Originally Posted by RiverMonkey

Ok, I'll give this the ol' college try. Please let me know if you agree with my reasoning. I'm just learning how to calculate hold'em odds and proabilities, so I'm posting this in hopes of getting feedback.

Also, I would love to see a simpler method. (I like to do these things the long-winded way at first, and then come up with a more elegant, quicker way afterwards - it allows me to apply first principles and see all the reasoning in gory detail).

As a side note, there are a grand total of nPr = 47_P_2 = 47!/2! = 2162 possible permutations of turn, then river cards. That includes all the ways you'll hit your draw. Including one card first, then the other afterwards and vice-versa. You can also get this from 47*46 = 2162 = 47C1 + 46C1. (I believe, although haven't thought through the details yet (maybe later), that you can come up with the answer I derive below in slightly different way by using the fact that there are 2162 possible turn then river possibilities and comparing the no. of ways to hit the runner-runner to this total sample space. There may be an advantage to going this route in that you can always work in terms of odds i.e. there is would be no need to work out probabilities first and then covert to odds ).

Let's say you need two cards card_A and card _B to complete your runner-runner draw.

e.g. you hold 7h8h, and the flop is 6c,Ah,Ks. card_A=10 remaining hearts, and card_B = 9 remaining hearts [13-(3 you can see)-1 (card_A hits on turn, thereby reducing the possibilities by one card)].

Note that card_A and card_B actually represent sets of cards however below, I'll refer to card_A and card_B as specific single cards selected from these sets.

First you need to calculate the probability of hitting one of your cards on the turn (call it card_A), and then, based on this, also calculate the probability of hitting card_B on the river. (call the overall probability of both these events occuring P_AB = P_A_Turn * P_B_River - why the multiplication of probabilities? Because, event B_River is dependent on event A_Turn occuring first).

P_A_Turn = no. of cards left in deck that 'card_A' represents/47 = [count them]/47

P_B_River = no. of cards left in deck that 'card_B' represents/46 =

[count them (knowing that card_A may have reduced the possibilities for card_B)]/46

But, I don't think you are done yet. You can also hit your runner-runner draw by getting one of the card_Bs on the turn and then one of the card_As on the river. i.e. the reverse of the the above = event_B happens on the turn and event_A happens on the river instead of the other way around).

Applying the above reasoning we also have:

P_BA = P_B_Turn * P_A_River

and, from above P_AB = P_A_Turn * P_B_River

The overall probability of hitting the runner-runner is then given by:

P(overall runner-runner) = P_BA + P_AB

(note that you add the probabilities here because BA, and AB are independent compound events).

And finally, Odds(overall runner-runner) are easily calculated from
P(overall runner-runner) by realizing that probabilities are actually just odds-for an event compared to 100 rather than the usual odds against an event compared to 1. Don't let the distinction between odds-for and odds-against evade you.

Originally Posted by RiverMonkey

P(overall runner-runner) = P_BA + P_AB