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What is the minimum number Lotto 7 / 39 tickets to match 3?

That's the question. I am not sure of the answer, but I think it should be 9139/35. That's right. That's what I thought, but I did not answer. : (

It seems that you play a lottery, where players choose 7 numbers from a total of 39 different numbers, without repetition. The draw selects a single "winner" 7 digit number (no repetitions). Your question might be: How many tickets do you need to buy to ensure that at least one of them corresponds to at least three winning numbers? Only 39Choose3 = 9139 possible triples match. You can buy a ticket for every possible (triplet 9139 tickets), but that leaves 4 random numbers on each ticket. You can at least use 3 of these 4 numbers to correspond to a triplet second. That gets all triples with only 4570 tickets. (Note that since the number of triples is odd, we can not put two triplets on each ticket without repeating. The number of Tickets for this strategy is more than 9139 / 2.) Moreover, there 7Choose3 = 35 triples on the ticket, you need only Game 1. So you can thanks at least 34 triplets. It notes at least 17 less. It is true that each ticket has 35 triples. However, you will NOT be able to organize a series of tickets without triplets twice, since 9139/35 is not an integer. I suspect you'll need more than 9139/35 tickets to this work. This seems linked to a "Hamming Code", except that in a Hamming code, you want as many tickets as possible without repeating a triplet. This as some tickets as possible, without missing a triple. (Except that you might miss a few, and is always at least one triple on the ticket.) This is not a solution, just some ideas. I'll try later. There are still a number wasted on each ticket and each ticket is actually 7Choose3 = 35 triplets. However, since 9139/35 = 261 4 / 35 is not an integer, it is impossible to take notes so that each triple is represented once and only once. The best that can can hope for is 262, which is well below 4570.

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