Question on poker odds?

User45701

i was thinking i know theres odds that u can work out depending whats on the turn flop and the river but what are the odds before any cards get shown?

Can someone tell me what the odds are in a 6 person game of more than 1 person having 2 cards of the same suit?

last nite when iwas running low i just decided to go all in on ace/2 clubs and beat pocket kings when flush comes up
but i wanted to know what the odds are of someone elce having 2 cards of the same suit ?

Brayruit

Starting the calc. off simply, with just 2 players... if you look down at two cards of the same suit, the prob that the other player has 2 cards in the same suit as you is (11/51) * (10/50) = 4.3% or about 1 in 23 times.

Now with 3 players, the prob. that the first player (say the one on your left) has 2 of same suit is still 4.3%. The prob. that the second player has it is also 4.3%. BUT the prob. that they BOTH have 2 cards of the same suit is (11/51) * (10/50) * (9/49) * (8/48) = about 0.1% and this has to be deducted to avoid double counting so that the overall calc is 4.3 + 4.3 - 0.1 = 8.5%

You could keep going like this but it is a bit tedious to make all the adjustments. A reasonable approximation 6 handed for 1 or more players to have 2 cards the same suit as your 2 cards would be 4% x 5 = 20% or about 4 to 1 against.

Mellor

Brayruit wrote:

Starting the calc. off simply, with just 2 players... if you look down at two cards of the same suit, the prob that the other player has 2 cards in the same suit as you is (11/51) * (10/50) = 4.3% or about 1 in 23 times.

Now with 3 players, the prob. that the first player (say the one on your left) has 2 of same suit is still 4.3%. The prob. that the second player has it is also 4.3%. BUT the prob. that they BOTH have 2 cards of the same suit is (11/51) * (10/50) * (9/49) * (8/48) = about 0.1% and this has to be deducted to avoid double counting so that the overall calc is 4.3 + 4.3 - 0.1 = 8.5%
You could keep going like this but it is a bit tedious to make all the adjustments. A reasonable approximation 6 handed for 1 or more players to have 2 cards the same suit as your 2 cards would be 4% x 5 = 20% or about 4 to 1 against.

Shouldn't that be (11/50) * (10/49). Seeing as you are holding the 51st and 52nd card. Unless of course you are playing with a jokers
So that would make it 4.5%.

If I was working out the odds I wouldn't subtract the 0.1%. I know what you mean that if one player has the suit then the next player has a decreased chance of getting the suit. But if he doesn't have them the next player has an increased chance of gettign the suit and this cancels the decrease.
I remember having to prove this cancelling out for a similar problem way back when in 6 year.

Brayruit

Mellor wrote:

Shouldn't that be (11/50) * (10/49). Seeing as you are holding the 51st and 52nd card. Unless of course you are playing with a jokers
So that would make it 4.5%.

If I was working out the odds I wouldn't subtract the 0.1%. I know what you mean that if one player has the suit then the next player has a decreased chance of getting the suit. But if he doesn't have them the next player has an increased chance of gettign the suit and this cancels the decrease.
I remember having to prove this cancelling out for a similar problem way back when in 6 year.

You are quite right on the first point of course... (shd be 11/50 not 51, etc.). I don't know what I was thinking...

No ... you are wrong on the 2nd point, you have to make the deduction to be accurate.

Mellor

I think i might be mixing this up with slightly different problem.
Where you are woorkign out the second players odds and to correct the effects of the first player.
In that case you are right. But .01% is really not that much adjustment.
About 22% in a six handed game. Never really thought of that.
To the original poster, that only works preflop, as new info on the flop will cause that to go up or down.

LuckyLloyd

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