We have this official rank of hands.

I have a theory (which I do not know how to prove) that the number of players in a texas-hold-em table will affect the probability distribution of the hands.

That is, depending on how many players are on the table, the chances of being dealt a certain hand will vary. In the extreme case that the probabilities vary a lot, this would imply that, depending on the number of players, we have the interesting situation that the game is being played with the established rank of hands, but the actual probabilities are not the ones implied by that rank.

Please note that I am not talking about odds, implied-odds, pot-odds or anything related to that. I am only talking about the probabilities of the different possible hands.

Let's keep the assumptions simple: we have N players, all playing until river. Here are my two simple questions:

  1. Does the number of players N affect the probability distribution of hands?
  2. In case 1. is affirmative: is there a number of players that will cause a "reversal" in the probability distribution (as compared to the official rank of hands). How many players, and what hands get "reverted"?

And to further clarify my questions. If the official rank is

Straight Flush > Quads > Full House > Flush > Straight > Set > Two Pair > Pair > High Card

Is there a number of players N where the actual probability is:

.... > Straight > Flush > ...

This is just an example, any other "reversal" would be interesting. Actually I am interested in evidence that the number of players has an effect (however small) in the probability of the hands, even if no actual "reversal" ever happens.

  • I don't think that will be any reversal, because the rank depends only by the number of cards and how many cards you can change (none in this case). Anyway you can test it by a simple montecarlo simulation. R has a library for texas holdem. You can measure the frequency for straight and flush on changing the number of players.
    – emanuele
    Commented Feb 11, 2013 at 10:22
  • 1
    @gonvaled: Although this statement is false: "Depending on how many players are on the table, the chances of being dealt a certain hand will vary.", the following statement is true: "Depending on how many players are on the table, the chances of SOMEONE being dealt a certain hand will vary." But this can never change the probability of making a particular hand.
    – TTT
    Commented Apr 6, 2013 at 14:43
  • 2
    What you suggest is impossible but I'm curious what leads to think that way.
    – DPM
    Commented Mar 24, 2014 at 5:02

17 Answers 17


This can not be true... A flush is always better than a straight, no matter how many players there are at the table.

Now, if you're talking about a higher probability of a player to have a flush instead of a straight... this depends on a lot of factors... and by a lot I mean a lot:

  • players are usually tighter when there are 8 players at the table vs. 3 players
  • the player's skill is also a factor
  • the structure of the game also has a big influence on what cards are being played. You may play K 9 suited if you're at a final table, but fold pocket Tens (a much better hand) if the bubble is about to burst.
  • and much more

All these factors influence what hands are folded and which are not. Which means that the probability of a player having a flush or a straight is greatly changed.

Now, if everyone plays their hand and goes to the river, like you said, then the probabilities stay the same. There are 4 cards of the same rank and 13 cards of the same color. If the dealer is shuffling properly, then there's the same probability for each player to get J♣ or K⋄ or 3♣ etc.

Let's say there are 3 players at the table. If you give them 3 random cards (1 to each), then it means that there is:

  • a 1/4 chance for each one that they'll get a club
  • a 1/4 chance for each one that they'll get a diamond
  • a 1/4 chance for each one that they'll get a heart
  • a 1/4 chance for each one that they'll get a spade.

25 % FOR EACH SUIT, right ?

This means that, on average, there will be X number of hearts left in the deck, X number of clubs, X number of spades and X number of diamonds. I don't know how much X is and it doesn't matter. All that matters is that X is the same for each suit.

Now, let's do it for 8 players. If you give 1 random card to each one, there will be:

  • a 1/4 chance for each one that they'll get a club
  • a 1/4 chance for each one that they'll get a diamond
  • a 1/4 chance for each one that they'll get a heart
  • a 1/4 chance for each one that they'll get a spade.

It's the same 25 %. In this case, on average, there will be Y number of hearts left in the deck, Y number of clubs, Y number of spades and Y number of diamonds.

This means that, just as in the case of 3 players, Y is the same for every suit.

Now here's the key point: The next time you want to give to each of your 3 OR 8 players a card, the probability for each of them of getting a club, diamond, heart or spade is still 1/4, EVEN THOUGH there may be more or less cards in the deck.

This also applies to the community cards. And to the different ranks of cards as well (not just the suits). Which means that there's the same chance of someone making a straight or a flush when there 3 players at the table or 8 players.

