probability of flush draw in texas hold em poker

How do I calculate the probability of getting flush in texas hold em poker after the 3-card flop?
Specifics: N players at the table (N = 2 to 10)

Review of rules:
Each player gets two cards face down. Then after pre-flop betting, a flop of 3 cards is dealt open faced to the table. If players are still active, there will be another card dealt to the table (the turn), and a final 5th card dealt to the table (the river card).

Scenario:

All but two players fold after the 3-card flop; so now it is heads-up (i.e., one on one).
One player has two hearts and there are two hearts on the flop.
What is the probability that in the either the turn or the river, another heart will show up producing a flush for the player having the two hearts?

Some say the answer is a function of N, but others say it is independent of N. I am looking for a math derivation rather than some numbers that can be found in any poker book. I don't play poker, but I think I gave wrong advice to someone years ago, and now I need to make amends before online poker is made legal in my state next year.

Thanks.
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phoffric\Asked:
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ozoCommented:
In principle it can be a function of the strategy of the other players.
If they are more likely to fold when they hold certain cards than when they hold certain other cards,
then the probability distributions on the turn and river card may be different for games in which all but two players fold after the 3-card flop and games in which different numbers of players fold after the 3-card flop.
d-glitchCommented:
There's math and there's poker.
Math says you've seen 5 cards and 4 hearts.  So you have two chances to get one (or two) of the 9 hearts remaining among the 47 unseen cards.  Calculate the probabilities.

In poker, especially with lots of players, other people get to see some of the unseen cards.  If they fold after the flop, it is quantifiable more likely than not that they don're hold hearts.
d-glitchCommented:
My Kindle is disconnecting my typing.
That should be unquantifiably more likely

And if you don't see the ace of hearts, it is more likely than not that people who folded before the flop did not hold it.
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ozoCommented:
I agree that it is unquantifiable without knowing the other players' strategy, which can be a function of N.
The other players' optimal strategy is in principle mathematically analyzable, but the analysis can be fiendishly difficult in practice, so with actual players it is likely that their strategy is not completely optimal, and there may also be non-mathematical indications of their strategy.
The strategy of the other player who did not fold can also be a factor, and it could depend on the face values of the flop cards as well as on how many others had already folded and how many had yet to fold at the point when that player made the decision not to fold.
phoffric\Author Commented:
Only want this to be a math question and not about poker strategy or likelihood of winning.

Clarifying the original scenario:

N players at table (N = 2..10); and N-2 players are sitting out. Dealer deals out two cards to each of the N players. (This actually does occur as a player may return before being required to call). But in this scenario, all sit-outs are forced to fold leaving only 2 players to see the flop.

One player has two hearts and there are two hearts on the flop. At least one of the two players goes all-in (e.g., maybe he is low-stacked). This means that two more cards will be put on the table with no further betting. What is the probability that another heart will show up in those two additional cards resulting in a heart-flush, and is this probability a function of N?
phoffric\Author Commented:
d-glitch wrote:
Math says you've seen 5 cards and 4 hearts.  So you have two chances to get one (or two) of the 9 hearts remaining among the 47 unseen cards.  Calculate the probabilities.
I take this to mean that d-glitch is in the camp where the probability is not a function of N.
d-glitchCommented:
No.  The math is very clear, but the reality of the poker game is murky.  Players who drop out after a 2-heart flop are less likely to have hearts.  Players who stay in are more likely to have them.
If ten out of twelve players drop out, pure math says there are 9 hearts in 47 unknowns.
Poker sense says it might be 7 in 27.
d-glitchCommented:
If the players fold before the flop, there is no information to be inferred about hearts.

If you want to ignore inferences, for N=10 you can ask what is the probability that either the 24th or the 25th cards or both will be hearts given the 5 cards you have seen.  For N=20 you would be asking about the 44th and 45th positions.  But the answer does not depend on N.
ozoCommented:
If the the strategy used by N-2 other players is to always fold regardless of the cards they are dealt, and the strategy used by 1 other player is to never fold, then the probabilities can be modeled as as d-glitch described in http:#a40828480 independently of N
If the other players use a different strategy, then the probabilities may depend on N.

In general, if all the other players employ a strategy in which the cards they hold have no bearing on whether or not they fold or what bets they make, then the number of such players should make no difference to the probabilities of the remaining cards.
(However, if you play against such players using a strategy in which the cards you hold does have a bearing on whether or not you fold and what bets you make, then you could have a good chance of coming out ahead of those players)
phoffric\Author Commented:
Clarifying again with a different game - same math problem - just clearer, I hope.

There are only two active players in a special kind of game. Each player is given two face-down cards. Player #1 turns over his cards showing everyone that he has two hearts. The other player rolls a pair of dice and N = sum of the two dice; So, N can take on values between 2 to 12 inclusive.

The dealer then puts 2*(N-2) cards in a pile that is called a dead-card stack. These dead cards are not going to be dealt at all.

