#1
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The odds of holecards both being suited
Hey everyone.
I like the maths that are in poker but unfortunatelly I am rather new to the game, and i havent gotten all the math I need. so here is my question for you: What are the odds that the two holecards are in the same suit? I would appreciate it if you gave me the answer in % *EDIT* I see now that the question was lacking an important piece of info... If I have two suited cards, what are the odds of someone other having it in the same suit to? (could be valuable info on flush draws, thats my thought) thank you |
#2
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Re: The odds of holecards both being suited
12/51= .2353= 23.53%
I'm not too good at maths either, though, so I'd wait for someone else to confirm this. Also, this is a pretty useless stat to know, there are lots of other more relevant things to learn. For further probability questions, you should try that forum. |
#3
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Re: The odds of holecards both being suited
[ QUOTE ]
12/51= .2353= 23.53% [/ QUOTE ] that is correct. |
#4
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Re: The odds of holecards both being suited
For them to have the same suit as you:
=(11/50)*(10/49)= .044897 So roughly 4.5% of the time |
#5
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Re: The odds of holecards both being suited
[ QUOTE ]
For them to have the same suit as you: =(11/50)*(10/49)= .044897 So roughly 4.5% of the time [/ QUOTE ] These are the odds that a specific opponent holds two cards of the same suit as you (given you are suited, and have not seen the flop yet). These are not the odds that _any_ opponent holds two cards of the same suit as you. |
#6
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Re: The odds of holecards both being suited
[ QUOTE ]
If I have two suited cards, what are the odds of someone other having it in the same suit to? [/ QUOTE ] Heads up, the chance the other player has 2 cards of the same suit is p = (11/50) * (10/49) = 4.490 % or once in every 22.3 attempts. Versus N opponents, an approximation that at least one player has 2 cards of the same suit is P(N) = 1 - (1-p)^N for N = 9 this is ~ 33.9 % As approximations go, its not too bad. Since the hands are not completely independent, I believe the second formula is only correct if You keep your two cards (leaving 50 in the deck) The other N guys randomly select 2 cards from the remaining 50. After each opponent selects 2 cards, he puts them back in the deck and you reshuffle for the next guy. |
#7
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Re: The odds of holecards both being suited
[ QUOTE ]
Heads up, the chance the other player has 2 cards of the same suit is p = (11/50) * (10/49) = 4.490 % or once in every 22.3 attempts. Versus N opponents, an approximation that at least one player has 2 cards of the same suit is P(N) = 1 - (1-p)^N for N = 9 this is ~ 33.9 % As approximations go, its not too bad. Since the hands are not completely independent, I believe the second formula is only correct if You keep your two cards (leaving 50 in the deck) The other N guys randomly select 2 cards from the remaining 50. After each opponent selects 2 cards, he puts them back in the deck and you reshuffle for the next guy. [/ QUOTE ] My probability is a little fuzzy, but doesn't P(N) = 1 - (1-p)^N calculate the probability that exactly one opponent has 2 cards of the same suit? If you wanted one or more players isn't it p * N? Here is another interesting one: Heads up, the chance the other player has 2 cards of the suit when 3 of the 5 cards on the board are the suit and your holding 2 cards of the suit is: p = (8/45) * (7/44) = 2.83% |
#8
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Re: The odds of holecards both being suited
I believe N*p is the average number of opponents that have 2 cards of the same suit as you. Let's pretend p = 0.20 (some other condition). 6*p = 1.20. This cannot be the percentage of times that at least one guy has met the condition. It is more the 100%. The probability is 1 - (no one meets the condition) The chance that one player *misses* (doesn't meet the condition) is 1-p. The change that N players miss is (1-p)^N . The probably this doesn't happen is 1- (1-p)^N |
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