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Hi I am seeking help trying to figure out how often a hold em board could statistically have four to one suit. Ie 1 out 20 boards or 5 %, something along those lines if possible. Thank you in advance

  • 5
    I closed your question because it's a rant, not an actual question. You have significantly higher chance of remembering those kinds of boards because you tend to lose in those situations or it kills your action. Therefore they seem more frequent than they actually are, which in turn gives birth to this rant. – Radu Murzea Mar 16 '16 at 7:54
  • Since @RaduMurzea closed this Q, rightfully so, Its been significanty edited and may provide more useful information now to players experiencing the same thing. For that reason I've reopened it. – Toby Booth Mar 18 '16 at 18:32
  • This question is not a rant, and should not have been closed. As far as I can see, it is asking what the chances are of 4 cards on the board (after the river) being of the same suit. – Jake Oct 2 at 7:12
  • @Jake Did you take a look at how the question was initially phrased? – Radu Murzea Oct 2 at 7:56
  • @RaduMurzea Have now looked at original post. Still doesn't seem too much of a rant to me, more a genuine question about probability. I'm here because I was playing with/against a player who qestioned the partilcular site's dealing and wanted to check what the probabilities acutally were, as I'm sure did the OP. So, no reason to close whatsoever. – Jake Oct 2 at 8:21
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Using combinations
I will use combin(n,k) as that is the syntax in excel
Buy board you mean the 5 cards up?

4*combin(13,4)*48 / combin(52,5) = 0.0528 = 5.3% = 18:1

above includes full flushes
it ignores the cards in your hand as you did not state if you were a card to the suit or not

exactly 4 is
4*combin(13,4)*39 / combin(52,5) = 0.0429 = 4.3% = 22:1

exactly 5 is what you see as the odds of a flush
4*combin(13,5) / combin(52,5) = 0.00198 = 0.198% = 504:1
this includes a straight flush

make flush on the river from a suited hand
combin(11,3) * combine(47,2) / combin(50,5) = 0.08418 = 8.42% = 10.9:1

  • The first one (at least 4) should give the sum of 'exactly 4' and 'exactly 5' shouldn't it? The figures for 'exactly 4' and 'exactly 5' I think are correct. – Jake Oct 6 at 19:19
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Extrapolating the working method from a post on twoplustwo.com about caculating the odds for 1-, 2- or 3-suited flops

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

  • 5 one suit0.198% – C(4,1) * C(13,5) = 4 * 1287 = 5,148
  • 4 one suit4.29% – C(4,2) * C(2,1) * C(13,4) * C(13,1) = 6 * 2 * 715 * 13 = 111,540
  • at least 4 one suit – 4.49% – 5,148 + 111,540 = 116,688
  • 3 one suit, 3-2 – 10.3% – C(4,2) * C(2,1) * C(13,3) * C(13,2) = 6 * 2 * 286 * 78 = 267,696
  • 3 one suit, 3-1-1 – 22.32% – C(4,3) * C(3,1) * C(13,3) * C(13,1) * C(13,1) = 4 * 3 * 286 * 13 * 13 = 580,008
  • 3 one suit, any combo – 32.62% – 267,696 + 580,008 = 847,704
  • at least 3 one suit – 37.11% – 5,148 + 111,540 + 847,704 = 964,392
  • max 2 one suit, 2-2-1 – 36.52% – C(4,3) * C(3,2) * C(13,2) * C(13,2) * C(13,1) = 4 * 3 * 78 * 78 * 13 = 949,104
  • max 2 one suit, 2-1-1-1 – 26.37% – C(4,1) * C(13,2) * C(13,1) * C(13,1) * C(13,1) = 4 * 78 * 13 * 13 * 13 = 685,464
  • max 2 one suit, any combo – 62.89% – 949,104 + 685,464 = 1,634,568

This concurs with @paparazzo for ‘exactly 4’ and ‘exactly 5’ – though not ‘at least 4’ which should simply be the sum of the two (around 21:1). The figures I’ve given add up to 100% which I hope should be convincing.

