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. 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.)

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: I will count just spades. If we want all suits we can simply multiply by $4$ since a hand cannot contain two s-f of length $\ge 9$ of different suits.

• $13$-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 $= 571130$.

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 $= 571130$.