# What equations for lottery?

• Feb 25th 2009, 03:28 PM
SoftwareTester
What equations for lottery?
I would like to know what equations are needed to calculate minimal tickets needed to garantee (at least) one '3-OK' in a 6/49 lottery.

You can find quite some info about 'minimal numbers of tickets required' (current record being 163 tickets), but I think rather then using a trial and error approach it is problem a mathematician can describe well.
• Feb 25th 2009, 04:24 PM
awkward
Quote:

Originally Posted by SoftwareTester
I would like to know what equations are needed to calculate minimal tickets needed to garantee (at least) one '3-OK' in a 6/49 lottery.

You can find quite some info about 'minimal numbers of tickets required' (current record being 163 tickets), but I think rather then using a trial and error approach it is problem a mathematician can describe well.

I think you will get a quicker response if you first define your terms.

What is a 6/49 lottery, and what is a 3-OK?
• Feb 26th 2009, 12:28 AM
SoftwareTester
In 6/49 lottery 6 numbers are drawn from 49 (1...49)
A person can fill in a number of tickets (number depending on how much he wants to pay) and if 3 of the 6 numbers on a ticket are amongst the 6 numbers drawn he will have a (small) price.
Example :
a person buys 4 tickets and fills in following numbers
1 12 18 26 27 29
10 11 32 33 41 49
8 12 26 28 32 48
3 4 14 23 26 35

suppose the numbers actually drawn will be
1 11 33 39 40 49
In this case only the second ticket would he 3 numbers occurring in both the numbers drawn and the (second) ticket, so he would have a '3-OK'

As there are 13983816 different tickets possible for drawing 6 numbers out of 49 people are interested in choosing numbers such that they have best chance of having a 3-OK if they buy a number of tickets.
This means the numbers on each ticket can best be chosen such that together they having maximum coverage of those alomst 14 million tickets.
Chosing
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
37 38 39 40 41 42
43 44 45 46 47 48
(or any other combination of those numbers each occuring just once) will have maximum coverage for 8 tickets. But it will be much mor difficult to find best coverage for more (say 50 or 100 etc) tickets.

Obviously there WILL be an optimum combination of tickets providing maximum coverage for each number of tickets (so : an optimal set of 25 tickets, 48, 100, 120 etc) and also obvious is that A combination of tickets will have coverage of 100% (so ALL almost 14 million possible drawns WILL provide at least ONE 3-OK). Currently 163 tickets are needed to have 100% coverage but it contains a lot of tickets that have several numbers in common and not every number has ben used the same numer of times in those tickets.

I believe there MUST be a 'rule' to generate those optimal tickets.
And I believe it all comes down to formulating (and solving) the right equations for getting maximum coverage.
This is pure a mathematical problem and should have similarity with other problems regarding obtaining maximum coverage.
So, as I'm not a mathematician (but a softwaredeveloper), I ask help from people who "do know" and put the question here.

I hope my question becomes more clear now. If any questions remain : please ask