The dealer then deals face up three cards, which happen to be two hearts and a club.

The dealer will now deal two more cards face up. What is the probability that there will be at least one heart dealt in these last two face-up cards?
d-glitchCommented:
1st card is a heart:       (9/47)
and 2nd card is a heart:   (9/47)*(8/46)

1st card is not a heart:   (38/47)
and 2nd card is a heart:   (38/47)*(9/46)
==========================================
Add lines 1 2 and  5

Open in new window


And not to mess things up, but if the other player gets to look at his cards then he has more information and gets a different answer.
ozoCommented:
1 - 38/47*37/46
phoffric\Author Commented:
ozo's answer (i.e., 0.35) is what I had given to someone years ago. Likewise, d-glitch's answer (with a small correction) is another way to get to that answer.

I now believe that the answer is a function of N, and your two answers do not indicate this.
Does anyone else think the probability = f(N)?
ozoCommented:
For the game in http:#a40829235 you might call it a function of N in that  f(N) is constant over its range,
or if you define f(N)=0 for N>24, or if you use a larger deck for N>24, or if the deck was incompletely shuffled after having previously been used to play the game of http:#a40828480
d-glitchCommented:
I don't see my mistake, but I do see that ozo's method (the birthday problem again) is better and I am sure her answer is correct.

I believe the answers we gave (tried to give) are mathematically correct according to the restrictions you placed on the problem, i.e. no inferences are allowed.

But this is certainly not the correct answer when you are at the table, playing a hand of poker with real people.

I don't play poker, but I am quite sure that if you play according to precise probabilities you will lose to someone who knows the probabilities and can augment that information with inferences.
d-glitchCommented:
Given N and the five cards you know, you know which card positions will be the turn and river (2N+4 and 2N+5).

No matter what the value of N is, given the five known cards, the probability of any other two cards being hearts is as we have calculated.  It does not depend on N.
mccarlIT Business Systems Analyst / Software DeveloperCommented:
I now believe that the answer is a function of N, and your two answers do not indicate this.
Does anyone else think the probability = f(N)?
No, it's not a function of N. Maybe think of it this way... Because the game is determinant, which you made clear in your last clarification, ie. no one is making any decisions that affect the order of dealt cards, etc, then the outcome of this game is determined at the time the deck was shuffled. And the only information that you have is that you have seen 5 cards (therefore there are 47 unknown cards) and 4 of those 5 are hearts (therefore there are 9 other hearts somewhere in those 47 cards).

Therefore the probability is the same (ie. 9/47) that if you turn over ANY of the remaining 47 unknown cards you will see a heart. It doesn't matter if you turn of one of the other players 2 cards, one of the 2*(N-2) folded cards, OR one of the cards from the deck, which is obviously the case in the game. The probability is still the same.

(And so forth for the river, N is irrelevant)


And just in case, here is some more food for thought... A totally different game. There are 3 stones in a bag, a RED, WHITE and BLUE stone, and there are 2 players. Player 1 randomly draws a stone and shows it; it is the WHITE stone. Now consider two different scenarios, if player 1 were to immediately draw another stone what is the probability that they draw the RED stone? Pretty easy... 0.5 (it's either the RED or the BLUE, so 1 in 2). Ok, but consider a different scenario, after Player 1 drew the WHITE stone, then Player 2 draws a stone and looks at it but DOESN'T show the 1st player. It is the RED stone but only Player 2 knows this. Now, as far as Player 1 is concerned, if they pull out the last stone from the bag, what is the probability it is the RED stone? Well, they know nothing about what Player 2 drew, so all they know is that it is still a 1 in 2 chance of being RED, the probability is still 0.5. But Player 2 knows that the probability of Player 1 drawing a RED stone is 0, obviously because they have it.

The above shows 2 things... firstly, that it doesn't matter if Player 2 removed a stone of not, the probability (to Player 1's knowledge) is still the same. The same in the poker game, it doesn't matter if and how many cards are in the folded pile of cards, the probability of turning a heart is the same. The second thing it shows is that probability is not a global fact, it is dependent on the observer and what knowledge the observer has. And if different observers have different knowledge about the game, then they can have wildly different answers for the probability of an event occurring.

Hope this helped!
mccarlIT Business Systems Analyst / Software DeveloperCommented:
d-glitch said:   I don't see my mistake, but I do see that ozo's method (the birthday problem again) is better and I am sure her answer is correct.

Someone correct me if I'm wrong (in which the probability is quite high ;) that that is the case) but I believe the difference between d-glitch and ozo's answers are due to "Independent" vs "Dependent" events. In d-glitch's answer the result is assuming that the drawing of the second card was affected by what the first card was. This would be the case if you were randomly drawing these 2 cards from the 47 cards left. But I think that in this case, the result was determined at the time the deck was shuffled. The turn card and the river card are the next two in the deck and their values are pre-determined, ie. the value of the turn has no affect on the river.

