A lottery game consists of scoring 15 numbers out of 25. The chance to hit 11 points with a single ticket in this lottery game is 111. What is the minimum amount of tickets to score 11 points? And how to choose these tickets?
To win the main prize with one ticket:
{25 \choose 15} = \frac{25!}{15!(25−15)!} = 3268760
Probability of hitting 11 numbers with one ticket:
\frac{{15 \choose 11} \cdot {10 \choose 4}}{{25 \choose 15}}
=\frac{\frac{15!}{11!(15−11)!} \cdot \frac{10!}{4!(10−4)!} }{\frac{25!}{15!(25−15)!}}
=\frac{1365 \cdot 210}{ 3268760}
=\frac{286650}{3268760}
\frac{1}{ 11.4033141462 \cdots}
\approx \frac{1}{11}
So the chance to hit 11 points with a single ticket in this lottery game is \frac{1}{11}.
I think that the minimum amount of tickets to score 11 points is 11, because \frac{1}{11} \cdot 11 = 1 but I am not sure. And I dont know how to choose these tickets.
Answer
First, let’s solve an easier problem. There are 25 total numbers, 15 of them are winning, and we still want at least 11 points, but this time the lottery tickets are only 11 numbers long. In order to buy enough tickets to be guaranteed 11 points on one of them, in the worst case scenario {25\choose 11}-{15\choose 11}=4456035 of the tickets would be duds.
What would happen if the tickets were actually 15 numbers long instead of 11? Well then this would make things easier because instead of having to test 4456035 combinations of 11 numbers individually, we can now test {15\choose 11}=1365 of them at once!
This means we now only have to fail at most 4456035/1365=3264.49 times. So we need to buy 3265 tickets with the following strategy. For the first 3264 tickets, ensure that no two tickets have the same combination of 11 numbers. For the last ticket, make sure that it includes as many combinations of 11 numbers as possible that haven’t already appeared in the other tickets.
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