So, bottom line:

  • if everyone plays every hand and goes to the river, the probabilities are the same
  • if not, then they are greatly influenced by a huge number of (sometimes subtle) factors, which means probabilities are kind of useless in this situation....
  • I see you understand my question. I am indeed talking about pure probability, not about "psicology-influenced" probability (which, as you correctly explain, is indeed affected by countless factors). But I am still not convinced: to compute the probabilities of me having a certain hand, I must take into account that some cards are not on the deck, but belong to a player. I do not know which cards have been dealt, but I know that the number of available cards on the deck is directly affected by the number of players on the table. Surely that must have an effect on the mathematical computations?
    – blueFast
    Commented Feb 8, 2013 at 10:36
  • @gonvaled See my edited post. Commented Feb 8, 2013 at 11:24
  • Thanks, I'll think about that and reply if any question is still open
    – blueFast
    Commented Feb 8, 2013 at 12:09
  • Very good answer. Commented Apr 6, 2013 at 6:59

Radu Murzea already gave a pretty thorough answer, but I wanted to chime in with another way to think about this that might help.

Lets imagine two games:

In the first game I deal hands to 9 players and then deal out the board.

In the second game I deal a hand to 1 player, muck 16 cards, and then deal out the board.

It should be obvious that player one gets exactly the same hand every time so the probability of any given hand is identical.

It should also be obvious that in the second game if I didn't muck 16 cards the probability of any given hand would still be the same - all I've done is perform a minor permutation on the deck - if the deck was random before it's just as random now.

Finally it should be obvious that there's nothing special about the first player, each of the 9 players should have the same probabilities of getting hands (I could muck hands for 4 players, deal player 5, and then muck the last 4 hands).

So the probability of any given hand on the river must be unaffected by the number of players.


No matter how many players are at the table, dealt hands are completely random.

After hands are dealt, some players decide to fold their weaker hands so as not to lose money. This means that stronger hands make it to the river.

The more people who are playing, the more likelyhood that someone has a very strong hand, which means that even middling hands will fold.

If you were playing 50-handed poker (ignoring the fact that there aren't enough cards to go around), the only cards worth playing would be connected broadways, ace-high flush draws, and pocket pairs above ~8 or 9. BECAUSE: a) connected broadways can make the nuts if there is no flush or pair on the board, b) ace-high flushes will make the nuts if the board isn't paired, and c) sets and full houses will often make the best hand on wildly disconnected boards.

  • 1
    Your last paragraph may not be relevant to the original question, however that is an interesting point you bring up, and I'd like to expand on it: A 10 handed Omaha game is kind of like being dealt 60 hold'em hands. (6 ways to make 2 card combos when you are dealt 4). And what you described pretty much falls in line with the approach for Omaha starting hand strategy, exactly for the reason you described.
    – TTT
    Commented Apr 6, 2013 at 14:39

I think I know how to explain this in a way you can understand. In a 10 player game of hold'em everyone gets two cards, so there are 20 cards missing from the deck. That obviously affects the odds of what will come on the board, but only if we know what their cards are. Since we can never know what our opponents hold then we can never make assumptions about the odds.

Another way to look at it is like this: You have 5,6 and you want to make a straight, however, everyone around you has 7,8 and 3,4. You have no chance to make a straight because all of your straight cards are in players hands. However, there is the opposite case where you have 5,6 but no one else has a 3,4,7, or 8. Your chances to make a straight are much better because the ratio of your straight cards left in the deck is much higher.

Since the pre-flop odds of getting cards are always the same, we can assume that everything will even out over time, no matter how many players are playing. There would be more deviation from hand to hand in a 10 player game than a 2 player game, but over all it will even out.

I think that makes sense...

So anyway, in a single hand you may have higher or lower chances to make the hand you want, but it would not affect the probability of a straight vs a flush for everyone. It only changes with regard to the hand you have.

That is my understand any way.


If I understand your question right then yes the number of players in the hand does change your odds of hitting a hand. If you have 2 suited cards in a nine handed card game then there is over a 95% chance of one of the other nine players having one of your outs, in fact there is over a 70% chance that 3 or more of your outs have been dealt to other players on the table. I have run this out on a thousand hands and only had the cards pan out once where one of the other 8 players didn't have at least one of the outs, and I didn't include the two burn cards. I am not that great in math but I had a moment of stupid brilliance at the table once and figured this out, but since I could not recall how I came up with the calculations, I ran it out as mentioned above. If someone does know the math involved I would love to see an example as I would like to apply it to other scenarios. I have not read all of these other answers but of the ones I have read they all talk about pot odds which is based on unknowns, but it is possible to determine true odds I'm just not smart enough to do it and I'm not friends with Caro who I am sure has a book of these calculations.


