The other day at an online bridge site I got dealt a $13$-card hand containing $3,4,5,6,7,8,9,10,J$ of spades, i.e. a $9$-card-long straight-flush (s-f). This is so rare that I decided to count how many $13$-card hands contain a s-f of length $9$ or longer.
For the purpose of this question:
A s-f is a consecutive sequence in the same suit, and aces always count high (since I was playing bridge!) so $A,2,3,4,5$ of spades is not a $5$-card s-f for this purpose.
Let us just count s-f in spades. If we want to count all suits we can simply multiply by $4$ since a hand cannot contain two s-f's of different suits each of length $\ge 9$.
I was trying to use inclusion-exclusion but got very confused. Finally I decided to count explicitly as follows. My questions:
(1) Can someone verify my answer below?
(2) Can someone show how this can be counted via inclusion-exclusion, or explain why that's not a good idea?
My attempt:
$13$-long s-f: No. of hands $=1$.
exactly $12$-long s-f: There are two such ($2-K$ or $3-A$), and each must be completed with an extra card (to make a $13$-card hand) but the extra card must not complete a $13$-long s-f. No. of hands $= 2 \times {39 \choose 1}$.
exactly $11$-long s-f: Three such; the $2-Q$ hand must not include the $K$ (leaving $40$ valid cards to choose $2$ to fill out the hand), the $3-K$ hand must not include $2$ or $A$ (leaving $39$ valid cards), and the $4-A$ hand must not include the $3$. No. of hands $= {40 \choose 2} + {39 \choose 2} + {40 \choose 2}$.
exactly $10$-long s-f: Similar to above, no. of hands $= {41 \choose 3} + {40 \choose 3} + {40 \choose 3} + {41 \choose 3}$.
exactly $9$-long s-f: Similar to above, no. of hands $= 2 \times {42 \choose 4} + 3 \times {41 \choose 4}$.
According to wolfram alpha total $= 571130$.