# If I pick a random sequence of non-repeating 8 letters, what's the probability that the letters are in alphabetical order?

For instance, abcdefgz works, but redecfed does not.

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"redecfed" has repeated letters (3 'e's, and 2 'd's). So it's not something that's possible in "If I pick a random sequence of non-repeating 8 letters", and hence not relevant to the probability. – ShreevatsaR Sep 9 '12 at 6:59
Would the experiment stop at the first letter that is out of sequence or not? – NoChance Sep 9 '12 at 7:05

## 2 Answers

I assume that by "non-repeating", you mean "distinct". And by "random" I assume you mean that you're picking a sequence uniformly at random (i.e., all sequences are equally likely).

You can imagine decomposing the process of picking a sequence into two steps: first picking the set of $8$ letters, and then putting these $8$ letters into some order. There are $8!$ orderings (permutations) of these $8$ letters, and all of these are equally likely, and only one of these orderings has the letters in alphabetical order, so the probability that the letters are in alphabetical order is $$\frac{1}{8!} = \frac{1}{40320} \approx 0.000025$$

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Whatever letters we pick, there are $8!$ equally likely orders they could appear in. So the probability they will be in the correct order is $\dfrac{1}{8!}$.

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