Newbie Poker Math Question

[Erasa 0]

If this has been discussed before here or elsewhere, I would appreciate a pointer...I know the answers (from using odds calculators) but I am interested in the exact formula (no "rule of thump" or approximations plz)1) Let's say I flopped a four card flush; how do I calculate the probability that I will fill my flush after the river. I understand that the probability is 9/47 to do it on the turn and 9/46 on the river, but what is the combined probability - it is obviously not the sum of the two.2) How do I calculate the probability of hitting runner-runner after the flop (i.e. a pair becoming quads; 3 card straight getting filled; 3 card flush getting filled etc etc)?Everywhere I look it just gives you the result or an approximation on how to calculate it, but I haven't seen the exact formula (and unfortunately math classes are too many years ago).Thanks in advanceErasa

[juggo 0]

If this has been discussed before here or elsewhere, I would appreciate a pointer...I know the answers (from using odds calculators) but I am interested in the exact formula (no "rule of thump" or approximations plz)1) Let's say I flopped a four card flush; how do I calculate the probability that I will fill my flush after the river. I understand that the probability is 9/47 to do it on the turn and 9/46 on the river, but what is the combined probability - it is obviously not the sum of the two.2) How do I calculate the probability of hitting runner-runner after the flop (i.e. a pair becoming quads; 3 card straight getting filled; 3 card flush getting filled etc etc)?Everywhere I look it just gives you the result or an approximation on how to calculate it, but I haven't seen the exact formula (and unfortunately math classes are too many years ago).Thanks in advanceErasa

why aren't approximations good enough? its not rocket science.....all i do it is count my outs and multiply by 4(before the turn) and 2(before the river). this gives you a good idea about what % of the time you will hit your hand.probability of A * probability of B = probability of A and B happening

[TheBoobster 0]

all i do it is count my outs and multiply by 4(before the turn) and 2(before the river). this gives you a good idea about what % of the time you will hit your hand.

Why 4 before turn and 2 before river?

[LongLiveYorke 38]

Well, unfortunately, there is no real answer to your question. The only exact way in poker to determine the probability of one hand beating another is to run out every possible combination of cards and define the probability of winning by the fraction of the total number of combinations that leads to a "win."In other words, if we're on the flop and we have a flush draw versus a pair, the only exact way is to run every possible turn and river (based on the cards that we haven't seen), to collect every turn and river that wins for us, to figure out how many total turns and rivers we ran, and to divide the number of turns and rivers that win for us by the total amount of turns and rivers.Obviously the only way to actually do this is with a computer program, and there are many simple programs that do this for you (pokerstove, for example). The reason that one must simply enumerate all possibilities to come up with an exact answer is that all poker hands are different and it would be impossible to come up with a formula that takes into account all back door draws, all unclean outs, etc etc. Bit if you only want to make a good approximation, you can do this by ignoring back door, tricky two card scenarios and only considering those cards which will win the hand for one player (which we call "outs").It is simple math (knowing the amount of outs and the amount of cards left in the deck) to determine the percentage of hitting an out after one card or after two cards.

[juggo 0]

Why 4 before turn and 2 before river?

hahah.....i dont know the exact mathematical reason for it. but give it a shot.....it gets you close enough.Ex. For chasing a flush. you have 2 hearts in your hand and 2 hearts on the flop. there are 9 left for you to hit a flush. after the flop if you plan on seeing both the turn and river.....9x4=36. so you have a 36% chance of hitting a flush IF YOU SEE BOTH THE TURN AND RIVER. after the turn you have 9X2=18. so you have 18% chance on hitting your flush with one card to come. you also only have that 18% chance to hit if you are only going to see the turn.

[LongLiveYorke 38]

Why 4 before turn and 2 before river?

Well, considering that you've seen the flop and you know his hand, you have knowledge of 2 +2 + 3 =7 cards.So, if you've concluded that you have n outs, then the odds of hitting one of the n outs on the turn is n/(52-7) = n/45The approximation made is that 45 is about 50, so n/45 is about n/50, so our odds as a percentage are about2n/100 for hitting on the turn. Now, one can win by the river in one of three ways: One can hit an out on the turn, one can hit an out on the river, or one can hit an out on the turn and the river. The odds of winning by the river, therefore, are 1- the odds of missing on the turn and on the river= 100%- (45-n)/45 * (44-n)/44 *100% = 4.49 n + .0505 n^2In the approximation that n is small (we have few outs), we can estimate this to be about 4n (really, it's closer to 4.5n). Since the term of order n squared has a very small coefficient, it is somewhat suppressed. However, as n gets large, the term becomes more important and our estimation breaks down for large n. For example, for n=10, the term of order n squared is equal to about 5% (so by ignoring it we'd be about 5% off).

