What is probability of flop having two of the same suit?

[cianosheehan]

Does anyone know the statistical probability of this happening? Disregarding your own hole cards etc.
Thank you.

[aloiz]

Flops of only two of the same suit:
4 * C(13,2) * 39 / C(52,3) = .55

Flops of all of the same suit happen:
4 * C(13,3) / C(52,3) = .052

Rainbow flops:
C(4,3) * 13^3 / C(52,3) = .40

aloiz

[Tharpab]

Could you explain this math?
I didnt learn at school what this C means =)
Isnt the probability
(50/50) * (12/50) * (11/50) *3???????

[Precision1C]

C in the Aloiz post represents the formula for finding out the number of combinations there are for choosing the second number of cards from a group the size of the first number. For instance C(52,3) is the number of combinations of choosing 3 cards, a flop, from 52 cards. The formula uses factorial where 5!=5*4*3*2*1 and C(N,K)= N!/[K!(N-K)!].

[aloiz]

C(x,y) is just the number of different ways you can "choose" y from a group of x. Say you have 5 different colored balls in an urn, and you want to know the number of different ways you can pick three balls from the urn. In other words how many distinct groups of three balls can be made from the five (note that the order in which you pick them does not matter, just the resulting group). So C(5,3) = 5! / (3! * 2!) = 10.
More generally C(x,y) = x! / (y! * (x - y)!)

So back to the original problem. If we were to try and figure out the odds that a flop comes with two cards of the same suit a single card at a time, we would need to account for all of the different ways a two suited flop could occur. XYY, YXY, and YYX. We could then figure out the probability of each occurring, and then add those probabilities together (note that each of the three will be close, they won't be exactly the same).
(39/52 * 13/51 * 12/50) +
(13/52 * 39/51 * 12/50) +
(13/52 * 12/51 * 11/50) ~= .138
Now we multiply that by four to account for all of the suits to get .55

My original way is much quicker and easier. If you can count the total number of possible two suited flops and divide by the total number of possible flops then you get your odds of a two suited flop occurring.

You have C(52,3) possible flops that can occur.
To count the number of two suited flops we “choose” a suit: C(4,1), then we choose two cards of that suit that will be in the flop: C(13,2), and finally we choose the last card from the remaining 39: C(39,1). Multiplying those together and dividing by C(52,3) gives us the our probability.

Hope that all makes sense.

aloiz

[Tharpab]

Got it. Much more precise than the old hacks =).
I just realized the windows calculator stinks and dont have C, my friend told me a hack to calculate it by hand but do you know an software the allows more hardcore math for download?

[Lost Wages]

Microsoft Excel has the "COMBIN" function for calculating C(r,n).

Lost Wages

[aloiz]

Excel has a COMBIN function that you can use. Also http://www.calculator.org/download.html has a free scientific calculator that can do combinations.

aloiz