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vkotlyar
05-04-2003, 06:06 PM
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
05-06-2003, 12:37 PM
"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
05-06-2003, 04:24 PM
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... (http://www.twoplustwo.com/forums/showflat.php?Cat=&Board=mediumholdem&Number=251534 &page=0&view=expanded&sb=5&o=14&fpart=)

if you hold T/forums/images/icons/spade.gifJ/forums/images/icons/spade.gif, there are 11 /forums/images/icons/spade.gifs left.

the number of flops that contain two of them are:

number of two card /forums/images/icons/spade.gif combos = (11 choose 2) = 110
*TIMES*
the number of non-/forums/images/icons/spade.gif 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/forums/images/icons/spade.gif9/forums/images/icons/spade.gif2/forums/images/icons/diamond.gif 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
05-06-2003, 05:41 PM
50 Choose 3 is 19600.

irchans
05-06-2003, 05:52 PM
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
05-06-2003, 06:03 PM
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
05-07-2003, 12:46 PM
"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) /forums/images/icons/wink.gif
C(50,3)=19,600 ...117,600 is the # of possible Permutations

The archived thread also has wrong figures.

SittingBull
05-07-2003, 09:10 PM
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, /forums/images/icons/laugh.gif
SittingBull

Cyrus
05-08-2003, 12:30 AM
Greetings, Sitting Bull.

"If U have a scientific calculator..."

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

"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
05-08-2003, 09:16 PM
on my computer.I'm a computer illiterate. /forums/images/icons/frown.gif No,I haven't considered the Str. cases.
Happy Pokering, /forums/images/icons/laugh.gif
SittingBull

Cyrus
05-11-2003, 08:58 AM
"I haven't considered the Straight cases."

You need to consider those cases too. As well as lots of others. The calculation of the exact probability of flopping only a 4-flush draw (not a flush, not a SF, not a straight, not a straight draw), when holding JTs, is extremely tedious! (Pairing your hand or a pair on the flop are acceptable.)

"I do not know how to use "EXCEL" on my computer. I'm a computer illiterate."

So am I! But I know how to use Excel a bit. Here's how to get the program to do Combinations for you: Say you want to calculate the number of ways you can combine 10 items 2 at a time without respect to order.

Open a new File. Write 10 in cell A1 and 2 in cell A2. Then click on cell A3.

On the Command Toolbar, at the top of the screen, there must be a command icon titled fx. Click on it.
(If the Toolbar doesn't contain that command icon, click View > Toolbars > Customize. Then from the box that opens, choose Commands > Insert > fx Paste Function. Left-click the mouse, keep it clicked and drag the icon of "fx Paste Function" on your toolbar. Presto, you have the fx icon on your Toolbar. Click CLOSE.)

When you click on the fx icon, another box appears. It contains all the functions you can use. Click on ALL in the left frame of the box, scroll down in the right frame and click on COMBIN. Then click OK.

Another box appears. (Bear with me because it's all a matter literally of two seconds as soon as you orient yourself!)

Click on the top row, titled "Number", to activate it and write in it

a1

then click on the second row, titled "Number_chosen", and write in it

a2

(Doesn't matter if you use proper caps or not.) This way, what you did is you ordered the program to accept the value that appears in cell A1 as the number of all the items you choose from, and the value that appears in cell A2 as the number of items in each combination. The answer should appear in cell A3, which is the cell that you have clicked on right before calling up the fx icon to perform for you.

Now, if it all went well (and I haven't forgotten anything!), you should click on OK, and then see the number 45 appearing in A3. That's truly C(10,2).

Change the numbers in cells A1 and A2 to fool around. Writing 15 in cell A1 and hitting ENTER, gives 105 in A3, which is C(15,2). Try also using some other functions from the fx icon in this manner, for example PERMUT.

Take care.

SittingBull
05-12-2003, 05:29 PM

BruceZ
05-12-2003, 05:54 PM
4-straight = 2048./19600. = 10.4%

This is 3*16*40 + 4*4*4*2 = 2048, but that overcounts cases where we pair the board. It must be

[ 3(16*34 + 6*4*2) + 4*4*4*2 ]/19,600 = 9.7%.