[TheMathProf 0]

Basically, the general rule of thumb for probability is:

• If you want event A OR event B to happen, and the possibilities are mutually exclusive (both events cannot happen simultaneously), you add the probabilities.
• If you want event A AND event B to happen, and the possibilities are independent (the occurrence of event A does not influence the likelihood of event B, or vice versa), you multiply the probabilities.

1) Let's say I flopped a four card flush; how do I calculate the probability that I will fill my flush after the river. I understand that the probability is 9/47 to do it on the turn and 9/46 on the river, but what is the combined probability - it is obviously not the sum of the two.

It depends. Do you only want one flush card, or is two OK? If you have the Ace of the suit, you'd probably be fine either way; with suited connectors, you almost certainly only want one card to come.For one flush card, you want the probability that it either:(A) does come on the turn (9/47) AND does not come on the river (38/46), OR(B) does not come on the turn (38/47) AND does come on the river (9/46).So, the probability that exactly one flush card comes is (9/47 * 38/46 + 38/47 * 9/46) = 31.637%.If two flush cards are OK, you want the probability that it either:(A) does come on the turn (9/47), in which case the river card is irrelevant (46/46), OR(B) does not come on the turn (38/47) AND does come on the river (9/46).So, the probability that at least one flush card comes is (9/47 + 38/47 * 9/46) = 34.968%Notice that these calculations are pretty much in line with the 2:1 odds (33.333%) that the rule of thumb suggests.

2) How do I calculate the probability of hitting runner-runner after the flop (i.e. a pair becoming quads; 3 card straight getting filled; 3 card flush getting filled etc etc)?

This obviously depends on what you want to have happen, as the probabilities for each of these events is different, but the basic idea is that you find the probability of hitting the first runner, the probability that you hit the second runner, and then multiply them.So in your pair becoming quads example, the probability that you hit your first runner is 2/47, and the probability that you hit your second runner is 1/46. You want both the first runner AND the second runner, so the probability is 2/47 * 1/46 = 0.093%.The three card-straight example is even more complicated, because it depends on which three cards to the straight that you have:

• If you have three directly connected cards (i.e. 7, 8, 9), then you can hit your straight by hitting the 5 and 6, the 6 and 10, or the 10 and Jack. The probability of hitting one of these three straights is 8/47 * 4/46 = 1.480% (it doesn't matter which of the two straight cards come first for calculating the probability that it occurs, although clearly one card or the other might be more beneficial for betting purposes), and since the possibilities are mutually exclusive and equally likely to occur, 3 * 1.480% = 4.440%.
• If you have one gap in your straight cards (i.e. 7, 8, 10), now you only have two possiblities to get the straight (the 6 and the 9 or the 9 and the J), so the probability is 2 * 1.480% = 2.960%.
• If you need both cards in the inside (i.e. 7, 8, J), now you only have one possibility to get the straight (the 9 and the 10), so the probability is just the 1.480% we calculated earlier.

I'm presuming you could take the three-card flush example from here.Also, as another poster has mentioned, this does not calculate the probability that your hand will actually be a winning hand, merely the probability that the hand occurs.Hope that helps.

[nutzbuster 7]

Here's your answer.Quit poker.

[juggo 0]

omg.....too much math!

[fleung22 1]

I before E...something something

[CBass1724 1]

I watched an Unsolved Mysteries a few years back that talked about people spontaneously combusting and catching fire for no apparent reason.I think I solved the reason why after reading this thread.I smell smoke.

[hblask 1]

For practical purposes, you only need a couple numbers:1) If you have four to a flush on the flop, you have about a 1 in 3 chance of hitting by the river if you see both cards (around 1 in 5 for each card)2) Your odds of hitting runner-runner are never good enough. Fold.

[TheBoobster 0]

Basically, the general rule of thumb for probability is:

• If you want event A OR event B to happen, and the possibilities are mutually exclusive (both events cannot happen simultaneously), you add the probabilities.
• If you want event A AND event B to happen, and the possibilities are independent (the occurrence of event A does not influence the likelihood of event B, or vice versa), you multiply the probabilities.

