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 $\frac{1}{11}$. 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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