In my country, Brazil, we have a lottery game called "Mega-Sena".

You can choose from 6 (cheapest set) to 15 (most expensive set) numbers from a total of 60.


*Blue: Chosen numbers;
*Green: Amount of chosen numbers.

Every week they have a new contest where are drawn 6 numbers from 1 to 60, no repeated numbers (give a look).

Any gambler who hits 6 numbers wins.

Gambling with only 6 numbers, odds are:



Gambling with 15 numbers, odds are:

$\frac{\binom{60}{6}}{\binom{15}{6}} \approx 1 : 10,003$

The cost of the set of 06 numbers is \$1.00.
$\binom{60}{6}$ * \$1.00 = 50,063,860 * \$1.00 = $50,063,860.00

The cost of the set of 15 numbers is \$5,005.00.
$\frac{\binom{60}{6}}{\binom{15}{6}}$ * \$5,005.00 = 10,002.7692308 * \$5,005.00 = $50,063,860.00

Based on the cost, no gambler have any advantage over another one.
The odds vs cost are the same.

When you choose a set having over 6 numbers (7 to 15), the lottery understands that you want all combinations of 6 numbers based on the chosen numbers.



Gambling with the above set of 7 numbers is the same as gambling with all the sets of 6 numbers below:


The question:

I know that I can win for sure choosing 50,063,860 different sets with 6 numbers each. But can I guarantee winning this lottery by choosing 10,003 different sets of 15 numbers each? How?

If no, what is the minimum required sets/combinations to have a guaranteed victory (remember, you can choose sets from 6 to 15 numbers each)?

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    $\begingroup$ "Is it possible to choose $10003$ different sets with $15$ numbers each to win this lottery?" - I think you need to rephrase the question, because it is possible to win this lottery even by choosing a single set of $15$ numbers (if you're really lucky). So the question should be rephrased as: "Can we guarantee winning this lottery by choosing $10003$ different sets of $15$ numbers each?". Nice question BTW (+1). $\endgroup$ Dec 17, 2015 at 8:15
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    $\begingroup$ @barakmanos Thanks. Question edited. $\endgroup$
    – Lucas NN
    Dec 17, 2015 at 14:59
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    $\begingroup$ Look at en.wikipedia.org/wiki/Steiner_system. $\endgroup$
    – Aryabhata
    Dec 18, 2015 at 2:02
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    $\begingroup$ @Aryabhata Thanks. I will give a look. $\endgroup$
    – Lucas NN
    Dec 18, 2015 at 2:56
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    $\begingroup$ You are asking if there is a Steiner system $S(15,6,10003)$ in the Wikipedia notation. I would guess not, just because $\frac {{60 \choose 6}}{{15 \choose 6}}=10002\frac{10}{13}$ is not an integer, so there will be at least one set of six that is present more than once, so there are likely to be too many that are present more than once. That is just a guess. $\endgroup$ Dec 19, 2015 at 17:16

1 Answer 1


I can't answer the minimum, but I can suggest a computer process that will give you an upper bound, which I suspect will be close. There are ${60 \choose 6}=50,063,860$ possible draws. Each ticket covers ${15 \choose 6}=5005$ of them. Start by dividing the numbers into batches of $15$ with as even overlap as possible. Your first four should be $1-15, 16-30, 31-45, 46-60$ Then split each group of $15$ into $4,4,4,3$ and make new tickets with that many from each group, so things like $1-4,16-19,31-34,46-48$ and so on. Keep an array of length $50,063,860$ showing all the combinations you have accounted for by making the corresponding entry $1$. After you get tired of specifying "smooth" combinations, start picking random tickets-choose $15$ random numbers and count how many new combinations each ticket accounts for. Pick, say, $100$ tickets and keep the one that covers the most new combinations. Try it for a few different starting combinations (maybe even start doing random tickets initially). Toward the end you may have to shift to starting with a set of six that is not yet covered and see what you can add best. When you have covered them all, you have a set of tickets which you have proven will cover all the possibilities. It will not prove that your group is minimum, but it may not be too bad, especially if you try a number of times.


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