Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 5 down vote accepted

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 $$ \frac{{m\choose k}{n-m\choose m-k}}{n\choose m}$$

share|cite|improve this answer
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 $\binom{n}{m}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.