Disclaimer:

I am not a mathematician, but I pretend well.

You are already part way there.

From flop to turn, you have

*n* outs, and there are 47 cards you don't know.

So the chance of hitting one of them is n/47.

The chance of missing all of them is 1 - (n/47)

From the turn to the river (assuming you missed going from flop to turn):

You still have

*n* outs, but now there are only 46 unknown cards, so the chance of hitting one of them is n/46.

The chance of missing all of them is 1- (n/46)

The chance of missing all of them both times is:

(1-n/47) * (1-n/46)

And the chance of hitting any out from flop to river is:

1 - ((1-n/47) * (1-n/46))

This is beginning to look ugly.

Let's use a standard 9-out flush draw.

9/47 = 0.191

9/46 = 0.196

Chance of missing on the flop: 1 - 0.191 = 0.809

Chance of missing on the turn: 1- 0.196 = 0.804

Chance of missing both times: = 0.809 * 0.804 = 0.650

Chance of hitting on one of the streets: 1 - 0.650 or 0.35

Notice I said "chance of" each time.

"Odds" is the ratio of chance of missing of to chance of hitting

so the odds of hitting at least one of your outs on two streets is:

0.65/0.35 : 1 or 1.85:1 against.

I hope that helped.

If not, have a look at:

http://www.pokerstrategy.com/strategy/bss/1563/1/
Best of luck,

--VS