Odds to flop a draw? - Page 4

BruceZ · #31 09-23-2003, 03:25 AM

What do you all see when you click on the link marked independent method in the above post? I'm wondering if it might be something different for you, perhaps depending on if you're in threaded mode or flat mode. What I see, and the post I intended to link to, starts with this paragraph:

The above calculation differs from the one I provided earlier in that mine allowed gut shots, while this one attempts to exclude gut shots. If we add back in the 540 gut shots which were subtracted by the above calculation, the result is (1326+540)/19,600 = 1,866/19,600 = 9.5% which is exactly to the flop what I calculated earlier.

The independent method I was referring to was Cyrus' calculation of the probability of flopping a clean flush draw with JTs. This calculation is in the post at the top of the thread I linked to, except that his calculation attempted to exclude all of the flush draws which also happened to be gut shots. All of the various flush draw cases which Cyrus counted in this thread are correct except for the 540 gut ghots. I pointed out in my post quoted above that if you add back in the 540 gut shots which he subtracted, then you get the 9.5% which I computed earlier by my method. When gut shots are correctly excluded, his method produces 7.4%. My method includes gut shots, and we are including them in the question being answered in this thread. To state this question again, we want to know the probability of flopping an 8 out straight draw or a 9 out flush draw holding suited connectors which can make the maximum number of 4 straights (such as JTs). The only cases we are excluding from this are those that flop a flush or a straight.

One thing that might be confusing as you read these different methods is that Cyrus and I have used JTs as our prototype hand, and Copernicus used 98s. This doesn't change the numbers or the method, only the wording.

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.

That should be 6 of these complete a straight, and 4 of them will pair one of the other straight cards. There are only 6 that make a straight since 2 were already subtracted as cards of the original suit. This same typo is made in the 4th case as well. This doesn't affect my equations. Also, the original solution actually counted each the paired flops 3 times, not just twice. It counted each of the 6 permutations 3 times.

Total straight draws not 2 or 3 flushes = .0886

I didn't mean to include that in the stuff I was agreeing with. The probability of a clean straight draw is still 8.4%. It didn't change from my original post.

C Dubya · #32 09-23-2003, 03:49 AM

I see the correct link.

Thanks for the education. It's very clear to me now.

Cyrus · #33 09-23-2003, 03:52 AM

I forgot you had taken out the gutshots.

BruceZ · #34 09-23-2003, 11:52 AM

In my method, for counting the straights:

[ 3*(15*34 + 1*27 + 2*6*4) + 4*4*4*2 - 2 ] / C(50,3) = 9.60%.

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

That should be, "there is 1 2-card combination that is suited in our suit".

BruceZ · #35 09-28-2003, 12:55 PM

Here is a new version with all of the typo fixes incorporated. The answer to one of the parts had a decimal point off, but the final answer was unaffected. -Bruce

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 = 10.94% and agrees with H'eA.

You have to subtract off the two flushes 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 subtract 4*3*3/C(50,3) to get 10.76%. 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 * 37/48 * 3 card combos * 3 (position the non straight card is drawn in) = .0509 (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. 6 of these complete the straight, and 4 of them will pair one of the other straight cards. The 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 again. 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*27/48*3 + 2*3*3/C(50,3) ]*3 = .0400

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 were doing this from scratch, I would have replaced the first term with C(6,2)*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 that format. There is no order problem in this term since the 6*3 generates all permutations of these 2 cards, and the other 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 original suit:
6/50 * 3/49 * 11/48 * 3 card combos * 3 position of the non straight card = .0152

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*7/48*3*3 = 0.964%

With one straight card in the original suit
2/50 * 3/49 * 37/48 *3 card combos *3 positions = .0170

Similar to the first case, there are not 37 cards for the last card, which in this case is the suited card, because 6 of these make straights, and 4 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*27/48*6 + 2*(3*3 + 3)/C(50,3) ]*3 = .0285

For the paired 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, .0028

dbb one of the suit

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

This is fine, .0028

Total straight draws not 2 or 3 flushes = .0886 -> .0836.

[Grand total = 19.80% -> 19.1%

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 9.52%, 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) = 9.60%.

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 suited in our suit, 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% = 19.1%

In agreement with your method.

Software conundrum

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.