gain?

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 $20 prize would be 0(.174) because the cost of the ticket is 20 and so on... thus the expected gain would be: = 0(.174)+20(.087)+30(.065)+480(.00133)+9980(.0001)+999980(.000001) Is that correct? The question continues to ask: Suppose that I continue to buy tickets until I win a prize (of any amount). After I win a prize, I will not buy any more tickets. Let T be the number of tickets that I will buy. 1. Find the pmf of T if the tickets are not independent of each other 2. Under the assumption of independence, what is the distribution of T? Write the pmf. How many tickets would I expect to buy? Any suggestions for these last two questions? • "Question continues to ask:..." should begin another post, with a link back to this one. Feb 22 '17 at 2:05 ## 1 Answer Mostly. You forgot to include "not winning any prize", which still costs the ticket price so has a very probable negative gain. • how to I include this loss in the equation? Feb 22 '17 at 0:53 •$+(-20)(\tfrac{982,303}{3,000,000})$or simply don't subtract the ticket price until after weighting all the prizes. Feb 22 '17 at 0:55 • It's easier to calculate$\mathbb{E}[\text{Gain}]=\mathbb{E}[\text{Winnings}]-20$, first because you won't forget your 0, then because knowing$\mathbb{E}[\text{Winnings}]$independent of price is always useful. Feb 22 '17 at 1:00 • Precisely so.$\ddot\smile\$ Feb 22 '17 at 1:03
• so it would be E= 20(.174) + 40(.087) + 50 (.065) + 500(.00133)+ 10000(.0001) + 1,000,000(.000001) - 20 Feb 22 '17 at 1:19