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.

  • $\begingroup$ These answers are definitely worth $1$ million as they are not trivial. The algorithm that generates the set of tickets with $100$% coverage to obtain a prize pool exists, but not even super computers could process in a timely manner to find such a set. As Super Sjoerdie explained, it is not worth investing money in this, as the return will almost always be negative in relation to the initial amount invested. $\endgroup$ Commented Mar 23, 2023 at 16:14

1 Answer 1


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.

Just for the record, I'm not using the 17-tickets because they're useless. Nor should he. Unless they become relatively cheaper, they should not be used, as they only interfere with the simple calculations that otherwise could be made. Therefore, the third question also would not be to his advantage, as he should not buy those 17-tickets.

EDT: Also, how he could calculate the amount of 17-tickets he would have to buy if he'd do that, he could divide the price using normal tickets by 19448.

  • 3
    $\begingroup$ I didn't make the minor grammar edits, but why did you roll them back? They seem to be improvements. The word "amount" is spelled with one M, not two, for example. $\endgroup$
    – Brian Tung
    Commented May 14 at 0:57

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