Asked to give a couple of more examples of using this little Scala program I posted, so here they are. Same "A-diamonds, Q-clubs" or "AdQc", but with 2, 4, 7, and 10 players at the table:

scala> e("AdQc", 2) 64%, with 2 players, for the hand AQo

scala> e("AdQc", 4) 37%, with 4 players, for the hand AQo

scala> e("AdQc", 7) 23%, with 7 players, for the hand AQo

scala> e("AdQc", 10) 16%, with 10 players, for the hand AQo

Varies from 64% for 2 players down to 16% for 10 players. That's 4 TIMES difference ...


  • You haven't really explained what the output means. Does this show the probability of getting a particular hand, or the probability of your hand being the best?
    – BowlOfRed
    Commented Jan 4, 2021 at 22:45

The issue is what is the probability distribution of the hands that go to the river.

If it is ten-handed, there are ten hands, and the probability distribution (ex ante) of hands is the same from one round to another.

Whether people play "tight" or "loose" determines how many hands go to the river, rather than being folded earlier. In this regard, the number of people playing will determine the probability distribution of hands that go all the way to the end.


To add to the other answers I think you're confusing two probabilities here.

The first one would be the odds of being dealt any two given cards. That probability is constant and doesn't get affected by the number of players.

Now when the game goes through the different streets usually the probability that matters is "given my current hand, what are the odds of other player having a better hand than I do now and what are the chances another player will have a better hand than me by the river" which lously are the pot odds and implied odds. Those do get affected by the number of players.

**So, in summary, the odds of being dealt an specific ser of cards are independent from the number of players. Everything that happens after that is, in fact, pretty dependant on the number of players.

If I'm the only player I still have the same odds to get deal AKs, but I'm certain to habe the best hand since there're no more players.**


Probability is about information. It is a measure of certainty: how certain can you be that a future event will or will not happen. It is dependent only upon the information you have.

You are dealt two cards. The other 50 are unknown. Whether they remain in the stub, or are dealt to 9 other players, or 20 other players, is irrelevant unless you can somehow get information about them. If, for example, on the first betting round 8 players paid money to see the flop, you might surmise that the likelihood of those 8 players having high cards is somewhat higher than normal, and so the stub might be relatively depleted of high cards to come.

But absent such information, the probability of you ending up with a certain hand based on your two cards does not change based on the number of other players. Imagine this scenario: you and an opponent are dealt into a heads-up game. Now the dealer takes 16 cards from the bottom of the deck--that would never come into play--and sets them aside. Does that affect your probabilities? Of course not. And it would make no difference if those 16 cards happened to be in the hands of 8 more opponents. You don't know what they are, so you can make no assumptions about them.

Having different numbers of players does affect the probabilities of what a typical winning hand might be, and it affects the relative value of hands. With more players, winning hands are typically stronger. That doesn't change the odds of what any particular hand might become--it just makes it more likely that one of the 10 players will catch something higher.

As far as the rank of hands goes, no number of players will affect that. Other things will: for example, if you use a stripped 32-card deck (8 per suit), then a flush should beat a full house (as it does in Mexican Stud and Manila). With any 7-card game, one pair is more likely than no pair, but we don't rank no pair hands higher--and that's fine. With 8 cards, straights are more common than trips. Whether you choose to recognize that in an 8-card game is up to you.

  • «You are dealt two cards. The other 50 are unknown. Whether they remain in the stub, or are dealt to 9 other players, or 20 other players, is irrelevant» I don't think that is true. Bringing it to the extreme, in a table with 26 players the probability of anyone having an ace giving two cards to each one is 100%, for all the cards of the deck have been dealt. Obviously, such probability is not the same when there are only two players on the table. Commented Feb 21, 2016 at 19:35
  • Of course. But the probability of you having any particular hand doesn't change. Of course the probability of "someone out of three" and "someone out of ten" are very different. Commented Feb 21, 2016 at 20:05
  • I see. You are utterly right on that. Commented Feb 22, 2016 at 16:06

You have a theory you cannot prove because that theory is false

The number of players has no effect on the probability of a hand in any way

2 players or 8 players you have two random cards from the deck of 52
The down cards dealt to players before you are random cards
The burn card is random
You still have 2 random cards from a deck or 52

2 players or 8 players the flop is 3 random cards from 50

2 players or 8 players the turn is 1 random card from 47

2 players or 8 players the river is 1 random card from 46

The down cards in a players hand or muck are random cards

Your chance of drawing queens is not effected by your position on the table nor number of players on the table. At the point you have 2 random cards from a deck of 52. Now your chance of queens being best is effected by number of players at the table.