It depends. Do you only want one flush card, or is two OK? If you have the Ace of the suit, you'd probably be fine either way; with suited connectors, you almost certainly only want one card to come.For one flush card, you want the probability that it either:(A) does come on the turn (9/47) AND does not come on the river (38/46), OR(B) does not come on the turn (38/47) AND does come on the river (9/46).So, the probability that exactly one flush card comes is (9/47 * 38/46 + 38/47 * 9/46) = 31.637%.If two flush cards are OK, you want the probability that it either:(A) does come on the turn (9/47), in which case the river card is irrelevant (46/46), OR(B) does not come on the turn (38/47) AND does come on the river (9/46).So, the probability that at least one flush card comes is (9/47 + 38/47 * 9/46) = 34.968%Notice that these calculations are pretty much in line with the 2:1 odds (33.333%) that the rule of thumb suggests.This obviously depends on what you want to have happen, as the probabilities for each of these events is different, but the basic idea is that you find the probability of hitting the first runner, the probability that you hit the second runner, and then multiply them.So in your pair becoming quads example, the probability that you hit your first runner is 2/47, and the probability that you hit your second runner is 1/46. You want both the first runner AND the second runner, so the probability is 2/47 * 1/46 = 0.093%.The three card-straight example is even more complicated, because it depends on which three cards to the straight that you have:

• If you have three directly connected cards (i.e. 7, 8, 9), then you can hit your straight by hitting the 5 and 6, the 6 and 10, or the 10 and Jack. The probability of hitting one of these three straights is 8/47 * 4/46 = 1.480% (it doesn't matter which of the two straight cards come first for calculating the probability that it occurs, although clearly one card or the other might be more beneficial for betting purposes), and since the possibilities are mutually exclusive and equally likely to occur, 3 * 1.480% = 4.440%.
• If you have one gap in your straight cards (i.e. 7, 8, 10), now you only have two possiblities to get the straight (the 6 and the 9 or the 9 and the J), so the probability is 2 * 1.480% = 2.960%.
• If you need both cards in the inside (i.e. 7, 8, J), now you only have one possibility to get the straight (the 9 and the 10), so the probability is just the 1.480% we calculated earlier.

I'm presuming you could take the three-card flush example from here.Also, as another poster has mentioned, this does not calculate the probability that your hand will actually be a winning hand, merely the probability that the hand occurs.Hope that helps.

Very nicely done! Clearly explained with great examples!! GG sir!!PS> This is a great thread!

[Jeepster80125 0]

Why 4 before turn and 2 before river?

From Phil Gordon's Little Green Book.The rules of four and two.I have found a quick and easy way of figuring out how often I will draw to a winning hand after the flop. First, I count my outs, or the cards that will give me a winning hand. For example, let's say I have T 9 and I put my opponent on A-K (as it turns out, he has A K ). The flop comes A T 7 . My opponent is is front, of course, having flopped a pair of aces, but there are five cards- the two remaining tens and the three nines- that will put me in front. In other words, I have five outs.I can calculate the approximate odds of catching one of my cards on the turn or the river by multiplying the number of outs I have by four. In this case, 5 X 4 - 20%.According to this 'Rule of Four', I have about a 20% chance of catching a winning card on the turn or river. The actual odds turn out to be 21.2%, a tiny difference that is irrelevant for most purposes.With on the river card to come, the 'Rule of Four' becomes the 'Rule of Two'. Let's say the 8 comes on the turn. Not one of the five outs we're looking for, but it turns our hand into an open ended straight draw that can be completed by any jack or six. The additional 8 outs gives me thirteen in all. Using the Rule of two, 13 X 2 = 26%.The actual percentage turns out to be 29.5%, but once again that is close enough.

[Socrates 0]

For the basics you can use this methodAfter the flop1-8 outs: multiply by 49-12 outs: multiply by 4 and subtract 113-16 outs: multiply by 4 and subtract 4Ex:6 outs you will make your hand 24% of the time10 outs 39%15 outs 56%After the turn, multiply your outs by 2 and add 2Ex: 9 outs gives you 20%These numbers are within 1%This is the same as what Jeepster is telling you, just in a different formatRemember that while you X% of making you hand, you need to make sure you are getting the correct odds to call. This area will be your next step to learn, implied odds etc. You will also want to learn to control pot odds by putting your opponents on cards and betting in order to make sure they are wrong to call.

