The other side of the odds

I’ve been looking for the odds for the board going four to a flush, All I could find is the odds for the board going five to a flush, which is dependent on how many cards of that suit you have in your hand. Not knowing any, it’s 0.198%(cardchat). I don’t have the math skills to work out four to a flush hitting the board, but don’t mind asking for help.
At any rate, with the odds being low, and the number of times i see this happen, there must be people who never see it. this is what it looks like
OK. I’m done wining about it, but would still like to know the odds, if anyone has them worked out.

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This is not all that uncommon, really.

There are four suits. There are 13 ranks. So there are 13 cards of each of the four suits.

Since every card has a suit, the odds of the first card on the board being one of the four suits is 1:1.

Not considering the cards that have already been dealt, the odds of the second card on the board matching the suit of the first card is about 1 in 4, but it is actually 12 in 51, or about 23.52%.

The odds of the third card on the board matching the suit of the first two cards is therefore 11 in 50, or 22%.

The odds of the fourth card onthe board matching the suit of the first 3 is 10 in 49, or 20%.

If that one misses, we get one more chance to have a fourth card of the matching suit, and that would be 10 in 48, or 20.83%

The total odds:

1 * (12/51) * (11/50) * (10/49) * (10/48) = 0.00220088035214085634253701480592 = 0.22% or about twice in 1000 hands.

Of course, the cards removed from the deck by dealing out the hole cards and the burn cards can change these odds quite a bit, depending on which cards have been removed from the deck in this manner.

With four to the flush on the board, the odds of your opponent having the 5th for the flush when you don’t have it approach 100% :slight_smile:

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Yoiur math there is off. The last term should be applied to the turn or the river, not to both.

In a game of imperfect information, the other hole cards and burn cards don’t make a difference. We can illustrate that…

Deal 6 places with 2 cards each. Of these 12 cards, we would expect 3 to be spades, for example, because 25% of all cards are spades.

Of the remain 40 cards, 10 will be spades. This too, is 25%

If we look at the whole deck, we see 13 spades, which is also 25%,

We can only be certain of our own hole cards and the cards on the board. For the “unseen” cards, we have to assume a standard distribution, and the odds remain the same. In this example, 10 in 40 is the same as 13 in 52.

Thinking about it some more, I believe you are right. I usually get it wrong on the first try.

So then, is this the correct calculation:

1 * (12/51) * (11/50) * (10/49) = 1.5% to have 4 cards of the same suit, but then if one of the first four cards misses, we have another chance at 10/48 of hitting it on the river.

To get to three-suited, it’s 1 * 12/51 * 11/50 = 0.05176470588235294117647058823529 ~= 5.18%

To hit the fourth of the suit, then it’s 10/49 + 10/48 = 0.2083333 times 0.051764… ~= 0.0107824412, so about 1.08%, which is about 5x more frequently than I calculated initially.

If I’m still wrong, I hope someone just gives the correct answer so I can quit embarrassing myself.


Let’s say the first card is a spade and the second a heart. There are now 2 possible 4 flushes by the river.

I don’t know how to go about calculating it, but I think the answer is somewhere between 3% and 5%, depending on your hole cards. I don’t remember where I saw that, and my memory isn’t what it used to be, so I dunno.

As far as being embarrassed, meh. If, at the end of this thread, we understand exactly how to calculate it, that should be all that matters. It should be more embarrassing to be too afraid to ask than to get something wrong.

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Let’s compute the probability that (exactly) four spades appear on the board. That means we need to draw 4 cards out of 13 spades and 1 card out of 39 non-spades. The probability that this happens is described by the hypergeometric distribution. I cannot write binomial coefficients here, so I’ll write [n, k] for the binomial coefficient “n over k”. The probability of exactly four spades (without the knowledge of any hole cards) is
[13, 4] * [39, 1] / [52, 5] = 1.07%

The probability that (exactly) four cards of the same suit are dealt is then four times this probability, about 4.29%.