8 players compared to 2 yes the chance of someone having straight goes up. But 2 or 8 a flush is still more random then a straight.

The rank of hands is based on 5 random from 52 and is not effected by number of players in the game.


You can only include known cards in your calculations. If you have two hearts in your hand, your chances of making a flush by the river are the same whether you're playing against one opponent or ten. If somehow you know that four other players are also holding two hearts, there's a much lower chance, because there are exactly three hearts left in the deck, and you need all of them. You may also know that your flush will be no good if you make it, but that's another issue. On the other hand, in poker variants such as Manila which are played with cards lower than 7 removed from the deck, it's much harder to make a flush because there are less cards of each suit - a flush draw has four outs, not nine; and with less ranks in the deck, it becomes a lot easier to make pairs, trips, and full houses. The hand rankings reflect the probabilities, and so a flush actually beats a full house, not because of the cards you can't see that are being held by other players, but because there are less cards in the deck to begin with.


How about this - I take the four Aces from the pack, shuffle them, place them on the table and ask you to pick one.

Your odds of picking the Ace of Spades is 1 in 4, 25%, 3/1 however want to call it, yes?

But before you get the chance, I remove a random card - I don't know which Ace it is.

Your odds of picking the Aces of Spades are still the same, 1 in 4, although you only have 3 cards to pick from - does that help or hinder the discussion?

  • Now somebody looks at the card that was removed. Did the probability just change? Nothing changed in the distribution of cards. Does it matter which person looked?
    – user1934
    Commented Mar 24, 2017 at 5:21
  • By the way, I call this thought experiment Schrödinger's Ace...
    – user1934
    Commented Mar 24, 2017 at 5:23

import scala.collection.mutable.HashMap def ee( h: String) : String= { var suit = "" if( h(1) == h(3)) suit = "s" else suit = "o" h(0).toString + h(2).toString + suit } var tab = HashMap("AAo" -> "" ) tab += ("JJo" -> "78 61 49 40 34 28 25 22 19") tab += ("JTs" -> "58 42 34 28 25 22 20 18 16") tab += ("JTo" -> "55 39 31 25 22 19 16 14 13") tab += ("J9s" -> "55 40 31 26 22 20 18 16 15") tab += ("J8s" -> "54 38 29 24 20 18 16 14 13") tab += ("TTo" -> "75 58 45 36 30 25 22 19 17") tab += ("T9s" -> "54 39 31 26 22 20 18 16 15") tab += ("T9o" -> "52 36 28 22 19 16 14 13 11") tab += ("T8s" -> "53 37 29 24 21 18 16 15 14") tab += ("99o" -> "72 54 41 33 27 22 19 17 16") tab += ("98s" -> "51 36 28 24 20 18 16 14 13") tab += ("88o" -> "69 49 37 29 24 20 18 16 14") tab += ("77o" -> "66 46 34 27 22 19 16 15 14") tab += ("66o" -> "63 43 31 25 20 17 15 14 13") tab += ("55o" -> "60 40 29 22 18 16 14 13 12") tab += ("44o" -> "57 37 26 21 17 15 14 13 12") tab += ("33o" -> "53 34 24 19 16 15 14 13 12") tab += ("22o" -> "50 31 22 18 15 14 13 12 12") tab += ("KKo" -> "82 69 58 50 43 38 33 29 26") tab += ("KQs" -> "63 47 38 33 28 25 22 20 19") tab += ("KQo" -> "61 44 35 30 25 22 19 17 15") tab += ("KJs" -> "63 46 37 31 27 24 21 19 18") tab += ("KJo" -> "61 43 34 28 24 20 18 16 14") tab += ("KTs" -> "62 45 36 30 26 23 20 18 17") tab += ("KTo" -> "60 42 32 26 22 19 17 15 13") tab += ("K9s" -> "60 42 32 27 23 20 18 16 15") tab += ("QQo" -> "80 65 54 45 38 32 28 25 22") tab += ("QJs" -> "60 44 35 30 26 23 21 19 17") tab += ("QJo" -> "58 41 33 27 23 20 17 15 14") tab += ("QTs" -> "60 43 35 29 25 22 20 18 17") tab += ("QTo" -> "57 40 31 26 22 19 16 14 13") tab += ("Q9s" -> "58 41 32 26 22 20 18 16 14") tab += ("A8o" -> "60 41 31 24 20 16 14 12 11") tab += ("A8s" -> "62 45 34 27 23 20 18 16 15") tab += ("A9o" -> "61 42 31 25 20 17 15 13 11") tab += ("A9s" -> "63 45 35 28 24 21 19 17 15") tab += ("ATo" -> "63 44 34 28 23 20 17 15 13") tab += ("ATs" -> "65 47 37 31 27 24 21 19 17") tab += ("AJo" -> "64 46 35 29 25 21 18 16 14") tab += ("AJs" -> "65 48 38 32 28 25 22 20 18") tab += ("AQo" -> "64 47 37 31 26 23 20 18 16") tab += ("AQs" -> "66 50 40 34 30 26 23 21 19") tab += ("AKo" -> "65 48 39 32 28 24 22 19 17") tab += ("AKs" -> "67 50 40 35 30 28 25 23 21") tab += ("AAo" -> "85 73 64 56 50 44 39 35 31")