[rgold79 0]

From Phil Gordon's Little Green Book.The rules of four and two.I have found a quick and easy way of figuring out how often I will draw to a winning hand after the flop. First, I count my outs, or the cards that will give me a winning hand. For example, let's say I have T 9 and I put my opponent on A-K (as it turns out, he has A K ). The flop comes A T 7 . My opponent is is front, of course, having flopped a pair of aces, but there are five cards- the two remaining tens and the three nines- that will put me in front. In other words, I have five outs.I can calculate the approximate odds of catching one of my cards on the turn or the river by multiplying the number of outs I have by four. In this case, 5 X 4 - 20%.According to this 'Rule of Four', I have about a 20% chance of catching a winning card on the turn or river. The actual odds turn out to be 21.2%, a tiny difference that is irrelevant for most purposes.With on the river card to come, the 'Rule of Four' becomes the 'Rule of Two'. Let's say the 8 comes on the turn. Not one of the five outs we're looking for, but it turns our hand into an open ended straight draw that can be completed by any jack or six. The additional 8 outs gives me thirteen in all. Using the Rule of two, 13 X 2 = 26%.The actual percentage turns out to be 29.5%, but once again that is close enough.

Yeah... he shouldn't be saying "I have found" - these ideas go back way further than Phil Gordon. Also, good to see there are some genuine math geeks hiding in this forum. Come out into the light, fellas! I just bought Bill Chen's book and am looking forward to getting a view into how he sees the game.

[strivewind 0]

if you like math, get william chen's book "The Mathematics of Poker"http://www.amazon.com/Mathematics-Poker-Bi...TF8&s=books

[Erasa 0]

Basically, the general rule of thumb for probability is:

• If you want event A OR event B to happen, and the possibilities are mutually exclusive (both events cannot happen simultaneously), you add the probabilities.
• If you want event A AND event B to happen, and the possibilities are independent (the occurrence of event A does not influence the likelihood of event B, or vice versa), you multiply the probabilities.

It depends. Do you only want one flush card, or is two OK? If you have the Ace of the suit, you'd probably be fine either way; with suited connectors, you almost certainly only want one card to come.For one flush card, you want the probability that it either:(A) does come on the turn (9/47) AND does not come on the river (38/46), OR(B) does not come on the turn (38/47) AND does come on the river (9/46).So, the probability that exactly one flush card comes is (9/47 * 38/46 + 38/47 * 9/46) = 31.637%.If two flush cards are OK, you want the probability that it either:(A) does come on the turn (9/47), in which case the river card is irrelevant (46/46), OR(B) does not come on the turn (38/47) AND does come on the river (9/46).So, the probability that at least one flush card comes is (9/47 + 38/47 * 9/46) = 34.968%Notice that these calculations are pretty much in line with the 2:1 odds (33.333%) that the rule of thumb suggests.This obviously depends on what you want to have happen, as the probabilities for each of these events is different, but the basic idea is that you find the probability of hitting the first runner, the probability that you hit the second runner, and then multiply them.So in your pair becoming quads example, the probability that you hit your first runner is 2/47, and the probability that you hit your second runner is 1/46. You want both the first runner AND the second runner, so the probability is 2/47 * 1/46 = 0.093%.The three card-straight example is even more complicated, because it depends on which three cards to the straight that you have:

• If you have three directly connected cards (i.e. 7, 8, 9), then you can hit your straight by hitting the 5 and 6, the 6 and 10, or the 10 and Jack. The probability of hitting one of these three straights is 8/47 * 4/46 = 1.480% (it doesn't matter which of the two straight cards come first for calculating the probability that it occurs, although clearly one card or the other might be more beneficial for betting purposes), and since the possibilities are mutually exclusive and equally likely to occur, 3 * 1.480% = 4.440%.
• If you have one gap in your straight cards (i.e. 7, 8, 10), now you only have two possiblities to get the straight (the 6 and the 9 or the 9 and the J), so the probability is 2 * 1.480% = 2.960%.
• If you need both cards in the inside (i.e. 7, 8, J), now you only have one possibility to get the straight (the 9 and the 10), so the probability is just the 1.480% we calculated earlier.

I'm presuming you could take the three-card flush example from here.Also, as another poster has mentioned, this does not calculate the probability that your hand will actually be a winning hand, merely the probability that the hand occurs.Hope that helps.

