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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.

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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 Feb 11 '13 at 10:22
    
@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 Apr 6 '13 at 14:43
    
What you suggest is impossible but I'm curious what leads to think that way. –  Jubbat Mar 24 at 5:02
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5 Answers 5

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....
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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? –  jeckyll2hide Feb 8 '13 at 10:36
    
@gonvaled See my edited post. –  Radu Murzea Feb 8 '13 at 11:24
    
Thanks, I'll think about that and reply if any question is still open –  jeckyll2hide Feb 8 '13 at 12:09
    
Very good answer. –  Bogdan Apr 6 '13 at 6:59
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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.

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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.

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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 Apr 6 '13 at 14:39
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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.

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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.

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