Odds to flop a draw?

[Net Warrior]

I realize that my original question was not precise. What I wanted to know was how often you flop 8 outs or better. So, double belly buster str8s should be included in the calculation as well as all better hands (my original thought was 2 pair or better for a made hand to qualify)

[Copernicus]

And I think that was answered that way early on.

I still stand by the 19.8% for all 8 and 9 out draws, BruceZs objections notwithstanding. Hold em Analyzer uses a precise calculation, not a simulation for this type of calculation, and I was able to reproduce it above, and have never found an error in HeA.

[BruceZ]

I still stand by the 19.8% for all 8 and 9 out draws, BruceZs objections notwithstanding.


I don't know if you saw this post a few back.

list

There's no question that you're overcounting the paired flops. These need to be separated out. This is a very common issue with this type of calculation. You also have overlaps between cases, straights being included, and a place where order isn't being handled right. My calculation for the flushes was verified a completely independent way. I don't know what you're asking the program to compute, it seems to be including something else like straights. I'll try to bring your answer inline with the other two methods if I have time, it's kind of a pain, but we can definitely do it. It took me awhile to fix Cyrus' too.


Will: "Do you have any idea how easy this is for me? I mean this is a f*ckin' joke! And I'm sorry you can't do this, I really am, because then I wouldn't have to sit here and watch you fumble around and f*ck it up!"

-Good Will Hunting

[ccwhoelse?]

middle size connectors would give you a straight draw 8.4% of the time but what about higher and lower connectors that can make less then four possible straights?

[Copernicus]

I'm not sure what you mean "overcounting the paired flops". If a flop gives 2 of the suit, and the third card happens to pair, it is counted because it still gives a flush draw. the fact that there are other draws to better or worse hands, doesnt matter. Also, in my enumeration I only see them counted once.

I would not be surprised if there was an order problem, I screw that up frequently. However, since it matches the software, there is likely to be an offsetting error.

As far as what I am "asking the software to do", you simply input the two starting cards and it enumerates all of the flop and turn drawing possibilities. Unfortunately the evaluation version only lets you see the turn and river possibilities and not the flop, or I would suggest trying it out.

when I get to my other computer I will copy down all of the flop possibilities, which may give you a flash of insight as to where you differ from it. (I tried to post a .bmp of the screen output here, but cant figure out how to do it. The image code below only lets you link to an image on the web, not a file. Attaching files is apparently disabled here. I could email the .bmp to anyone interested in all of the output for this problem.

(I am not a shill for HeA or its author Tristan Steiger, though i wish he had the time to write some software for me!)

One thought I had re offsetting errors might be (without looking back in detail right now) double counting double belly busters that are also flush draws?

[slider77]

I've seen the number 10.94% for a 4-flush flop.

Here's how I do the math:

11/50*10/49*39/48 X 3 = .1094

[BruceZ]

Copernicus,

I reworked your solution, and it now agrees with mine to all decimal places. The final answer is 19.1%. Your answer was previously 19.8%. I also corrected a problem with my solution which previously gave 19.3%.

Below is your original solution in italics with your original numbers that I'm changing marked in red. My changes are explained beneath each step with numbers or formulas I've changed marked in blue. Hopefully this will allow you to see very clearly exactly what I have modified.

Your Method

All two flushes are easy C(11,2)*C(39,1)/19600 = <font color="red">10.94%</font> and agrees with H'eA.

You have to subtract off the ones that also make straights. There are 4*3*3 of these since there are 4 straights, 3 ways to pick the rank of the non-flush card, and 3 ways to pick the suit non-flush card. So we <font color="blue">subtract 4*3*3/C(50,3)</font> to get <font color="blue">10.76%</font>. Some of the flops enumerated by the software must have been straights.