This is great; I was looking for the exact numbers - not what my chances of actually winning the hand are (that requires a subjective look at the numbers and obviously can't be computed). I was thinking about writing my own odds calculator as an exercise in picking up another programming language (time to make the jump from C++ to .NET/C#).Thanks TheMathProf!Erasa

[11 to 1 0]

First, let me say I am completely math deficient and those who are not will be happy to call me an idiot. But - I think the way poker probabilities are calculated in some ways is wrong. For instance: a flush draw. The way this is done now is you have two hearts, there are two hearts on the board, so, 13-4=9. You have seen five cards, there are 47 unseen, so the probability is 9/47 or about 19% of catching a heart on the turn or river. You add them up and that's a 38% probability of making a four-flush. This makes no sense to me. If what we do is assume even distribution, and we can't know that but we have to assume something, the probability is 25% the turn will be a heart and a 50% probability that either the turn or the river will be, adding the independent probabilities. If you need runners then you have to multiply the probabilities and then it's .25 X .25 and you fold.But if you have a straight draw along with a flush draw, and only need one card for either, you do the probability of each and add them. You can get over a 70% probability of hitting one or the other with some draws.

[psujohn 0]

But if you have a straight draw along with a flush draw, and only need one card for either, you do the probability of each and add them. You can get over a 70% probability of hitting one or the other with some draws.

Not quite because you'd be counting some of your outs twice. With a OESFD you have 9 outs to the flush plus 6 outs to the str8. There are 2 cards that would give you a straight flush but you can count 9 outs for the flush plus 8 outs for the str8.

[TheMathProf 0]

First, let me say I am completely math deficient and those who are not will be happy to call me an idiot. But - I think the way poker probabilities are calculated in some ways is wrong. For instance: a flush draw. The way this is done now is you have two hearts, there are two hearts on the board, so, 13-4=9. You have seen five cards, there are 47 unseen, so the probability is 9/47 or about 19% of catching a heart on the turn or river. You add them up and that's a 38% probability of making a four-flush. This makes no sense to me. If what we do is assume even distribution, and we can't know that but we have to assume something, the probability is 25% the turn will be a heart and a 50% probability that either the turn or the river will be, adding the independent probabilities. If you need runners then you have to multiply the probabilities and then it's .25 X .25 and you fold.But if you have a straight draw along with a flush draw, and only need one card for either, you do the probability of each and add them. You can get over a 70% probability of hitting one or the other with some draws.

Keep in mind two other small but "hidden" assumptions:(1) The probability of catching one of nine hearts is ever so slightly better on the river than on the turn, because there is one less non-heart in the deck.(2) What you're really saying when you say "catch a heart on the turn or the river" in typical layman's usage translates mathematically to "catch a heart on the turn OR if I don't catch a heart on the turn, catch a heart on the river". If we catch the heart on the turn, we usually don't think to point out, "Oh, and the river was a heart also" unless we're rubbing it in to someone who is now drawing dead. However, it's the underlined portion of this phrase that is a major assumption that affects probability, even though we often don't say it or sometimes even consider it.So when you talk about that 19% on the river to catch the heart, the scenario only comes up about 81% of the time. When you consider that 81% of 19% is approximately 15% or so, and you add that to the 19% probability of catching the flush on the turn, you get 34% to catch the flush on the turn or the river, which turns out to be pretty darned close to the actual computations performed previously in this thread.[EDIT: Also keep in mind that if for some reason, you wanted to calculate the probality of the heart on the river given the heart on the turn that you now have one less heart to deal with, so you still couldn't straight add the 19%.]

[hblask 1]

From Phil Gordon's Little Green Book.The rules of four and two.I have found a quick and easy way of figuring out how often I will draw to a winning hand after the flop. First, I count my outs, or the cards that will give me a winning hand. For example, let's say I have T 9 and I put my opponent on A-K (as it turns out, he has A K ). The flop comes A T 7 . My opponent is is front, of course, having flopped a pair of aces, but there are five cards- the two remaining tens and the three nines- that will put me in front. In other words, I have five outs.I can calculate the approximate odds of catching one of my cards on the turn or the river by multiplying the number of outs I have by four. In this case, 5 X 4 - 20%.According to this 'Rule of Four', I have about a 20% chance of catching a winning card on the turn or river. The actual odds turn out to be 21.2%, a tiny difference that is irrelevant for most purposes.With on the river card to come, the 'Rule of Four' becomes the 'Rule of Two'. Let's say the 8 comes on the turn. Not one of the five outs we're looking for, but it turns our hand into an open ended straight draw that can be completed by any jack or six. The additional 8 outs gives me thirteen in all. Using the Rule of two, 13 X 2 = 26%.The actual percentage turns out to be 29.5%, but once again that is close enough.

You forgot a key part of this:If you are on Pokerstars, multiply your final result by 0.10 to get the Pokerstars probability of hitting your card.