# Find the probability of winning at this lottery.

So, the problem I found goes like this:

You have $n$ different numbers, numbered from $1$ to $n$. You can randomly choose $m$ (different) of them. The computer also randomly selects $m$ (different) of them. If you and the computer have exactly $k$ common numbers, then you win a certain amount of money.

The problem asks us to find the probability of winning.

I have solved some easier problems involving probabilities. But here, the only thing I could think of was that the probability for a certain sequence of $m$ numbers to emerge is:

$$\frac{1}{\dbinom{n}{m}}$$

How do you solve it? I'm on my way of getting used to this type of problems and I could really use some help.

-

Good divided by total; or multiply your result with the number of matching sequences. There are $m\choose k$ ways to pick $k$ of the $m$ winning numbers and $n-m\choose m-k$ ways to pick the remaining numbers as non-winners. Divided by the total ways to pick $m$ numbres, we find $$\frac{{m\choose k}{n-m\choose m-k}}{n\choose m}$$

-
Thank you so much! It wasn't that hard after all! – Bardo Aug 26 '14 at 7:22

Let us assume you have picked your $m$ numbers. Now it's the computer's turn. It has to match $k$ of your numbers. Which $k$? These can be chosen in $\binom{m}{k}$ ways. Then it has to produce $m-k$ numbers which do not match any of yours. This can be done in $\binom{n-m}{m-k}$ ways.

So the number of ways the computer can match $k$ of your numbers is $\binom{m}{k}\binom{n-m}{m-k}$.

For the probability, divide $\binom{m}{k}\binom{n-m}{m-k}$ (the number of "favourables,") by the number of (equally likely) choices the computer can make. This is $\binom{n}{m}$.

-
Thank you for your answer! I've designated Hagen's answer as the winner since it appeared on top of the list. But thank you too!! – Bardo Aug 26 '14 at 7:23
You are welcome. – André Nicolas Aug 26 '14 at 8:43