The regular open ended straights of 6 and 9 5 and 6 or 9 and 10 ie (3 sets of card combos):

With non in the original suit 6/50 * 3/49 * <font color="red">37</font>/48 * 3 card combos * 3 (position the non straight card is drawn in) = <font color="red">.0509</font> (the numerators take into account the order of the straight cards, by allowing all 6 first and only 3 second)

There are not 37 choices for the non-straight card. 8 of these complete the straight, and 2 of them will pair one of the other straight cards. The 2 that pair the board cannot be combined with the others because the multiply by 3 positions assumes this is a non-straight card. When we multiply by 3, this will count the same flops twice. For example, we could start with 9s Tc Td, and also 9s Td Tc, but when we multiply by 3, each of these will get transformed into the other and counted again. This is what I meant by over counting paired flops in my above post, and it is related to order type problems. To fix it, we have to only consider the 27 cards that don’t pair the board or make a straight, and then handle the paired flops in a separate term. So the expression becomes:

[ 6/50*3/49*<font color="blue">27</font>/48*3 <font color="blue">+ 2*3*3/C(50,3)</font> ]*3 = <font color="blue">.0400</font>

since for the paired board there are 2 ranks that can pair, 3 ways to pick the paired cards, and 3 ways to pick the unpaired card. If I was doing this from scratch, I would have replaced the first term with 3*3*27/C(50,3) so the two terms could be combined over a common denominator. You used fractions for the first term, which is equivalent to using permutations instead of combinations in numerator and denominator, and I maintained this format. There is no order problem in this term since the 6*3 generates all permutations of these 2 cards, and the last card is always from a separate rank. This technique is sometimes easier, but it can lead to the order problem or the over counting problem if we’re not careful. To avoid that, it’s always safer to make each term of the product represent a distinct set of cards so there is no possibility of overlap.


With one non-straight card in the orginal suit:
6/50 * 3/49 * <font color="red">11</font>/48 * 3 card combos * 3 position of the non straight card = <font color="red">.0152 </font>

There are not 11 non-straight cards in the original suit because 2 of them make straights, and 2 of them pair one of the other straight cards. In this case, we don’t have to consider any paired boards as we did in the last case since the card in the original suit must be a non-straight card, so it can’t pair the board. The case where the card in the original suit pairs the board will be considered in the next case. So removing the 2 straight cards and the 2 paired cards changes the expression to:

6/50*3/49*<font color="blue">7</font>/48*3*3 = <font color="blue">0.00964</font>


With one straight card in the original suit
2/50 * 3/49 * <font color="red">37</font>/48 *3 card combos *<font color="red">3</font> positions = <font color="red">.0170</font>

Similar to the first case, there are not 37 cards for the last card, which in this case is the suited card, because 8 of these make straights, and 2 of them pair the board. The two that pair the board again must be separated out for the same reason as in the first case. Additionally, we have to multiply by 6 different permutations not 3 since all cards are unique. The expression becomes

[ 2/50*3/49*<font color="blue">27</font>/48*<font color="blue">6 + 2*(3*3 + 3)/C(50,3)</font> ]*3 = <font color="blue">0.00285</font>

For the paird board term, there are 2 ways to pick the rank of the card of the original suit. If that rank pairs the board, then there are 3 ways to pick each of the other 2 cards. If that card does not pair the board, then there are 3 ways the remaining offsuit cards can pair.


double belly buster, none of the suit

9/50 * 6/49 * 3/48 * 2 (sets of ddbs) = .0028 (position is already incorporated in the numerators)

This is fine, <font color="blue">0.0028</font>

dbb one of the suit

3/50 * 6/49 * 3/48 * 2 (sets of dbbs) * 3 (order of the suited card) = .0028

Total straight draws not 2 or 3 flushes = .0886

This is fine, <font color="blue">0.0028</font>


Grand total = <font color="red">19.80%</font> -&gt; <font color="blue">19.1%</font>


My method

Your method counts the straight draws that are not flush draws (clean straight draws) and adds the number of all flush draws. My method does the opposite. It counts the number of clean flush draws, and adds the number of all straight draws. In a previous post , I computed the probability of a clean flush draw to be <font color="blue">9.52%</font>, and this was verified by an independent method . My computation of the probability of a straight draw did not properly eliminate flushes and straights. The correct expression for this probability should be

[ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ] / C(50,3) = <font color="blue">9.60%</font>.

That is, there are 3 2-card combinations, 15 of these are unsuited, and there are 48-8-6 = 34 ways to pick the last card so that it doesn’t complete the straight and doesn’t pair the board. There is 1 2-card combination that is a pair, and there are 34-7=27 ways to pick the last card so it doesn’t complete the straight, doesn’t pair the board, and doesn’t make a flush. We subtract 7 instead of 9 since the 2 flush cards that make a straight were already subtracted. The third term is for the paired boards. There are 2 ranks that can pair, 6 ways to make the pair, and 4 ways to choose the other card. Then we add the 4*4*4*2 double belly busters, and subtract 2 for the double belly buster draws that are flushes. Then we just add this to the number of clean flush draws to get the final answer.