BTW, if we ignore the double belly busters, the number of 4-straight flops is 1776, a nice number.

BruceZ
05-12-2003, 09:12 PM
Or you could just click in a cell and type =combin(10,2). Done.

The calculation of the exact probability of flopping only a 4-flush draw (not a flush, not a SF, not a straight, not a straight draw), when holding JTs, is extremely tedious! (Pairing your hand or a pair on the flop are acceptable.)

It need not be extremely tedious if you know what the frick youíre doing, and choose an appropriate partition for the sample space. I claim you are not a proper judge of what must be tedious and what is not tedious, as I will now prove:

First of all, since we are allowing pairs, Iím assuming you want to allow 2-pair and 3-of-a-kind as well. If you want to disallow any of these, including pairs, let me know and I will do that just as easily, I would just have to change a few numbers in what follows. The probability we are after is equivalent to the probability of flopping exactly a 4-flush draw without the cards for a straight draw, since if we donít flop the cards for a straight draw, then we donít flop a straight either. We can always flop a 4-flush draw without the straight draw cards if we flop a 4-flush draw with no more than 1 of the 4 denominations 8,9,T, or K (more than one of the same denomination is OK, like 888 since we are allowing this). We can also always do it when we flop only the denominations 8,K. We can also always do it if we flop only 8,Q or 9,K so long as we donít flop exactly 8,Q,A or 7,9,K, since these are the only double belly buster draws. These are all the cases, since the other combinations of two cards 8,9; 9,Q; and QK are straight draws. So now we count the cases.

No 8,9,Q,K: C(7,2)*27 = 567
Exactly 1 denomination of 8,9,Q,K: 4*7*30 + C(7,2)*12 = 1092
8,K: 2*3*7 + 1*33 = 75
8,Q but not 8,Q,A: 2*3*6 + 1*30 = 66
9,K but not 7,9,K: same as above = 66

Where I have added two terms, the first is for exactly 1 of the listed cards being a flush card. The second term is for either exactly 0 or 2 of the listed cards being flush cards, as appropriate to the case. Exactly 1 denominaton means multiple of same denomination still OK.

(567 + 1092 + 75 + 66 + 66)/C(50,3) = <font color="red">9.5%</font color>. This is the probability of a clean flush draw, exact to within the probability of a blunder, which in the limit should go to zero. /forums/images/icons/laugh.gif

Now the probability of a flush draw including the times we also make a straight draw or straight is
C(11,2)*39/C(50,3) = 10.9%. So the probability of making a flush draw while at the same time also making either a straight draw or straight is 10.9% - 9.5% = 1.4%. The probability of making a straight draw or straight including the times we also make a flush draw is [ 3(12*42 + 4*33 + 6*4*2) + 4*4*4*2 ]/C(50,3) = 11.1%. Therefore, the probability of a straight draw or a straight is 11.1% - 1.4% = 9.7%. The probability of a straight including straight-flush is 4*4*4*4/C(50,3) = 1.3%, so the probability of a clean straight draw is 9.7% - 1.3% = <font color="red">8.4%</font color>. The probability of a straight draw or a flush draw or both is 9.5% + 11.1% - 1.3% = <font color="red">19.3%</font color>.

This is in agreement with the figures given earlier for the clean straight draw, but not for the clean flush draw. Since we used one to get the other, the earlier figures do not appear to be consistent with each other.

So in summary:

P(clean flush draw) = 9.5%
P(clean straight draw) = 8.4%
P(clean straight draw OR clean flush draw not both) = 17.9%
P(clean straight draw OR clean flush draw OR both) = 19.3%
P(both clean straight AND clean flush draw) = 1.4%

Now was that so tedious?

BruceZ
05-12-2003, 11:36 PM
You can't flop those hands with either a 4-flush or 4-straight draw, so we don't have to worry about whether or not they are allowed, and these are being handled correctly above. 888 is not one that is possible, the most you can get is 2 of a denomination.

