Considering I've been overflushed three times recently, I'm wondering what the odds for any given hand are of getting to the river and being overflushed on a 3-flush board with no pairs? I.e., this will happen once every ?? hands.

To simplify, suppose that I always call and everybody else always calls at a 9-handed table, and we all get to the river.

What if I (and everybody else) only play the top X% of suited combos?

More realistic case, IMO (and easily solvable). You and a villain have suited hands- same suit and both flush. Prob you are out-flushed given both flush is same as Villain high card > Hero high card. With your two known cards, there are 11 remaining cards of the same suit.

Pr(V high card > H high card) = (14- R)/11, where R is rank of hero high card (Ace = 14)

H high card= King, Pr = 9%;
H high card= 8, Pr = 55%;
H high card= 3, Pr = 100%

Thanks... I'm good with counting combos and figuring out all of that... My question was after a discussion about KK vs AA preflop. I was basically trying to figure out how often you can expect to get into a flush over flush cooler situation vs getting into something like a KK vs AA cooler situation (or set over set or whatever).

Similar to another recent thread, it seems like you are looking for a table of the following. Assume that Hero holds two suited cards. Assume that the board has exactly three cards of that suit (I think further assume no straight flush is possible) and is not paired.

(1) What is the prob of being over-flushed if facing N opponents, for N=1-9, and my highest card is X. Assume players play 100% of range and never fold (all deals go to a complete (N+1)-player showdown).

(2) Same question if players only play (hold) certain suited hands. Something like A2s+, K5s+, Q8s+, J7s+, T8s+, 97s+, 86s+, 76s, 65s.

I think these questions are doable. I may be able to get to them later.

Q1 in particular seems eminently amenable.

Edit: It obviously also depends upon how many suited board cards are higher than your highest card.

For starters I'll tackle it for 9max and 100% ranges.

P(3-flush unpaired board) = 4*C(13,3)*(30*27/2) / C(52,5) = 17.827%

Edit: the rest of what I did was useless because I forgot we're calculating the probability of a higher flush rather than just any flush.

What comes to mind is to add
P(Hero bottom flush)*P(higher) + P(Hero 2nd-low)*P(higher) + ... P(Hero 2nd-nut)*P(higher)

If we ignore straight flushes, we needn't know the board.

P(Hero bottom flush) = 1 / C(47,2)
P(Hero 2nd-low) = 2 / C(47,2)
...
P(nut) = 9 / C(47,2)

P(at least one higher flush | Hero bottom flush) =
8 * C(8,2) / C(45,2) -
C(8,2) * C(8,4) / C(45,4) +
C(8,3) * C(8,2) / C(45,6) -
C(8,4) / C(45,8)
= 21.33%

When Hero has higher than bottom flush, it's trickier. For the villains, you have to filter out the lower flushes and you must take partitions into consideration.

I piggybacked off the simulation I made in the set over set thread, and I made a few changes to create a simulation for flush over flush situations. There is absolutely nothing in my simulation about hand ranges, but it does do the job of answering OP's question from a broad view.

Assumptions:
1. Hero is dealt 2 suited cards at random (for purposes of this model, I'm only looking at making flushes with suited hole cards. I did not look at 1 hole card from unpaired holding + 4 board card situations, unless the board ran out with 4 of the hero's hole cards' suit anyway).
2. 9-handed table where the other 8 players all were dealt two random cards (not necessarily suited).
3. A random flop, turn, and river is dealt from the remaining cards in the deck (it's measuring the odds of hitting a flush by the river).
4. The algorithm then checks for 4 possible outcomes: no flushes made, hero made the only flush, hero made the bigger flush, or hero made the smaller flush.
5. I ran the simulation 500,000 times.

Results:



The probability of getting over-flushed assuming you have 2 random suited cards is ~.00675, or about 1 in every 148 hands. The probability of any flush over flush situation assuming you have 2 random suited cards is ~.0131, or about 1 in every 77 hands.

Edit: I should also add that another flaw in my simulation was I did not create any functionality for straight flushes. It's entirely possible (if not probable) that my simulation's results included a few straight flushes that were misdiagnosed as under-flushes. So it's likely my number is actually a bit off, but it's not likely the straight flushes made a huge impact either way.