9.5% + 9.6% = <font color="blue">19.1%</font>

In agreement with your method.

Software conundrum [img]/images/graemlins/confused.gif[/img]

I don’t understand how the software could have agreed with your calculation. Even for the simple case of total flush draws for which you said the software agreed, your number included straights, so there must have been straights among the enumerated flops. If it included straights for all the cases, then your number would have been even larger, since the probability of a straight is 4*4*4*4/C(50,3) = 1.3%, so the final answer would have been 19.1% + 1.3% = 20.4%. Also, if you intended to count straights in your calculation, then you cannot multiply by 3 for the 3 2-card combinations, as this would count the same straights multiple times.

-Bruce

[Cyrus]

..Really.

You wrote : "In a previous post [1], I computed the probability of a clean flush draw to be 9.52%, and this was verified by an independent method[2]."

The link [1] takes us indeed to a post in this thread whereby you gave the following (my emphasis):

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%

By "clean flush draw", I take it we mean "drawing to a flush without drawing to any straight and without flopping a flush or a straight".

In a post outlining some basic probabilities of max suited connectors, I gave this figure to be 7% (6.94897959183673%). This was followed by my explanation for that figure in a rather long post.

Your link [2] takes us to a post whereby you corrected this figure as being 7.4% (1452/19600=0.0740816326530612).

So, when did we go from 7.4% to 9.52% ?? I'm missing something here, possibly on definitions. (Is this thread not about Maximum Suited Connectors? NetWarrior asked about "meduim suited connected hands like 87s" which I take it to be exactly that.)

Irrespectively, once again, great post, terrific work put into it.

--Cyrus

PS : If by "independent method" you meant my posts above, I do not recall coming up with a figure close to 9%. Of course, I could be wrong and you are referring to somebody else's verification -- or I could be simply forgetful since I haven't revisited this thread until now.

[C Dubya]

Bruce -

I have been following this thread from the beginning and have been trying to calculate the answer using 4 different approaches: 1. All flush draws plus all straight draws eliminating flush draws that make straights, straight draws that make flushes, straight flush draws, and draws containing both flush and straight draws that are not straight flush draws. 2. All flush draws plus straight only draws eliminating flush draws that make straights. 3. All straight draws plus flush only draws eliminating straight draws that make flushes. 4. Flush only draws plus straight only draws plus straight flush draws plus draws containing both flush and straight draws that are not straight flush draws.

I have not yet been able to make these agree (they're close) nor have I been able to find a solution that agrees with you or Copernicus. Like you, I have been working with outcomes that are divisible by C(50,3). Your work is always spot on so I was hoping that you could explain where my error is in the following:

Let's look at my case 2 (Copernicus' first computation), all flush draws plus straight only draws eliminating flush draws that make straights. I'm with you through the first three steps here:

1. C(11,2)*39 = 2,145. 2,145/19,600 = 10.94%.
2. 4 straights*3 ranks for the nonflush card*3 offsuits = 36. 36/19600 = .18%.
3. Step 1 less step = 10.76%.

The next step is calculating straight draws with no flush cards on the board. It seems to me that:
3 possibilites for the 1st card * 3 possibilities for the 2nd card * 31 possibilities for the 3rd * 3 straight draws = 837 is correct. Card 3 possibilites are calculated as follows: 48 remaining cards less 8 cards that make a straight less 9 remaining cards that make a flush (2 of the remaining flush cards have already been accounted for in the 8 straight cards).

While I realize that 4 of the remaing 31 cards are offsuit cards that will pair the board, I don't understand the adjustment for paired boards. You come up with 783 ways while I calculate 837 ways. It seems to me that in this instance, we are considering three categories, 2 defininte ranks (e.g. K and Q if we hold JTs) and all other ranks, a category which the offsuit K's and Q's fit into.

Maybe you can help me see the error of my ways. Thanks for all the time and work you've put into your enlightening posts.

