Math regarding flush

How do you calculate the exact probability of following:

1. Getting two suited cards on the flop (holding 2 of the same suit in the pocket)

2. Getting AT LEAST two suited cards on the flop (holding 2 of the same suit in the pocket)

3. Getting three suited cards on the flop (holding 2 of the same suit in the pocket)

Would be great if anyone could help me out with this.
Thanks !

[disjunction]

[ QUOTE ]
How do you calculate the exact probability of following:

1. Getting two suited cards on the flop (holding 2 of the same suit in the pocket)Would be great if anyone could help me out with this.


[/ QUOTE ]

How would *YOU* approach #1 ? If you give your thoughts I can walk you through it.

[JinX11]

I believe the following is correct (post in Probability for a sure answer):

1) Odds of two suited cards pre-flop: 23.5%, or 3.25-to-1.

Calculated as: 12/51.

2) Odds of flopping a 4-flush: 10.94%, or 8.1-to-1.

Calculated as: There are 11 cards of your suit left, you want exactly two to match your suit. So, this is described as "11-choose-2" or C(11,2), the number of combinations of cards that match your suit. The formula for calculating C(x,y) is x! / ((x-y)! * y!), which would be 55 combinations that meet your criteria. There are 52-13 = 39 other cards that are NOT of your suit.... Finally, there are "50-choose-3" possible flops - C(50,3) = 19600.

So, of all of the possible flops, the chance that two will be of your suit and one will NOT be of your suit is described as:

(C(11,2)*39) / C(50,3) = (55*39) / 19600 = 10.94%


3) Odds of flopping a flush (which, thankfully, is much easier to calculate than a 4-flush): 0.8%, or 118-to-1.

Calculated as:

C(11,3) = 165 combinations that all match your suit.
C(50,3) = 19,600 possible flop combinations

C(11,3)/C(50,3) = 0.0084, or 0.8%


I think I have that all correct.... [img]/images/graemlins/confused.gif[/img]

Your calculations looks right I think - thanks alot for helping me out.
Disjunction, could you confirm that his calculations are correct? (u seem to know about this)

Thx alot.

[JinX11]

No problem - seriously, those dudes over in Probability live for deriving this stuff. You should post there to be sure....

[K C]

I'll walk you through this to so that you can get enough of a feel for how these things are calculated to be able to figure out similar problems yourself.

You have two of the suit. First we'll figure out the chances of flopping the flush, which is very easy to do. You have 50 unknown cards, of which there are 11 of the suit. So we've got 11/50 * 10/49 * 9/48

Getting at exactly two suited cards on the flop is a bit different. We need to take the three ways that this could occur and add them up, but we must be careful to calculate the whole series here, including the probability for the card that didn't hit. This will leave us 3 possible combinations - A and B, A and C, and B and C:

A and B but not C will be: 11/50 * 10/49 * 39/48
A and C but not B will be 11/50 * 39/49 * 10/48
B and C but not A will be 39/50 * 11/49 * 10/48

If you look closely what we've done is accounted for the cases where we didn't hit in all of the outcomes. For example, in the last one where it's B or C, A didn't hit, so we have to use the probability it didn't. The same goes for the other two cards when they don't hit. You take these probabilities and now you can just add them up.

As for getting at least two, this means either two or three. Now that we know the probability of getting exactly two and exactly three, we can just add the two together.

KC

Great explaination, very easy to follow. Thank you very much for quick and good answer !

BR

[srm80]

i have an idea...play 100,000 hands online and make a chart for each time you get suited cards and whether it flops 1,2, or 3 of your suit! Of course this would take awhile...but what the hell!

[disjunction]

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Your calculations looks right I think - thanks alot for helping me out.
Disjunction, could you confirm that his calculations are correct? (u seem to know about this)

Thx alot.

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Yeah, looks good to me. Of course there's more than one way to skin a cat, personally, I would say, what's the chances of the 1st flop card being my suit? 11/50. Given that, chances of the 2nd being my suit is 10/49. Given that, the chances of the 3rd not being my suit is 39/48. Multiply those together, it's 3.64%. But there are two other symmetric cases (Case2: the first and third flop card are my suit, Case3: the second and third flop card are of my suit). So multiply 3.64% x 3 is 10.9%.

This method avoids the combinatorials and is a little easier to do in your head if you are so inclined. Incidentally, as my public service announcement, if you can not do these calculations in your head someday, it may not be possible to ever be that good at poker.

i think if your ready to play for real you need to memorize every statistic.this would only be the beginning.so have fun on your journey of the numbers.
by the way if you have two suited whole cards you have a 3 percent chance of completing a flush. so 97 times you wont make your flush. after you memorize every statistic you will then need to engage in several phsycology books . but if you love the game as much as i do youl love the whole process. good luck dude.