# Mad billionaire wants to win Lottery 7/39

I'm writing a character for a book. He got rich by inventing holodeck. He's a billionaire! But, his true desire was always to win his country's lottery! Then do it again, just to prove a point. What makes mad billionaire mad? He believes that lottery draw is crooked. Because the lottery organiser can pick the 1st prize winner/ choose that there are no 1st prize winners. But they can't fix 2nd and 3rd prize! (or so the billionaire believes) So his definition of winning changes to winning second or third prize. What is his one true desire? To have a system of consistently winning that prize so he can "win" lottery constantly, just to prove a point that he can!

The setup: It is 7/39 lottery. You choose 7 numbers on a ticket out of possible 39, (pool 1-39). If those seven numbers are drawn in any order, the first prize is won. Price of playing one 7 number ticket is 1p. You cannot choose more, you cannot choose less on a single ticket. The prizes are, as follows:

Correct guesses = prize

7 = 6.000.000p

6 = 600.000p

5 = 220p

4 = 10p

3 = 1p

2 and 1 = you get nothing!

Now, the lottery allows you to choose more than 7 numbers in one way: to buy all combinations of them. For example, if you have 8 numbers continually rattle in your head and you can't choose just 7, you buy so called system ticket with all unique combinations of those numbers. For above example, you would buy 8 tickets at the price of 8p. Lottery allows buying systems up to 17 numbers. System number= number of combinations (price in p)

8 = 8

9 = 36

10 =120

11 = 330

12 = 792

13 = 1716

14 = 3432

15 = 6435

16 = 11440

17= 19448

That's it for setup. Now come the questions:

1) What is the minimal number of 17th type system tickets billionaire would need to win 2nd prize (6 guesses correct)? What would his winnings be (considering numbers of 3,4 and 5 guesses he would get by guessing 6 correctly)?

2) What is the minimal number of 17th type system tickets billionaire would need to win 3rd prize (5 guesses correct)? Also, what would be his total winnings.

(optional question)3) How would he go about calculating those 17-type system tickets, so they do not repeat and he wins 2nd/3rd prize with a degree of certainty of 100%? He has supercomputer on his disposal.

To get at least 6 correct, he needs to buy $38$ of all $39$ numbers' full possibilities. That'd cost him! Let's see, $\binom{38}{7} = 12620256$ ammount of 7-combination tickets (the 17-ones make no sense, they're not cheaper nor more expansive in any way). That would be $12 620 256$p. Only winning 600.000p that would be a huge loss ($12620256 - 600000 = 12020256$p). For third prize it's about the same: $\binom{37}{7} = 10295472$p, losing $10295472 - 220 = 10295252$p.