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Let's say we have a set of digits $\lbrace{0,1,2,3,4,5,6,7,8,9}\rbrace$.

What are the odds of choosing $7$ unique digits at random in order (i.e. least to greatest) twice in a row?

I've never taken a class in probability, so forgive me if this is embarrassingly bad work:

There are $\pmatrix{10\\3}$ different combinations of $7$ numbers, which simplifies to $5!$ (which I thought was cool), or $120$ possible arrangements.

For each of the $120$ arrangements, there is exactly $1$ in $7!$, or $5040$, ways to arrange the set from least to greatest.

Therefore, if the odds of choosing $7$ of the $10$ digits in order are $1$ in $5040$, then the odds of doing it again with the same digits would be the following:

$$\left(\frac{1}{120}\right)\left(\frac{1}{5040}\right)^2=\frac{1}{3048192000}$$

Thanks for any help!

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    $\begingroup$ Please clarify your question. Are you allowed to choose the same digit twice in one try (e.g., $\{5,4,3,3,7,1,7 \}$? Or is there no replacement? Can the first selection of seven digits be anything, so that the real challenge is determining the chance of matching those seven on your next try? Does the order of selection of the digits matter? Please clarify! $\endgroup$ – David G. Stork Dec 17 '15 at 23:37
  • $\begingroup$ Edits made. The digits are unique (i.e. no replacement). The order definitely matters, as it must be from least to greatest on both attempts, and the second attempt must repeat the results of the initial attempt. $\endgroup$ – Lanier Freeman Dec 17 '15 at 23:39
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    $\begingroup$ With the clarifications, that all looks good to me: it's the product of three independent probabilities: second list has same elements as first list (1/120); first list in order (1/5040); and second list in order (1/5040). $\endgroup$ – Rob Arthan Dec 17 '15 at 23:49
  • $\begingroup$ Alright. I'll wait for one more person to agree with the work before I submit the answer to a friend. Thank you $\endgroup$ – Lanier Freeman Dec 17 '15 at 23:55
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Here is a simpler way to obtain your result: There are $$10\cdot9\cdot8\cdot 7\cdot6\cdot 5\cdot 4=604\,800$$ possible drawing sequences in the first round, and ${10\choose 7}={10\choose 3}=120$ of them are good. The probability that in the second round you obtain the same sequence again is ${1\over604\,800}$. It follows that the probability $p$ of the combined event comes to $$p={120\over 604\,800{}^2}={1\over3\,048\,192\,000}\ .$$

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  • $\begingroup$ I like this approach. Upvote given. $\endgroup$ – Lanier Freeman Dec 18 '15 at 18:20
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    $\begingroup$ @VelvetUndergrad: personally, I preferred your original analysis, and I don't see why the above is simpler. $\endgroup$ – Rob Arthan Dec 20 '15 at 23:34

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