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A game consists of picking 6 different numbers from a set of 36. Each player completes one or more tickets, each with 6 different numbers. After the draw is made, the winner(s) are any players who do not have these 6 numbers on their ticket. Show that with the right combination of multiple numbers in different tickets, it is possible to have a guaranteed win with 9 tickets, while it is impossible to achieve this with 8 or less.

The probability of not guessing any number in a draw of 6 out of 36 different numbers is: $P=\frac{30\choose 6}{36\choose 6} = 0.3048$. The 9 tickets must definitely contain all numbers uniformly distributed, to achieve symmetry. That is, since we have $6x9 = 54$ numbers, of which we must have correctly guessed the 30, each ticket must have about 3 common numbers with each other. I can't think of any way to distribute the numbers so as to have a guaranteed win.

RobPratt
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Wang Xiu Ying Zhang
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  • Maybe inclusion-exclusion could help: 1 ticket can avoid $\binom{30}6$ choices of $6$ loosing numbers. 2 tickets with disjoint sets of numbers can avoid $\binom {30}6+\binom{30}6-\binom{24}6$ numbers. 3 tickets can avoid…. (our goal is to reach $\binom{36}6$ avoidances using $9$ tickets, but I am not sure if this approach is an easy one). – Maximilian Janisch Apr 20 '22 at 17:44

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