Calculating runner runner odds.

[a500lbgorilla]

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.

I'm not much of a statistics guy and I was wondering how poker calculators take into account runner runner situations. Or if they just go through the 500 odd million possible boards and see which hand would win in every situation and tabulate the results. Just wondering...

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.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.

-'rilla

[AllinLife]

i think you calculate odds of it hitting on turn, then again on river and

multiply those fractions for instance

if you are chasing a runner runner flush, and say you have 1/5 of hitting that suit on turn, and a bit less on river

1/5 x 1/5 = 1/25 = odds of hitting runner runner flush

correct me if im wrong someone

[Fnord]

For post-flop computations, enumerate all of the possible boards and figure out who won at the end.

In the case of pre-flop, just sample a *lot* of boards.

[LockLow34]

If you're wondering about the outs you have to catch runner-runner, and how that figures into pot odds, best thing to do is add 1/2 an out to your calculations.

So if you have AKs, and the flop comes Q72r with 1 spade, figure you have 7 or so outs. 3 aces, 3 kings, runner-runner spade flush, and runner-runner broadway.

[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.

[elipsesjeff]

If you're wondering about the outs you have to catch runner-runner, and how that figures into pot odds, best thing to do is add 1/2 an out to your calculations.

Locklow, Its 1.5 outs for a backdoor flush draw, a little less than that for backdoor straight.

So in your example you really have 9 outs and not 7.

[RiverMonkey]

Can you please explain how you arrive at this no. of outs? And, ...... how that number is then used to calculate the overall odds for runner-runner situations.

When I'm only gunning for *one card* to complete my hand, I fully understand how to use outs for calculating odds of hitting on the turn, or, not hitting on the turn but then doing so on the river. I'm still not necessarily seeing how I would use the concept of outs in a similar manner to calculate runner-runner draw odds. (see my earlier post for my thoughts on how to to this calculation - In that post, I don't necessarily use 'outs' like I would if I were doing a "one-card to hit calc").

In other words, what's your line of reasoning, how did you derive these numbers? Please show a derivation/proof if you can.

Man, I sound like a high school teacher ... "you aren't going to get full marks unless you show all your work sonny" LOL

[elipsesjeff]

Basically, 1.5 outs is a general rule found in a lot of books (I got mine from Sklansky's SSH, page 102). Its a combination of estimating the likelyhood of getting your cards plus the added value of you being able to fold/raise the river if you don't/do get your card. Thus, if you don't get your card you can fold, saving bets. But if you do get your card, you can raise the pot, giving you an extra bet.

Hope this helps.

Calculating other odds check the link on the homepage.

[RiverMonkey]

OK .........? So ........???? That doesn't help me any more than the previous two posts that simply quote a certain number of outs and do not explain how that came about, or what to do with that info.

Let's just say you failed your exam sonny; not only for not showing your work, but also for cheating off another students exam paper. However, I will give you some credit for plagerizing the one of the better student's work LOL

I'm just pulling your chain ... any and all feedback is appreciated. However if you can give me more to go on, that would help even more. THANKS!

I'd look in the reference you mentioned, but I don't own that specific Sklansky book.

[whileone]

P(overall runner-runner) = P_BA + P_AB

I think you are done.

If I hold 2 harts, and there is one on the flop, there are 10 hearts left out of 47 cards. If and only if I hit that, do i need to continue. If I do catch a heart, then I have 9/46 on the river. We account for the heart that hit on the turn, by deducting from the count on the river.

so, 10/47 * 9/46 ~= 4.1%

if i just have 2 hearts, with no flop, then i have to be more tricky.
there are 3 ways to get just one heart on the flop:
hxx, xhx, xxh, so i think i have to handle each case separately.
(11/50 * 39/49 * 38/48) + (39/50 * 11/49 * 38/48) + (39/50 * 38/49 * 11/48)
13.86 + 13.86 + 13.86 =~ 41.8%

41.8 * 4.1 =~ 1.7% chance of getting exactly one heart on the flop, then runner runner flush.

[RiverMonkey]

I shouldn't try doing math with this head cold I am only a monkey after all.

You are right ......... There's no need to calculate the probability of hitting the cards in the reverse order (i.e. B then A). Those reverse events were already included in the A then B scenario calculation. That clears that up.

I'm still not understanding this 1.5 or 1/2 an out business that ElipseJeff was alluding to.

Can anyone shed some light on this? I'd love to know how to calculate/derive these numbers and more importantly how to put them to use.