Distribution of suits question

[elitegimp]

okay, you have 2 spades (for example), and there are 2 spades in the flop. At this point, you can not account for 9 of the spades, and there are 47 cards you haven't seen. From this, you should think that there is a 9/47 (19%) chance of turning a spade.

What you are asking about is conditional probability: since there are 9 spades un-accounted for, and 18 cards dealt to your opponents, there is a 0.003% chance that all the spades are in you opponents hands and that would give you a 0% of getting your flush on the turn.

There is also a 0.09% chance that 8 of the spades are dealt out, giving you a 1/29 chance (3 or 4%) of hitting your flush.

Anyway, you find the probabilities of X spades being dealt to your opponents, multiply that by the chance of hitting your flush with 9-X spades left, sum it all up, and you should get 9/47, just like if you don't know any of your opponents cards. Statistics is cool like that.

edit:
Here's a handy table I made... note that probabilities are not given as percents, and are rounded to 5 decimal places.
<font class="small">Code:</font><hr /><pre>
N P1 P2 P
- ------- ------- -------
0 0.00735 0.31034 0.00228
1 0.05670 0.27586 0.01564
2 0.17525 0.24138 0.04230
3 0.28446 0.20690 0.05885
4 0.26668 0.17241 0.04598
5 0.14934 0.13793 0.02060
6 0.04978 0.10345 0.00515
7 0.00948 0.06897 0.00065
8 0.00093 0.03448 0.00003
9 0.00004 0.00000 0.00000</pre><hr />

N is just a counter,
P1 is probability that your 9 opponents account for N of the spades,
P2 is probability that you hit one of the 9-N spades left in the deck on the turn
P is P1*P2

If you sum up all the P values, you do get 9/47