Probability of opponents with flush draw
May 7th, 2016 at 6:06:53 AM
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Texas Holdem
7 opponents in the hand
Flop has 2 cards of spades
My two hole cards are not of spades
What's the probabilitity that at least one has a flush draw?
I'm having trouble to find the correct formula... Can you help me?
7 opponents in the hand
Flop has 2 cards of spades
My two hole cards are not of spades
What's the probabilitity that at least one has a flush draw?
I'm having trouble to find the correct formula... Can you help me?
May 7th, 2016 at 6:25:18 AM
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Assuming all hands are random...
The probability that any individual opponent has two spades is:
(11/47) * (10/46) = 0.05088
The probability that any individual doesn't have two spades is therefore:
1 - 0.05088 = 0.94912
The probability that all seven opponents do not have two spades is:
0.94912 ^ 7 = 0.69383
So the probability that at least one opponent has two spades is:
1 - 0.69383 = 0.30617 = 30.6%
Now the psychology comes into play. If some opponents folded before the flop, I would assume that they would be more likely to fold two random cards than they would fold suited cards. So the remaining hands are not actually random, and it is probably a bit more likely that someone is on a flush draw.
The probability that any individual opponent has two spades is:
(11/47) * (10/46) = 0.05088
The probability that any individual doesn't have two spades is therefore:
1 - 0.05088 = 0.94912
The probability that all seven opponents do not have two spades is:
0.94912 ^ 7 = 0.69383
So the probability that at least one opponent has two spades is:
1 - 0.69383 = 0.30617 = 30.6%
Now the psychology comes into play. If some opponents folded before the flop, I would assume that they would be more likely to fold two random cards than they would fold suited cards. So the remaining hands are not actually random, and it is probably a bit more likely that someone is on a flush draw.