C

[BruceZ]

The next step is calculating straight draws with no flush cards on the board. It seems to me that:
3 possibilites for the 1st card * 3 possibilities for the 2nd card * 31 possibilities for the 3rd * 3 straight draws = 837 is correct. Card 3 possibilites are calculated as follows: 48 remaining cards less 8 cards that make a straight less 9 remaining cards that make a flush (2 of the remaining flush cards have already been accounted for in the 8 straight cards).


While I realize that 4 of the remaing 31 cards are offsuit cards that will pair the board, I don't understand the adjustment for paired boards. You come up with 783 ways while I calculate 837 ways. It seems to me that in this instance, we are considering three categories, 2 defininte ranks (e.g. K and Q if we hold JTs) and all other ranks, a category which the offsuit K's and Q's fit into.


You have to handle the 4 cards that pair the board in a separate term or else you will end up counting all the paired flops twice. For example, your way will count Ks Qc Qh, and also Ks Qh Qc. You only want to count each combination once in your method. The problem is that when you say you're counting 3 possibilities for the first card, 3 possibilities for the 2nd card, and then 31 possibilities for the 3rd card, 4 of those 31 possibilities for the 3rd card are the same cards that were possibilities for the first two cards. Then you are going to count the flops where one of those 4 cards are chosen as the 1st or 2nd card, and the same 3rd card as the one you chose before as the 1st or 2nd card. Hence you are counting the same combinations of cards twice, the only difference being that the 3rd card was interchanged with one of the 1st two cards. Notice that you have counted 54 flops more than me, which is 3*18, and that is because you counted the 18 paired flops twice, and then multiplied by 3.

This is a very common mistake. To correct it, you have to make sure that when you are counting distinct *combinations* (not different orders) of cards, that the set of cards you say you are choosing from in each term must be separate and distinct from the set of cards you are choosing from in other terms. They must be a *known* set of cards, not dependent on what the other cards are. Where this confusion comes from is that when we work with fractions or permutations, this is *not* the case. For example, in Copernicus' first term, he said he was going to generate all *permutations* (all orders) of KQ excluding 1 suit, by choosing one of the 6 cards for the first card and one of 3 for the second card. In that case, the first card could be either a K or a Q, and the second card had 3 choices, but it could be either 3 kings or 3 queens, depending on what the first card was. The 3 does not represent a distinct set of 3 cards which is known to us at the time we are doing the calculation, they are only known to us after we actually pick the first card. This is fine as long as you are trying to generate all orders as he was, but it's not fine if you are only trying to generate all combinations.

Whenever one set of cards you choose from can potentially overlap with another set of cards you are choosing from, you will be counting some combinations of the same cards multiple times. That is what happened in your case whenever the 3rd card paired the board. Copernicus was deliberately trying to generate all orders of these cards by multiplying 6*3 and then by 3 to reposition the last card, but this already generated all 6 permutations of the paired flops even before he multiplied by 3, so when he multiplied by 3, he effectively counted all the paired flops 18 times, while all the other flops were correctly counted 6 times (for 6 permutations) which is the same as triple counting only the paired flops. This is what I was explaining here:

The 2 [meant 4] that pair the board cannot be combined with the others because the multiply by 3 positions assumes this is a non-straight card. When we multiply by 3, this will count the same flops twice. For example, we could start with 9s Tc Td, and also 9s Td Tc, but when we multiply by 3, each of these will get transformed into the other and counted again. This is what I meant by over counting paired flops in my above post, and it is related to order type problems. To fix it, we have to only consider the 27 cards that don’t pair the board or make a straight, and then handle the paired flops in a separate term.

Note that there are 4 that pair the board, not 2 as I wrote originally. There are 6 that make straights, not 8 as I wrote, because 2 of these were subtracted as suited cards. My calculation was still correct because I correctly removed 10 from the first term, it's just that it was 6+4, not 8+2. The 6 straight cards are ignored, and the 4 that pair the board are handled in the separate term. Also, where I say it will "count the same flops twice", it's not *just* twice, it actually counts each permutation of paired flops 3 times.


Anyone should feel free to ask me questions about this if it still is not clear. This error is epidemic among people of all levels, including Ph.D., professional actuaries, etc., so I would like to get everyone to understand it once and for all.