Prob. of flush in 7 card stud with three-flush after 4th street?

[mdb]

In The Theory of poker on page 95, Sklansky says that there is 9-to-1 chance against making flush (spades) when you have a 3-flush (spades) after 4th street. In the example sited, there is only 1 opponent, and you see two of his exposed cards- both cards are not spades.

I am not sure why my math is incorrect.

I have {10/46 * 9/45 * 44/44) *3! = 26.1% - a very different result!!

Anyone know the correct math.

[roggles]

There are many combinations which you count a lot of times in that.

[Siegmund]

Well, 10/46*9/45 is 4.34%, not 26%, for a start.

Counting more carefully to avoid duplicates:

You can get three spades, 10/46*9/45*8/44 = 0.79%

Or, you can get spade spade plain suit, or spade plain suit spade, or plain suit spade spade: 10/46*9/45*36/44 = 3.56% each of three ways, for a total of 0.79+3x3.56=11.5%. That's about 9 in 1 or 8 to 1. It will move by about 0.4% for each additional non-spade you see.

[pococurante]

(edit: the answer below provides the wrong answer... though I don't quite understand why)

10/46 * (9/45 + 9/44)

which is 8.79%, or 1 in 11.37.

[pococurante]

spade spade plain = 10/46 * 9/45 * 36/44 = 3.557%
spade plain spade = 10/46 * 37/45 * 9/44 = 3.656%
plain spade spade = 38/46 * 10/45 * 9/44 = 3.755%
spade spade spade = 10/46 * 9/45 * 8/44 = 0.791%

total: 11.047%, or 1 in 9.05.

Clearly this is the correct answer. Can someone point out to me why my earlier answer is wrong, though? Why doesn't that provide the correct result?

mdb: Does the book actually say "9 to 1"? It's 8 to 1, or 1 in 9... it's a mistake in the book if it says 9 to 1.