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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.

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    $\begingroup$ Interestingly, this is a variant on the Birthday Problem! $\endgroup$ – Newb Jan 31 '14 at 6:44
  • $\begingroup$ @Newb Ive only seen that problem with determining if there are no duplicates. $\endgroup$ – Deekor Jan 31 '14 at 6:53
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    $\begingroup$ The odds of one or more of them picking the same number is 100%. $\endgroup$ – Marc van Leeuwen Jan 31 '14 at 7:49
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Hint: What is the probability that none of them get the same number? Think complementary events.

Edit: You can think of this probability like this:

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: $$\frac{8\cdot 7}{9^2}$$

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  • $\begingroup$ (9!/(9 - 3)!)/9^3 ? $\endgroup$ – Deekor Jan 31 '14 at 6:52
  • $\begingroup$ Yes you are correct, although your result is written (and possibly thought of) in a rather complicated manner (I will elaborate in the answer). Now, do you know what a complementary event is and how to calculate it? $\endgroup$ – Dahn Jan 31 '14 at 7:01
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    $\begingroup$ It does lend itself better to a general formula: $\frac{\frac{n!}{(n-k)!}}{n^k} = \frac{n!}{(n-k)!\cdot n^k}$ $\endgroup$ – SQB Jan 31 '14 at 7:54
  • $\begingroup$ Yes, you're right. However, it seemed to me that since he is confused about a basic question like this, it is in fact understanding rather than a formula what he's really after. $\endgroup$ – Dahn Feb 2 '14 at 16:56

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