Analytically, assume Hero has [Xs, Ys] where X>Y. Board has exactly 3 spades (we will ignore straight flush possibilities throughout) and two other cards; but there is no pair on board (so flush is best hand that can be made). Of the 3 spades on board, H of them are higher than X. Clearly H can take values 0,1,2,3.

We are interested in determining the prob that Hero loses to a higher flush facing N opponents based upon values for X and H. In what follows, we assume 100% hand ranges and nobody ever folds (so every hand all players go to showdown).

I don't know whether this will ever be relevant, but we can determine the number of boards that are possible, based upon X and H:

= C(14-X,H)*C(X-2-H,3-H)*C(10,2)*C(3,1)*C(3,1)

N=1 Opponent

Let's now tally how many opponent hands have us beat, based upon X and H (we split into two cases depending upon if opponent holds one or two higher cards than X):

= C(14-X-H,1)*C(X-2-1-(3-H),1) + C(14-X-H,2)

Now tally how many hands are possible for the one opponent:

= C(45,2)

The prob of being over-flushed is clearly the ratio of these two tallies.

N=2 Opponents

Here we can use Principle of Inclusion-Exclusion (PIE). Prob of being over-flushed by at least one opponent = 2*Prob of being over-flushed by either opponent (given above) - Prob of being over-flushed by both opponents.

I have not yet attempted to derive the analytical formula for this in the general case of any values for X and H but (famous last words) it should not be too difficult.

Of course, PIE can be extended to any number of opponents, though presumably the derivation gets more convoluted as the number of opponents increases.

Note that this question would be straightforward to simulate (which may be the next thing to do). Smaller cases can be handled via brute force.

Edit: I didn't see mkind0516 posted overall simulation results while I was typing.

Can you show me combinatorically what the odds of making a flush are by the river with two suited cards? I saw a few websites say it's around 6.25%, but they didn't prove it (my sim model showed it to be 6.4% over 500k hands fwiw, but I just wanted to see it with combinatorics).

There's more than one way, but I'd do it as follows:

P(flush) = [C(11,3)*C(39,2) + 39*C(11,4) + C(11,5)] / C(50,5) = 6.4%

I follow what you were doing there, but I got 5.2% when you multiply that out. I also tried to do it using 7 card combos, but that got 3.1%. That's about half of the actual answer, which makes me think I'm just missing one term somewhere.

[ 4C1 * ( 13C7 + [13C6 * 39C1] + [13C5 * 39C2] ) ] / 52C7

I'd like to understand where I'm getting this wrong, so that I can improve my long-hand math here. What are the missing terms in our two equations?

Writing down these terms in detail:

= (165*741 + 330*39 + 462) / 2118760

= (122265+12870+462) / 2118760

= 135597 / 2118760

= 6.399828201%

Note: this is the Prob of a flush in NLHE in the suit of two suited hole cards.

This formula is correct for the Prob of getting a flush in 7 cards which is:

= 3.056576932%

(including royal and straight flushes in the tally).

Okay I see why I kept getting 5%. I had 52C5 in the denominator instead of 50C5. That makes more sense. Thanks.

Thanks for this.... I guess the probability of getting a suited hand is 12 in 51, or 23.5%, so mulitplying this with your .00675, I get about 1 in 630 hands. At 25-30 hands per hour live, that happens maybe about once every 20-24 hours of play. I'm guessing that as we remove junk from our range and from villains' ranges, fold flop bets when BDFDs are available, or fold FDs on paired boards or whatever, that this happens much less frequently, maybe once every 100-200 hours?

Alright, I analytically derived the probabilities of being over-flushed when facing 2 opponents (the prob when facing one opponent was given above).

Let's review. Hero is dealt two suited cards. Call them (Xs, Ys) where X>Y and the suit is spades without loss of generality. Board comes exactly three spades and flush is highest possible winning hand (no straight flush, quads, or full house is possible). What is probability of being over-flushed?

For the purposes of this post, we will assume 100% ranges and all players go to showdown on all deals.

When you are sitting at the table and contemplating this question, you see your hand and the board. The relevant information, clearly, is your highest card (X) and how many board cards are higher than your highest card (call this H).

After a moment's reflection, it is clear that the probability only depends upon the sum of X+H. Call this Y. The only variable that affects the over-flush probability is how many of the suit are "available" for your opponent(s) to have which would give them a higher flush. This is clearly 14-X-H or 14-(X+H) or 14-Y.

