Counting the frequency of a flush hand in $7$-card poker

Question

I'm trying to count the frequency of a flush hand in $7$-card poker. Since a flush hand could be thought of as having $5$ cards with the same suit while the other $2$ doesn't matter, I wrote down as follows. $$4\ _{13}C_{5}\ _{47}C_{2}=5564988$$ But this clearly is too much than the actual frequency of flush. What did I think wrongly?

Answer 1

The trick is to add the separate probabilities of an exact 5-card flush, an exact 6-card flush, and a 7-card flush. The probability is: $$\frac{4\cdot_{13}C_5\cdot_{39}C_2}{_{52}C_7}+\frac{4\cdot_{13}C_6\cdot_{39}C_1}{_{52}C_7}+\frac{4\cdot_{13}C_7}{_{52}C_7}.$$

Edit. As @Logophobic pointed out in a comment, this answer is not quite correct, because it counts straight flushes as flushes. Assuming you don't want to count straight flushes, you have to subtract the probability of a straight flush.

Answer 2

The problem with OP's original figure:

$$ {4 \choose 1} \cdot {13 \choose 7} \cdot {47 \choose 2} $$

is that it counts some flushes more than once. As an example, it counts both the following as distinct: 'A2345' + '67' all of hearts, and 'A2456' + '37' all of hearts, when they are actually the same hand.

This explains why we have to separately consider the cases where we choose 5, 6, and 7 of a suit