plz help....i cant figure this out

[vkotlyar]

what is the probability of flopping a 4-flush or an open ended str8 draw on the flop with 10Js. i know that you are going to complete the draw about 33% of the time. Can some1 please explain to me how to perform this calculation.
thanks
--vitaly

[Cyrus]

"What is the probability of flopping a 4-flush or an open ended str8 draw on the flop with 10Js?"

It comes down to a 7% probability of flopping a 4-flush draw (and nothing else) and to a 8.5% probability of flopping an open-ended (8-outs) straight draw (and nothing else).

You have a 16% probability of flopping a 4-flush draw or an open-ended straight draw (but nothing else).

If you have flopped the 4-flush draw, the probability of completing to a flush with 2 cards to come is 35%; if you have flopped the open-ended straight draw, the probability of completing with 2 cards to come is 31.5%.

[switters]

I think these numbers are too high. I posted a response to vkotlyar's similar post in the mid-high stakes section:
link to other post...

if you hold T[img]/forums/images/icons/spade.gif[/img]J[img]/forums/images/icons/spade.gif[/img], there are 11 [img]/forums/images/icons/spade.gif[/img]s left.

the number of flops that contain two of them are:

number of two card [img]/forums/images/icons/spade.gif[/img] combos = (11 choose 2) = 110
*TIMES*
the number of non-[img]/forums/images/icons/spade.gif[/img] cards remaining = 39
= 4290 ways to flop a four-flush.

there are (50 choose 3) possible boards, given whatever hand you have = 117600

which means you flop a four-flush 3.64% of the time.

I see now that I made a small error in my other post, in that there are more 8-out straight draws than I gave credit for...

ways to flop an 8-out straight:
true "open-enders": 89x, Q9x, KQx, where x doesn't complete the straight.
there are 4 * 4 * 40 of each of these (4 8's, 4 9's, and 40 remaining cards that don't complete the straight)
double-gut shots: 79K, 8QA
there are 4 * 4 * 4 of each of these

so:
3 * (4 * 4 * 40) + 2 * (4 * 4 * 4) = 1920 + 128 = 2048

dividing by the number of flops: 2048 / 117600 = 1.74%

also -- you can't just add these up to get

3.64% + 1.74% = 5.38%, because many of the hands are being "double-counted" (i.e., we have counted 8[img]/forums/images/icons/spade.gif[/img]9[img]/forums/images/icons/spade.gif[/img]2[img]/forums/images/icons/diamond.gif[/img] as both a flush draw and a straight draw). The real number works out to more like 5.10%

I'm pretty sure my math is solid here, but I'd love to hear about it if there are disagreements -- Cyrus, where did you get your numbers?

-switters

[irchans]

50 Choose 3 is 19600.

[irchans]

switters,

I think you reasoning is correct, but your calculator is broke. The computations involving the choose function seem to be using the permutation function instead. When we fix that, your probabilities become:

4- flush = c[11, 2]* (3*13) /c[50, 3.] = 10.9%
4-straight = 2048./19600. = 10.4%

[switters]

Thanks!

you're absolutely right -- I didn't have a calculator with choose[] handy, and I was accidentaly computing the permutation function ((50! / (50-3)!) = 50 * 49 * 48) rather than the choose function ((50! / ((50-3)! * 3!)) = 50 * 49 * 48 / 6) which is why I was counting 117600 flops instead of the real number, 117600 / 6 = 19600!

thanks, irchans! apologies, Cyrus!

-switters

[Cyrus]

"Number of two card combos = (11 choose 2) = 110 ... There are (50 choose 3) possible boards, given whatever hand you have = 117,600"

C(11,2)=55 ...You forgot to divide by (2*1) [img]/forums/images/icons/wink.gif[/img]
C(50,3)=19,600 ...117,600 is the # of possible Permutations

The archived thread also has wrong figures.

[SittingBull]

have a scientific calculator,use the combination formula:
nCr=n!/r!(n-r)!
n!=n(n-1)(n-2)(n-3)...1
Example:
5!=5(5-1)(5-2)(5-3)(5-4)=5(4)(3)(2)(1)=120
Example;
compute 6C4
Ans
6C4=6!/4!(6-4)!
6!=6(6-1)(6-2)(6-3)(6-4)(6-5)=6*5*4*3*2*1=720


If U have a scientific calculator,U will usually see nCr
on the calculator.
nCr means that the order is not important.
For example,a selection of Ac,9h,2d is the same as the selection 2d,Ac,9h.
Let's assume U want to find out how many ways of choosing 2 K's from a deck of 52 .
There are 4 K's in the deck--Kc,Kh,Ks,Kd
Ans:
4C2=4!/2!(4-2)! = 4*3*2*1/2*1*2!=4*3*2*1/2*1*2*1=6
Note that Kc,Kh and Kh,Kc is the same selection--so U just count it as 1 selection.
The order is not important--just the selection.
********************************
The probability of any event occurring is the total # of favorable outcomes/total # of outcomes
Example for the 4-flush problem

Since U have 2 of the 13 suits,there are 11 of your suits remaining.
The # of DIFFERENT ways of selecting 2 of the remaining 11 suits is given by 11C2=11!/2!(11-2)!
These selections are all favorable to your oucome.
If U subtract the 13 suits from the total # of cards in the deck,52,then there will be 39 of these that are non - suited of your type of suit. 11C2 gives U the # of ways of flopping two different suits ,that is ,two of your suit. But there is a 3rd card to complete the flop.
So for each way of selection 2 of your suited cards from the 11 that are suited,there are 39C1 ways of selecting the remaining one non-suited card to complete the flop.
Hence,U will have a favorable flop 11C2*39C1 times.
These are your favorable outcomes.
Since U have 2 cards in your hand,there are 50 unseen cards.
U can select 3 of these 50 unseen cards 50C3 different ways.
These are BOTH favorable and unfavorable outcomes(the total # of different outcomes)
The probability=total # of favorable outcomes/total # of both favorable and unfavorable outcomes.
So Probability=11C2*39C1/50C3 for a 4-flush flop.
Happy Poering, [img]/forums/images/icons/laugh.gif[/img]
SittingBull

[Cyrus]

Greetings, Sitting Bull.

"If U have a scientific calculator..."

What's wrong with using Excel? We are in front of our computer anyway. [img]/forums/images/icons/smile.gif[/img]

"Example for the 4-flush problem..."

Have you taken out the flops that result in either flopping a straight or flopping a 4-straight draw? We need to have the "clean" 4-flush flops only. (Pairs and/or pairing the board are acceptable.)

--Cyrus

[SittingBull]

on my computer.I'm a computer illiterate. [img]/forums/images/icons/frown.gif[/img] No,I haven't considered the Str. cases.
Happy Pokering, [img]/forums/images/icons/laugh.gif[/img]
SittingBull