Let's use Y to rewrite our previous result for the Probability of being over-flushed when facing exactly one opponent. Let's call this P1.

P1 = [C(14-Y,1)*C(Y-6,1) + C(14-Y,2)] / C(45,2)

Here is a table of P1 values (probability of being over-flushed when facing exactly one opponent) based upon Y=X+H (the sum of the value of your highest card and the number of suited board cards that are higher than your highest card).

High Card Value + Higher on Board	Numerator	Denominator	Probability of being Over-Flushed facing One Opponent
14	0	990	0.000%
13	7	990	0.707%
12	13	990	1.313%
11	18	990	1.818%
10	22	990	2.222%
9	25	990	2.525%
8	27	990	2.727%
7	28	990	2.828%
6	28	990	2.828%

Now, again by way of review, via PIE the probability of being over-flushed by at least one of exactly two opponents, call this P2, is given by:

P2 = 2*P1 - Prob(both opponents over-flush Hero)

As heehaww mentioned above, the Prob(both) is a little tricky and, essentially, involves partitions. But we can definitely handle it for the case of two opponents.

It is pretty straightforward to break the P(both) into the following cases: (i) both opponents have exactly one card higher than X; (ii) one opponent has exactly one card higher than X and the other opponent has two cards higher than X; and (iii) both opponents have two cards higher than X.

Once this logic is utilized, it becomes clear that:

Prob(both) = [2*C(14-Y,2)*C(Y-6,2)+C(14-Y,2)*C(12-Y,1)*C(Y-6,1)+C(14-Y,4)*3!!] / [C(45,4)*3!!]

We now have all the information we need to create a table of values for P2 (probability of being over-flushed when facing exactly two opponents) based upon Y=X+H (the sum of the value of your highest card and the number of suited board cards that are higher than your highest card).

High Card Value + Higher on Board	Numerator	Denominator	Probability of being Over-Flushed facing Two Opponents
14	0	446,985	0.000%
13	6,321	446,985	1.414%
12	11,709	446,985	2.620%
11	16,179	446,985	3.620%
10	19,743	446,985	4.417%
9	22,410	446,985	5.014%
8	24,186	446,985	5.411%
7	25,074	446,985	5.610%
6	25,074	446,985	5.610%

If I have time later on, I will try to analytically tackle the case of 3 opponents. Of course, anybody else is welcome to work it out too.

very clever.

its also useful in determining -'how many of the suit are "available" for your opponent(s) to have which would give them a'- lower flush, which could be another column (a 2nd numerator) in the tables.

Yes, the .00675 is almost certainly the ceiling instead of the mean. The simulation pretty much implies that every hand at the table is always seeing the river. In reality, most people are open folding their 94s and 93s type hands, so they're probably never going to be in a position to be over-flushed by T5s or Q2s or some slightly higher but still junk suited hand. It also assumes everybody is seeing through their backdoor flushes, when in reality many of those are folded out on the flop. There are clearly flaws in my model, but it was never designed to do anything more than give a jumping off point on estimating the real number (which probably has too many variables to ever really guess at).

I do think it is also worth noting that the .00675 is maybe a little closer to true at a live 1/2 or 2/5 cash game where a large percentage of the player field plays >50% VPIP. There's probably a nominal percentage of those fields who do play around 100% of suited hands up to a single raise. That may actually increase your odds in situations where you may take a cheap flop with a small suited hand (i.e. you over limp on the button with a 74s, 53s type hand), since your opponents probably aren't folding those Q2s type holdings pre-flop. On the flip side though, it does also presumably increase your implied odds when you get to see a flop with a suited ace, so a hand like A6s is stronger at lower stakes than it is at higher.

I would assume this balances itself out a bit at 5/10+ (or in select tougher 2/5 games), but it does seem a bit more open to exploitation at lower stakes. Then again, it happens so infrequently that it's not like you're ever going to really adjust your strategy to this type of analysis, which was probably more "fun" than it was actionable.

Next installment of this rip-roaring serial. I have now analytically derived the probabilities of being over-flushed when facing 3 opponents (the probs when facing either one or two opponents were given above).

(I apologize, but I am going to repeat the intro stuff so this post is somewhat self-contained.)

