# Odds of picking the same number

I'm trying to figure out how to calculate this:

If 3 people have to pick a number between 1 and 9. whats the probability of 1 or more of them picking the same number?

I don't necessarily need an answer as that was just an example, rather I am looking for an equation.

• Interestingly, this is a variant on the Birthday Problem! – Newb Jan 31 '14 at 6:44
• @Newb Ive only seen that problem with determining if there are no duplicates. – Deekor Jan 31 '14 at 6:53
• The odds of one or more of them picking the same number is 100%. – Marc van Leeuwen Jan 31 '14 at 7:49

It doesn't matter what number the first one chooses, so we don't have to incorporate them in our calculation. The second one has a chance of $\frac{8}{9}$ of hitting a different number and the third one has a $\frac{7}{9}$ chance of hitting a different number than the two previous ones. Thus the result is: $$\frac{8\cdot 7}{9^2}$$
• It does lend itself better to a general formula: $\frac{\frac{n!}{(n-k)!}}{n^k} = \frac{n!}{(n-k)!\cdot n^k}$ – SQB Jan 31 '14 at 7:54