Hello. I read the following in a poker article and cannot derive the math:

"If you are heads up post flop and the flop comes with three spades and you have no spades in your hand, the odds that your opponent has a made flush is 3.3% and the odds that he has a flush draw are 15.8%."

I can't follow the math to arrive at this answer.

To me, your opponent either has 0 spades, 1 spade, or 2 spades. And the probabliity of these three events must sum to 100%. So the probability of 0 spades + probability of 1 spade + probability of 2 spades = 100%.

There are 10 unseen spades, 37 non-spades, and a total of 47 cards unseen cards. Therefore:

probability of 0 spades = (37 / 47) * (36 / 46) = 62%

probability of 2 spades = (10 / 47) * (9 / 46) = 4%

probability of 1 spade = 1 - (62% + 4%) = 34%

So I have him with a made flush 4% of the time and with a flush draw 34% of the time.

Am I correct (and the article is wrong)? Iff not, where am I wrong in my thinking?

Thank you for your input!

Did the article make any assumptions about the range the player gets to the flop with? 72o can have 1 club in it but I wouldn't expect to find it on the flop often

@tjpoker: I just checked to verify and, no, the article didn't make any pre-flop range assumptions.

You are correct.

it assumes villain could have any 2 cards.

Add up all the combinations of cards villain could have (remember to remove the 3 spades from the flop) then add up how many of those combinations have either 2 clubs or 1 club. I guess you'll find 3.3% of combinations have 2 clubs and 15.3% have 1 club.

@dogarse: thanks for your reply.

However, if he has a made flush 3.3% of the time and a 4 card flush draw 15.3% of the time, it means that he has 0 spades 81% of the time. But that can't be since (37/47 * 36/46) equals 64%.

So does he indeed have a 4 card flush draw 34% of the time?

I haven't done the math but if 25% of all combination have a spade and then you remove 3 of the 13 spades it makes rough sense to me that now around 18.6% of the combinations have a spade.

I don't think this is true. Yes, Each suit should have an equal number of combinations in which there is at least one of that suit. But these combinations overlap, therefore greater than 25% of combinations contain a spade.

There is a small error in your calculation -- take any two cards out of 52 full deck, the probability of those two cards have 1 spade is around 38%, not 25%.

Haha, well I just makes rough sense, not that it's correct.

Okay, lets do it the old fashioned way.

(78 possible unpaired hands x 16 combinations) + (13 paired hands x 6 combinations)
equals 1326 possible combinations.

Each 16 card combination contains 7 combinations of a spade, 1 of which is suited.

Each 6 cards combination contains 3 combinations of spades.

If we remove the AKQ of spades (we can effectively remove all AKQ because we are not interested in the hands that contain no spade) there are now 45 possible unpaired hands (which each contain 1 suited and 6 off suit spades) and 10 possible paired hands (which each contain 3 spades).

So chances of having a flopped flush are 45/1326 or 3.4%

Chances of having a flush draw are (45x6) + (10x3)/1326 or 22.6%

Any mistakes in that math?

I think you're thinking is likely to be on the right lines, though I'm pretty sure that we should be dividing by less than 1326 since some of those hands are not possible (were holding two non spades)

Also you're not including hands like 9sAo which should be.

Yeah good point... Math is hard.