Let's review. Hero is dealt two suited cards. Call them (Xs, Ys) where X>Y and the suit is spades without loss of generality. Board comes exactly three spades and flush is highest possible winning hand (no straight flush, quads, or full house is possible). What is probability of being over-flushed?

For the purposes of this post, we will assume 100% ranges and all players go to showdown on all deals.

When you are sitting at the table and contemplating this question, you see your hand and the board. The relevant information, clearly, is your highest card (X) and how many board cards are higher than your highest card (call this H).

After a moment's reflection, it is clear that the probability only depends upon the sum of X+H. Call this Y. The only variable that affects the over-flush probability is how many of the suit are "available" for your opponent(s) to have which would give them a higher flush. This is clearly 14-X-H or 14-(X+H) or 14-Y.

Let's use Y to rewrite our previous result for the Probability of being over-flushed when facing exactly one opponent. Let's call this P1.

P1 = [C(14-Y,1)*C(Y-6,1) + C(14-Y,2)] / C(45,2)

Now, again by way of review, via PIE the probability of being over-flushed by at least one of exactly two opponents, call this P2, is given by:

P2 = 2*P1 - Prob(both opponents over-flush Hero)

As heehaww mentioned above, the Prob(both) is a little tricky and, essentially, involves partitions. But we can definitely handle it for the case of two (and three) opponents.

It is pretty straightforward to break the P(both) into the following cases: (i) both opponents have exactly one card higher than X; (ii) one opponent has exactly one card higher than X and the other opponent has two cards higher than X; and (iii) both opponents have two cards higher than X.

Once this logic is utilized, it becomes clear that:

Prob(both) = [2*C(14-Y,2)*C(Y-6,2)+C(14-Y,2)*C(12-Y,1)*C(Y-6,1)+C(14-Y,4)*3!!] / [C(45,4)*3!!]

For the case of three opponents, PIE tells us that the probability of being over-flushed by any of three opponents, call this P3, is given by:

P3 = 3*P1 - 3*P(both) + P(trio)

where P(both) is the probability that two specific opponents both over-flush Hero and P(trio) is the shorthand for the probability that all three opponents over-flush Hero.

Since P1 and P(both) were both derived above, all we need to compute P3 is P(trio). Again, breaking down into cases makes this straightforward: (i) each of the three opponents has exactly one card higher than X; (ii) one of the three opponents has two higher cards than X and the other two opponents have exactly one higher card than X; (iii) two of the three opponents have two higher cards than X and the other opponent has exactly one higher card than X; (iv) each of the three opponents has two cards higher than X.

Then it is clear that:

P(trio) = [C(14-Y,3)*C(Y-6,3)*3!+C(14-Y,2)*C(12-Y,2)*C(Y-6,2)*2!+C(14-Y,2)*C(12-Y,2)*C(10-Y,1)*C(Y-6,1)/2!+C(14-Y,6)*5!!] / [C(45,6)*5!!]

Putting this altogether the probabilities are given in the following table.

High Card Value + Higher on Board	Numerator	Denominator	Probability of being Over-Flushed facing Three Opponents
14	0	366,527,700	0.000%
13	7,774,830	366,527,700	2.121%
12	14,365,170	366,527,700	3.919%
11	19,808,100	366,527,700	5.404%
10	24,133,104	366,527,700	6.584%
9	27,362,205	366,527,700	7.465%
8	29,510,055	366,527,700	8.051%
7	30,583,980	366,527,700	8.344%
6	30,583,980	366,527,700	8.344%

The next case would be if Hero faced 4 opponents. Note that a general formula is probably obtainable using partition notation, but it seems to me that it is just as easy to simply derive each case separately.

Hero faces 4 Opponents:

P4 = 4*P1 - 6*P(both) + 4*P(trio) - P(quad)

where the new term, P(quad), is the probability that all four opponents over-flush Hero. P1, P(both), and P(trio) have been derived in previous posts.

As above, P(quad) is straightforward to derive once we break down the cases. Rather than write out the cases in words, it is probably simpler to write down how many overcards each of the four opponents have in each of the five cases: [1,1,1,1]; [2,1,1,1]; [2,2,1,1]; [2,2,2,1]; and [2,2,2,2].

