Timeline for $9$-card-long (or longer) straight flush in a $13$-card hand
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2020 at 14:28 | comment | added | antkam | A $13$-card s-f can be represented as a pair-of-sets $(X,Y)$ by picking any of the $5$ sub-sequences as $X$ and the rest as $Y$. Since $5 \times {43 \choose 4}$ counts such pairs-of-sets, any hand with a $13$-long s-f contributes a count of $5$ towards that total, hence $5 N_{13}$. | |
| Nov 24, 2020 at 9:29 | comment | added | true blue anil | I think you have written the coefficients in reverse order. How can there be multiple ways of representing a $13$ card s-f ? | |
| Nov 24, 2020 at 6:17 | history | edited | antkam | CC BY-SA 4.0 |
deleted 13 characters in body
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| Nov 24, 2020 at 6:11 | history | answered | antkam | CC BY-SA 4.0 |