val e = (st:String, pl:Int) => try{ var ans = new String() ans = tab( ee(st)).substring( 3*(pl - 2), 3*(pl-2) + 2) println("\n" + ans + "%, with " + pl + " players, for the hand " + ee(st) ) } catch{ case x: NoSuchElementException => {println("Don't play that") }}

scala> e("AdQc", 4) // table with 4 players

37%, with 4 players, for the hand AQo

scala> Enjoy.

  • Hi @user8349, can you explain the numbers and calculation in a more readable format please for those that don't program in this language, thanks.
    – Toby Booth
    Commented Jan 3, 2021 at 19:09
  1. Probability of flush (against straight) stays the same and does not depend on N
  2. Frequency of flush (against straight) on the showdown INSREASES with an insrease of number of players!

That's true, because range is strongly skewed to SUITED card, so many straights are being folded before she showdown.

Just an example, illustrating skew to suited hands in CO range - as you can see, most UNSUITED, which should make a straight, would be folded.

enter image description here


Old thread but still relevent, particularly with live attendence down and the resulting increased probability of finding an under subscribed table. The key here, as so eloquently explained previously is "blockers". Without going into the math, the general rule of thumb is that the higher the number of pocket cards dealt, the more powerful high pair becomes because players have fewer outs available for draws. The caveat here is that on a 10 player table like we have in Sydney, when facing a connected straight flop with both outside draws up for grabs, a trip set by the river becomes practically a coin toss against the straight. Should see the looks when I fold trips and they have the straight.

  • You should go into the math. I mean I get if you had a 22 handed holdem game not many cards left to make a hand, but it does not make much sense to me that the odds that are based on unknown cards of making any particular hand are going to change. All I see changing is that one is going to need a much stronger hand to drag a pot a showdown, so yeah laying down trips is likely always wise whenever there is a wet board.
    – Jon
    Commented Feb 27, 2021 at 4:26

I think it can be proven in a rather straight forward fashion. Though the probability of any one hand being dealt to a player doesn't diminish with number of players, the strength of a hand should increase as fewer players are available.

For example: a J7 hand would be pretty crap on a table with 10 players as there's a high probability that at least 1 other player has an equal or higher hand (it's an approximate certainty to be honest). However on a table with only 4 players the chances of their having that same advantage should be lower. Hence why playing by traditional probabilities works best with high numbers of players and at about 4 players the odds will favour those willing to bluff the crap out of everyone who are still playing the odds. (This is my experience) Hope it helps.


I think the most direct answer at the question is: yes, it absolutely affects the odds. But in most theoretical calculations relating to poker, we disregard that fact and simplify because we cannot for sure what the other cards are. Ergo, we regard the deck as containing 50 cards, namely all the cards in the original deck minus the two we're 100% certain are not in the deck. It's obviously an imperfect system, but likely over the long term the best system available ex ante.

The only exception to this rule - which can rarely be relied on - is if you know 3-4 players at your table play ultra conservatively (only 2 face cards KQ+), then your low suited connectors or low-medium pocket pair look fairly good and a call against a medium-sized raise appears not too shabby b/c those players odds of making a pair or higher are starkly reduced while yours are (this is an assumption) pretty much the same.

In short, your right, it affects the odds and the odds calculations, but we don't have a better system...

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.