Permutation Flush Draw Probabilities

[Splitrig]

What are the true odds of getting a flush with two suited hole cards?
Poker Indicator seems to suggest that you have a 3% chance.
When I calculate it based on a full table I get 98/10842 = 1% chance.
When looking at over 10 million online hands the EV of AKs is .99 and AK is .61.
What is a girl to think?

[BruceZ]

[ QUOTE ]
What are the true odds of getting a flush with two suited hole cards?
Poker Indicator seems to suggest that you have a 3% chance.
When I calculate it based on a full table I get 98/10842 = 1% chance.
When looking at over 10 million online hands the EV of AKs is .99 and AK is .61.
What is a girl to think?

[/ QUOTE ]

If you stay to the river, the probability that you make a flush in your suit is exactly

[C(11,3)*C(39,2) + C(11,4)*39 + C(11,5)] / C(50,5)

=~ 6.4%.

This is the sum of the probabilities of getting a 5, 6 and 7 card flush. There are 11 flush cards remaining, and 39 non-flush cards. The number of players at the table is irrelevant, but you will get a different answer if you know their hole cards and take them into account.

What did you compute with Poker Indicator? What is 98/10842?

[Splitrig]

Poker indicator gives you 3% higher odds when you add suite to the mix.

I assumed that you you would get Spade Spade. Then I assumed 10 people at the table. 20 cards are delt and 32 remain. Chances are that 5 Spades are out and I have two of them. There are 10,842 probably flop+turn+river (32!/27!). Given that there are most likely 8 spades remaining, this means that there are 98 possible hands that make a spade flush (8!/5!).

Permutation Formula nPr = n!/(n-r)!

[BruceZ]

[ QUOTE ]
Poker indicator gives you 3% higher odds when you add suite to the mix.

[/ QUOTE ]

What does this mean? Was the 3% for unsuited hole cards? That number should be 1.9% for making a flush in either suit. Did the program give 6.4% for suited hole cards? This is the correct number.


[ QUOTE ]
I assumed that you you would get Spade Spade. Then I assumed 10 people at the table. 20 cards are delt and 32 remain. Chances are that 5 Spades are out and I have two of them. There are 10,842 probably flop+turn+river (32!/27!). Given that there are most likely 8 spades remaining, this means that there are 98 possible hands that make a spade flush (8!/5!).

Permutation Formula nPr = n!/(n-r)!

[/ QUOTE ]

This is incorrect for a couple of reasons. If you assume that there are 8 spades remaining out of 32 cards, the probability of making a flush is not 8P3/32P5. This only counts permutations of the 3 spades, and you must multiply this by the number of possibilities for the 2 non-spades. If we are only interested in 5-card flushes, the correct way to compute this would be:

C(8,3)*C(24,2) / C(32,5) =~ 7.7%

where C(x,y) = x!/(x-y)!/y! = xPy / y!. There are C(8,3) ways to choose the 3 spades, times C(24,2) ways to choose the 2 non-spades, out of C(32,5) total boards. If we also consider 6 and 7 card flushes, this becomes

[C(8,3)*C(24,2) + C(8,4)*24 + C(8,5)] / C(32,5)

=~ 8.5%

Another problem is that unless you have a read that tells you which particular cards in your opponent's hands are spades, you should not assume that there is a particular number out, but instead treat all of the opponent's cards as unseen. This would be equivalent to computing the result for each possible number of spades from 0 to 11, and weighting the results in proportion to their probabilities. Also, the average number of spades in your opponent's hands is not 3, it is (11/50)*18 = 3.96. If we compute the result assuming 4, so that there are 7 remaining, this gives a number close to the exact answer I calculated earlier:

[C(7,3)*C(25,2) + C(7,4)*25 + C(7,5)] / C(32,5)

=~ 5.7%

The exact calculation gave approximately 6.4%.