The probability of making a straight draw or straight including the times we also make a flush draw is
3(12*42 + 4*33 + 6*4*2) + 4*4*4*2 ]/C(50,3) = 11.1%

This should subtract the 2 double belly buster straight flush draws, but it doesn't change the final answer.

3(12*42 + 4*33 + 6*4*2) + 4*4*4*2-2 ]/C(50,3) = 11.1%

Cyrus
05-13-2003, 03:59 AM
"Or you could just click in a cell and type =combin(10,2). Done."

I didn't know that!.. And here I was trying to entice SittingBull to go through the motions and getting familiar with Excel or even his computer. You know, as in trying to fish? Silly me, typing all that tedious stuff. Thanks for straightening me out, Bruce, appreciate it.

"It need not be extremely tedious if you know what the frick you're doing, and choose an appropriate partition for the sample space. I claim you are not a proper judge of what must be tedious and what is not tedious, as I will now prove."

When you calm down enough to lose the all-caps from your posts, you might just realize that tediousness is a highly subjective term, and what is tedious for me, may not be for you or someone else. You consider the exercise not tedious, and I do, because, if you want to know, the cost/benefit ratio is too high for me. We learn little of practical substance, through this many-step exercise. An exercise which you should have realized, even if "I don't know what the frick (frick?!) I'm doing", and haven't actually performed it (yeah, right..), I must have read it and understood what the "frick" was being said. That's why I gave the correct answer.

Unless you are confusing "tedious" with "difficult". I think you do, but I will refrain from following you down another long thread arguing the difference between the two. Our non-argument about the definition of EV (!) was enough.

Take care.

BruceZ
05-13-2003, 05:38 AM
We learn little of practical substance, through this many-step exercise.

I strongly disagree. It is very important to know the probability of flopping various draws, so that you know under what conditions suited connectors can be played, especially in no-limit or tournament play. Knowing how to perform these computations yourself allows you to perform them for other types of draws as well, and to verify stated figures. I have not seen these stats published. They are not in the extensive appendix of Super System. We also learn to become more proficient at performing combinatoric analysis in general, so that these problems need not be so "tedious".

That's why I gave the correct answer.

But it appears that you did not give the correct figures. Do you disagree?

Unless you are confusing "tedious" with "difficult". I think you do

Tedious: adj. Tiresome or uninteresting by reason of extreme length or slowness.

Did that calculation look extremely long to you? It took 6 lines to get the probability of a clean flush draw. You said there were many cases. There were 5, and 2 were the same. I agree that it might be tedious for YOU due to your extreme slowness, but that judgement would not apply to everyone else.

Cyrus
05-14-2003, 03:38 AM
"It appears that you did not give the correct figures."

Since we seem to have, for some weird reason, a problem in communication, let me make it "perfectly clear", for your benefit, but more so for anyone following this soap opear and geting confused about what this is all about: I'll be putting up later on a post with the probabilities for suited connectors, and I will be more than grateful if you were to check them out for errors. Should keep one of us occupied, at least.

"Did that calculation look extremely long to you? I agree that it might be tedious for YOU due to your extreme slowness, but that judgement would not apply to everyone else."

More congenial words. Thanks, Bruce.

To be honest, I haven't checked out your calculations in that post. I had gone through the motions of calculating for myself the probabilities of the suited/non-suited connectors some time ago, and saw them verified in the book, later on. I know you think it must have taken me ages, since I'm "slow" and "don't know what the frick" I'm talking about, but what can I say, I managed to do it /forums/images/icons/grin.gif.

About tedium : The calculations were tedious for me, because
(a) some of them were done for completion's sake but didn't provide answers of any substantial practicality, and
(b) I'm not too fond of repetitious work.

I'm sure you are, but there's nothing wrong with that, honestly. I mean it. I know a lot of Accountants who positively love their entries; I know Engineers who work at Architects' offices and are only too happy to get down to the tedium of doing the statics and whatever of the building; I know Managers who are fond of routine, because what to others looks like routine, to them it looks like life's order; and so on. Someone has to do that work.

Bless your heart for enjoying your work. Either way, and as you said, that work is being appreciated by most everyone around here, including me.

--Cyrus