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I have been reasearching the internet for a few days now trying to figure out what we call here in México "Ruedas o Reducciones" of combinations, these are simple ways of reducing the numbers of tickets you need to buy to get "the most" out of your money.

Basic example:

In the lotto I play you pick 5 out of 28 numbers. But I pick 8 numbers, and instead of buying 56 tickets, and I make 5 combinations:

1,2,3,4,5

1,2,3,4,6

1,2,3,5,6

1,2,3,7,8

4,5,6,7,8

This guarantees that if the 5 winning numbers are in my 8 numbers, at least one of my tickets will have 4 hits.

I came across this post, and it is kind of similar to what I'm looking for, but not as complex.

What I want to know is 2 things:

1. If I want to make a "reduction", picking 14 numbers (instead of 8) out of the 28, while guaranteeing at least 1 ticket with 4 hits (if the 5 winning numbers are in the 14 I picked), how can I calculate the minimum ammout of tickets I have to buy and which are the combinations?

2. Is it possible to make it generic? Having "X" being the total lotto numbers, "Y" the total numbers you pick, "W" the total winning numbers and "Z" the guaranteed numbers of hits in at least 1 ticket.

I hope it is clear.

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A Covering Design $C(v,k,t,m,l,b)$is a pair $(V,B)$, where $V$is a set of $v$ elements (called points) and $B$ is a collection of $b$ $k-$subsets of $V$ (called blocks), such that every $m-$subset of $V$ intersects at least $l$ members of $B$ in at least $t$ points. It is required that $v \geq k \geq t$ and $m \geq t.$

You want $l=1,k=5,$ $V=\{1,2,\ldots,14\} \quad(v=14)$ etc. Google the "La Jolla Covering Repository" for extensive tables.

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  • $\begingroup$ This goes beyond my understanding of maths for me to develop an answer, but yeah, from what I researched this is the answer, thank you. $\endgroup$ – Lauro182 Feb 10 '17 at 1:10

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