# of unique preflop situations at a 10 player table (before UTG acts)

[Ludanto]

10 players are sitting at a table to play Hold'em. Every player gets dealt two cards. The player under the gun didn't act yet.How many unique situations are possible because of the different cards dealt to eacher player?
Here are my thoughts on this:

Calculation of possible hole card combinations:
[Number of possible first hole cards] times [the number of possible second hole cards]. The result is divided by 2 as the order of the cards does not matter.

1. Player: (52x51)/2 = 1326 (50 cards left in the deck)
2. Player: (50x49)/2 = 1225 (48 cards left in the deck)
3. Player: (48x47)/2 = 1128 (46 cards left in the deck)
4. Player: (46x47)/2 = 1035 (44 cards left in the deck)
5. Player: (44x43)/2 = 946 (42 cards left in the deck)
6. Player: (42x41)/2 = 861 (40 cards left in the deck)
7 .Player: (40x39)/2 = 780 (38 cards left in the deck)
8. Player: (38x37)/2 = 703 (36 cards left in the deck)
9. Player: (36x35)/2 = 630 (34 cards left in the deck)
10. Player: (34x33)/2 = 561

Number of possible unique Hold'em preflop situations:
1326 x 1225 x 1128 x 1035 x 946 x 861 x 780 x 703 x 630 x 561 =
299,348,225,803,817,914,056,933,600,000. In words: ...a LOT!

Questions:
1. Is this calculation correct?
2. Does the way of dealing the cards (every player gets one card first before the first player gets the second card) matter?
3. Do I need to divide my end result by 10! ? (because there are ten players) Hmmm... or maybe I should just ADD all the possible combinations of each player?

I'm confused now! [img]/images/graemlins/wink.gif[/img]
Please help!

[Ludanto]

I am starting to think that the calculation for unique hold'em preflop situations on a ten player table goes like this:
1326 is the number of unique starting hands possible for one player.
There would be 26 starting hands in one deck.

[(26x25x24x23x22x21x20x19x18x17) / (10x9x8x7x6x5x4x3x2x1)] x 1326
= 7,043,360,610 In words ...a LOT ...less than before [img]/images/graemlins/wink.gif[/img]

What do you think of my new idea? [img]/images/graemlins/smile.gif[/img]
(I am still not sure at all if this is right)

[Dromar]

[ QUOTE ]

Questions:
1. Is this calculation correct?
2. Does the way of dealing the cards (every player gets one card first before the first player gets the second card) matter?
3. Do I need to divide my end result by 10! ? (because there are ten players) Hmmm... or maybe I should just ADD all the possible combinations of each player?


[/ QUOTE ]

1. Yes... I think, lemme take a look again.
2. No. They're all random face down cards when they're dealt.
3. No. You might do this is position didn't matter, but it does.

[Dromar]

Okay. The calculation is correct IF:

1) you assume that Jc 5h (for example) is the same as 5h Jc.
2) you assume that Jc 5h is DIFFERENT from Jd 5h.

If those are your assumptions, I believe your math is correct.

[1huskerfan]

Yours sounds correct. It's a basic P(52,20)/2.

[BruceZ]

[ QUOTE ]
Yours sounds correct. It's a basic P(52,20)/2.

[/ QUOTE ]

You mean P(52,20)/2^10 = 52*51*50*...*43/2^10 = 52!/(52-20)!/2^10 =~ 3 x 10^29.

To answer the other questions, no it doesn't matter what order the cards are dealt, and you would only divide by 10! if you don't care which players get which hands.

[1huskerfan]

You are right. Divide by 2! for every hand (aka the "BANANA" problem)

[Ludanto]

I am assuming the following(if the situation would be the same with only 3 players):

PlayerA (AsTd), PlayerB (Ks4h), PlayerC (Jh7s) is the same as PlayerA (Ks4h), PlayerB (Jh7s), PlayerC (AsTd).

So if you would give every possible pocket pair (out of the 1326 possible ones) a number then for me it is the same if 10 players get 10 arbitrary pocket pairs no matter who gets which pair.

BruceZ, so this means I have to divide by 10! ?

Basically this was my way of thinking:
If I am sitting with 10 guys at a table to play hold'em then how many possible unique preflop situations could I face?
In my version of looking at it I consider that position does NOT matter. After all every person can bluff at any point and any time they want. So I have nothing 100% certain by which I could make a decision if I consider positions of the players.
I COULD take position into consideration if I would know that "the player UTG does CERTAINLY raise with AA" but this person might slow play them also.
I just wanted to have a calculation which involves no unknowns.

[Ludanto]

1huskerfan:

P(52,20)/2 doesn't sound right as this would give you this:

P(52,20) = 125994627894135

125994627894135 / 2 = 62997313947067,5

I don't think there can be one half of a possibility [img]/images/graemlins/wink.gif[/img]

[BruceZ]

[ QUOTE ]
I am assuming the following(if the situation would be the same with only 3 players):

PlayerA (AsTd), PlayerB (Ks4h), PlayerC (Jh7s) is the same as PlayerA (Ks4h), PlayerB (Jh7s), PlayerC (AsTd).

So if you would give every possible pocket pair (out of the 1326 possible ones) a number then for me it is the same if 10 players get 10 arbitrary pocket pairs no matter who gets which pair.

BruceZ, so this means I have to divide by 10! ?

[/ QUOTE ]

Yes.

You might also want to consider deals the same if they simply transpose suits (hearts->spades, spades-> clubs, etc.). For deals with 3 or 4 different suits, there are 4! = 24 deals with transposed suits (except for rare deals where 2 suits are only on pairs, so that transposing suits makes no difference). The deals with only 2 different suits would usually only have 3! = 6 transpositions, but these are very rare, so dividing by 24 should be almost exact.


[ QUOTE ]
[ QUOTE ]
1huskerfan:

P(52,20)/2 doesn't sound right as this would give you this:

P(52,20) = 125994627894135

125994627894135 / 2 = 62997313947067,5

I don't think there can be one half of a possibility [img]/images/graemlins/wink.gif[/img]

[/ QUOTE ]

[/ QUOTE ]

P(52,20) is a lot bigger than that, it is ~3 x 10^32. It is divisible by 2 and also 2^10, and huskerfan already said that he meant to divide by 2^10. You computed C(52,20), not P(52,20). P(52,20) = C(52,20)*20!.