Total number of possible boards: C(52,5) = 2,598,960

  • 5 one suit – 0.198% – C(4,1) * C(13,5) = 4 * 1287 = 5,148
  • 4 one suit – 4.29% – C(4,2) * C(2,1) * C(13,4) * C(13,1) = 6 * 2 * 715 * 13 = 111,540

For other odds:


Yeah, that sounds more like what i almost sort of remembered. While I bow to your superior accounting kung fu, your first bipedal co-defendant , C(52,5) would seem to indicate that you’re dealing a board without giving me any hole cards!

Would you be a dear (BW or Warlock) and run through one of those fancy hyperspatial disruptions with a fella like me having 0,1, or 2 of suit?

Let’s assume that you hold k spades. Then there are 50 cards still in the deck, 13-k of which are spades and 37+k of which are non-spades. The probability that exactly 4 spades appear on the board is
[13-k, 4] * [37+k, 1] / [50, 5]

k = 0: 1.25%
k = 1: 0.89%
k = 2: 0.61%

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First of all, thanks BlackWidow! (and Warlock)

So if I’m suited in spades, for example, there is a 0.61% chance of 4 spades to the board, and still a chance of 4 to any of the other 3 suits. I’m guessing the other suits will 4 flush the K=0% times 3, or 3.75%?

Also,. in the K=1 case, would I be correct in assuming that the 0.89% is for each suit, so I would have twice this chance to 4 to the board in one OR the other?

And thanks for the link! I’ve never even heard of this kind of accounting magic before, but now feel I must go master this arcane form of accounting wizardry!

That’s correct.

Yes, the two events are disjoint, so the probability of one of them happening is just the sum of the two (equal) probabilities.

I see some great potential for accounting wizardry in you :slight_smile:

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Awwww shucks, thanks! (stuffs his hands into his pockets and kicks the dirt)

So I found this nifty free online course, “Intro to Statistics” being offered by Stanford University / Udacity. I have no idea why I never took a statistics course before! In case anyone else is interested, here’s the link…

Put another way, the odds of a 4-flush board at 4.29% is a bit more than 1 in 25 hands (out of hands that make it to the river). Most tables at Replay run something like 50-80 hands per hour. Not all of them make it to the river of course. Still, you should expect to see a 4-flush board on average 1-2 times every hour you play.

A while ago I counted the frequency of some different “unusual” boards like 4-to-a-flush, from about 2.2k hands on Replay:



This is useful & much more inline with @waidus OP comment & question. Kudos

Question: Does the River board with 4-flush also include the occurance of 5-flush too? It obviously makes very little difference but its still relevant IMO. Thanks


Kudos on your answer. Your answer is the best provided, even if at first you made a mistake. I’ve been discussing patterns or rare occurrences in hands & dealings with waidus in a seperate thread so I understand his question and basically the kind of answer he is looking for and why he is asking this question to begin with. If I’m wrong hopefully @waidus will correct & clarify. The question arises bc players notice strange dealings, patterns etc in online poker that don’t happen in live poker. I’m not agreeing or disagreeing with this conclusion, simply repeating what lots of players say in general.

Also its funny how you stated: “not all that uncommon…” and arrived at 2 in 1000 or 1 in 500 0.22% which I would consider very uncommon. Your second answer seems more reasonably or closer to as you said not all that uncommon at 1 in 100 or 1%.

We are trying to find how often this rare occurrence will be dealt to the board. The hole cards are irrelevant & unknown. The calculation needs to calculate how often the board will be dealt at least 4 to a flush or 5. There is no reason to exclude 5 to a flush bc we are trying to figure out an unusual occurrence & not a specific occurrence of only 4 or 5 to a flush. Also as stated whether or not any player actually has the flush is also irrelevant.

A royal flush dealt to the board for everyone still counts for each player, & I would presume the chance of getting a royal flush includes the possibility it could be dealt to the board.

As a side note I hope players can give a definitive & actual answer & maybe in the future @puggywug you might include this in analyzing cards & hands dealt as your doing currently for 3000 hole cards dealt.

Thanks @DogsOfWar. The 4.29% is specifically 4 cards to the flush on board and doesn’t include 5 to the flush.