Then it becomes clear that:

P(quad) = [C(14-Y,4)*C(Y-6,4)*4! + C(14-Y,2)*C(12-Y,3)*C(Y-6,3)*3! + C(14-Y,2)*C(12-Y,2)/2!*C(10-Y,2)*C(Y-6,2)/2! + C(14-Y,2)*C(12-Y,2)*C(10-Y,2)/3!*C(8-Y,1)*C(Y-6,1) + C(14-Y,8)*7!!] / [C(45,8)*7!!]

Putting this together with the earlier formulas for P1, P(both), and P(trio) gives us the probabilities reported in the following table.

High Card Value + Higher on Board	Numerator	Denominator	Probability of being Over-Flushed facing Four Opponents
14	0	90,532,341,900	0.000%
13	2,560,510,680	90,532,341,900	2.828%
12	4,718,776,920	90,532,341,900	5.212%
11	6,493,205,160	90,532,341,900	7.172%
10	7,898,342,616	90,532,341,900	8.724%
9	8,945,010,900	90,532,341,900	9.880%
8	9,640,394,820	90,532,341,900	10.649%
7	9,988,086,780	90,532,341,900	11.033%
6	9,988,086,780	90,532,341,900	11.033%

Given our assumed scenario (Hero holds two suited cards, board contains exactly three cards of that suit where a straight flush, quads or full house are not possible, and all players play 100% of hands and never fold) it turns out that Hero cannot be over-flushed by more than four opponents. There just aren't enough cards in the suit for that to occur.

This means that the PIE derivations for the cases of more than four opponents are pretty simple. And, more importantly at this time, all the information required for those derivations has already been presented in previous posts.

For completeness, here are the additional formulas we need in the general case.

P5 = 5*P1 - 10*P(both) + 10*P(trio) - 5*P(quad) + P(quint)

P6 = 6*P1 - 15*P(both) + 20*P(trio) - 15*P(quad) + 6*P(quint) - P(sexa)

P7 = 7*P1 - 21*P(both) + 35*P(trio) - 35*P(quad) + 21*P(quint) - 7*P(sexa) + P(hepta)

P8 = 8*P1 - 28*P(both) + 56*P(trio) - 70*P(quad) + 56*P(quint) - 28*P(sexa) + 8*P(hepta) - P(octo)

P9 = 9*P1 - 36*P(both) + 84*P(trio) - 126*P(quad) + 126*P(quint) - 84*P(sexa) + 36*P(hepta) - 9*P(octo) + P(nona)

As mentioned above, in our specific case P(quint), P(sexa), P(hepta), P(octo), and P(nona) are all zero.

Since P1, P(both), P(trio), and P(quad) were derived up-thread, we have all the information we need to calculate the probability of Hero being over-flushed when facing 1-9 opponents, based upon the sum of the value of Hero's highest card and the number of suited board cards that are higher than Hero's highest card.

The following table presents these probabilities (the first four columns appeared in the thread previously).

Table of Probabilities of Being Over-Flushed with 100% Hand Ranges

High Card Value + Higher on Board	1 Opponent	2 Opponents	3 Opponents	4 Opponents	5 Opponents	6 Opponents	7 Opponents	8 Opponents	9 Opponents
14	0.000%	0.000%	0.000%	0.000%	0.000%	0.000%	0.000%	0.000%	0.000%
13	0.707%	1.414%	2.121%	2.828%	3.535%	4.242%	4.949%	5.657%	6.364%
12	1.313%	2.620%	3.919%	5.212%	6.499%	7.778%	9.051%	10.317%	11.577%
11	1.818%	3.620%	5.404%	7.172%	8.924%	10.658%	12.377%	14.078%	15.764%
10	2.222%	4.417%	6.584%	8.724%	10.837%	12.923%	14.982%	17.015%	19.021%
9	2.525%	5.014%	7.465%	9.880%	12.259%	14.602%	16.910%	19.181%	21.418%
8	2.727%	5.411%	8.051%	10.649%	13.203%	15.715%	18.185%	20.614%	23.001%
7	2.828%	5.610%	8.344%	11.033%	13.675%	16.272%	18.823%	21.330%	23.792%
6	2.828%	5.610%	8.344%	11.033%	13.675%	16.272%	18.823%	21.330%	23.792%

A similar approach can be applied to the different scenario in which opponents play a more realistic hand range. I may be able to get to that sometime later.