As I said, I'm quite likely to be wrong, but that was just my thinking on the subject...
phoffric\Author Commented:
@d-glitch,

Years ago, I broke the problem into the following possible orderings for the two remaining cards as follows (H == heart; N == non Heart): (Order counts):
H H
N H
H N
N N

To get at least one heart, you compute the P(N N) case and 1 - P(N N) is the answer. (This is ozo's approach to get 0.35.)
Your approach seemed to be going along the lines of P(H H) + P(N H) + P(H N)

1st card is a heart:       (9/47)                    != P(H N);
and 2nd card is a heart:   (9/47)*(8/46)    P(H H)

1st card is not a heart:   (38/47)
and 2nd card is a heart:   (38/47)*(9/46)   P(N H)
==========================================
Add lines 1 2 and  5
When I added P(H N) instead of your line 1, then your answer agrees with ozo's answer.
BTW - I didn't think ozo's answer dealt with the Birthday problem.
phoffric\Author Commented:
@ozo,
you might call it a function of N in that  f(N) is constant over its range,
or if you define f(N)=0 for N>24
Just a 52 card deck consisting of 13 H, 13 S, 13 C, and 13 D.
Was thinking that f(N) is non-const for N between 2 and 22, and game is unplayable for N = 23, not 24 because of the small point of requiring 2 burn cards.
ozoCommented:
Or Add lines 1 and  5
Since line 2 is already part of line 1
phoffric\Author Commented:
@mccarl,
Thanks for the detailed explanation of why N is not a factor. This is what I was looking for since, as I said, I already knew how to calculate the standard answer of 0.35.

With everyone saying that P(at least one H in remaining two cards) = 0.35, then I guess f(N) = 0.35, and is not a function of N.

If N = 22, then 44 cards were dealt out, with 40 cards in the dead stack. After the flop shows 2 hearts, there are 9 hearts left in the deck. The dealer has 5 live cards left to show two cards.

Of the 9 remaining hearts, you would expect that of the 42 unknown cards, most of the 9 remaining hearts would be in the dead stack combined with the other player's two cards. Could you include this expectation in your calculations and still come up with a 0.35 chance of drawing at least one more heart out of the 5 live cards?
phoffric\Author Commented:
Or Add lines 1 and  5
Yes, since
1st card is a heart:       (9/47)        = P(H x) = P(H N) + P(H H);
(i.e., 2nd card is a don't care)
phoffric\Author Commented:
The second thing it shows is that probability is not a global fact, it is dependent on the observer and what knowledge the observer has.
Yes, and the militaries are very interested in observing as much as they can to make the probabilities run in their favor.
d-glitchCommented:
>> ozo and phoffric

Thank you for the corrections.
The deadly combination of  Sort of Understanding Probability and Habitually Careless Calculation is why is stay far away from poker tables.
mccarlIT Business Systems Analyst / Software DeveloperCommented:
Could you include this expectation in your calculations and still come up with a 0.35 chance of drawing at least one more heart out of the 5 live cards?
You don't need to include that expectation because the probabilities are still the same. Yes, it is likely that most of those hearts are in the in pile of folded cards, but it still doesn't mean that there are no hearts left.

Ok, another way to think about it, an extension to what I said above...

Consider N=2, and so after the flop you have 2 hearts in your hand, two unknown cards, and 2 hearts and a club in the flop. There are 45 cards left in the deck. I think from what you are saying, you are ok with the probability of the turn card being a heart = 9/47. That just happened to be the top card on the deck. Instead of dealing that card, say the dealer picked out the 41st card in the remaining deck, what is the probability of that card being a heart?? Exactly the same, 9/47. The card dealt is irrelevant.

But wait, this 41st card would have been the turn card in an N=20 scenario. It's just that the 40 cards before it, instead of being in the deck, would be in the folded card pile. (Ok, that's not strictly correct, as the cards are dealt in a slightly different order, but that is irrelevant to this discussion). What remains the fact is that the turn card is still that 5th card from the back of the deck and it has the same probability of being a heart... 9/47

I understand what you are saying, and that intuitively it feels like the probability should be different. But sometimes our minds play tricks.... Like the "gambler's fallacy" where you flip a coin 10 ten times and flipped 10 heads. You feel like it should be more likely that the next flip is a tail, but it is still a 50:50 (assuming that the coin isn't rigged)

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phoffric\Author Commented:
Thanks. Your comments got me thinking about the Monty Hall game, but with 1000 chairs to drive your points home.
ozoCommented:
And as with the Monty Hall scenario, it matters whether the doors/chairs/hands are revealed/folded conditionally or unconditionally.
mccarlIT Business Systems Analyst / Software DeveloperCommented:
You're welcome!!
ozoCommented:
Just noticed http:#a40829304 
constant over its range,
I meant of course domain
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