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I have a problem:

If you have 10 balls labeled from 1 to 10, is the probability that the number of the draw matches the number of the ball $$1 - d(10)/10!$$ where $d(10)$ is the number of derangements of 10 balls?

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This would be the probability that at least one ball number matches the order of the draw. The number of derangements is given in A000166 It is approximately $\frac{10!}{e}$

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So it would be $(9/10)(8/9)(7/8) \cdots (1/2)$? – Tony Nov 18 '11 at 21:11
@Tony: no - because the number of derangements of 10 is 1334961, not 9!. – Henry Nov 18 '11 at 22:17

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