Here’s a quote from the reference which explains the working method:

How many total flops are possible (ignoring anyone's hole cards)?

Clearly there are 52 cards in the deck and 3 come on the flop. So, in words, it is choose any 3 from 52.

In math, we write this as C(52,3). Either by looking up the formula for the "C" (Choose) function or by using a calculator, this is easily found to be 22,100.

Three suits on flop

There are four suits in all, and we need to choose three of them here. Of course, there are 13 cards in each suit, and we need to choose exactly one card from each of the three chosen suits.

So we have: C(4,3)*C(13,1)*C(13,1)*C(13,1) = 8,788

Two suits on flop

There are four suits in all, and we need to choose two of them here. And then among these two chosen suits, we need to choose one of them to have the two cards (and the other suit to have one card). Finally, we need to choose exactly two cards from the "two-suit" and to choose exactly one card from the "one-suit".

So we have: C(4,2)*C(2,1)*C(13,2)*C(13,1) = 12,168

One suit on flop

There are four suits in all, and we need to choose one of them here. And then we need to choose exactly three cards from that chosen suit.

So we have: C(4,1)*C(13,3) = 1,144

From these tallies, the true odds of several prop bets can be easily determined.

  • Worth noting that face-to-face the odds might be different, because the cards aren't shuffled properly :) – Jake Oct 13 at 6:54
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After the river there are 5 community cards on the board. Each of these can have one of four possible suits. So there are 4 x 4 x 4 x 4 x 4 = 1024 possible combinations of suits.

We are interested in the case where any 4 of the cards are of the same suit. This can happen in any of the following ways:

S S S S X
S S S X S
S S X S S
S X S S S
X S S S S

Where "S" is a card of a given suit, and "X" is a card not of that suit.

And since there are 4 suits, there are four ways each of these combinations can occur, so we have 5 x 4 = 20 different layouts out of 1024 where 4 of the cards are the same suit.

But not all layouts have an equal probability. Let's ignore the cards in any player's hand, as knowing our hole cards or the hole will alter the probabilities slightly.

For the first way, the suit of the first card of the flop must be the same as the rest of the flop and the turn, but not the river.

For card 2: 51 cards left, 12 left of the same suit, P=12/51
For card 3: 50 cards left, 11 left of the same suit, P=11/50
For card 4: 49 cards left, 10 left of the same suit, P=10/49
For card 5: 48 cards left, 9 left of the same suit, P=(48-9)/48=39/48

P1= 12/51 * 11/50 * 10/49 * 39/48 = 0.00858343337334933973

For the remaining ways the table can play out:

P2= 12/51 * 11/50 * 39/49 * 9/48
P3= 12/51 * 39/50 * 10/49 * 9/48
P4= 38/51 * 11/50 * 10/49 * 9/48
P5= 38/51 * 11/50 * 10/49 * 9/48

Adding all five probabilities I get around 3.59%

  • Pretty sure that number is not correct – paparazzo Mar 19 '16 at 20:32
  • @Paparazzi It's not the same as yours, but where is the mistake? – user1934 Mar 20 '16 at 1:38
  • Review your numbers. 9? – paparazzo Mar 20 '16 at 1:51
  • Thank you for the help. So if I understand you right the overall odds of having 4 to the same suit on any given board would be around 3.59%, or roughly 1 in every 27.8 boards? I really appreciate you taking the time to help me. When I know the answer it takes speculation out of the equation. – RLC72 Mar 20 '16 at 13:45
  • @RLC72, I'm pretty sure that this answer is off and Paparazzi's answer is correct. The probabilities of P1, P2....P5 should all be the same: they are the same outcomes just in a different order and the order the cards come out is irrelevant here. Michael, if you take your value for P1 and simply multiply by 5 it is the same as Paparazzi's so you were on the right track. – Dr.DrfbagIII Mar 21 '16 at 13:23

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