Flopping a 4-Flush with max suited connectors

[Cyrus]

We hold JhTh and we want the probability of flopping a 4-Flush draw, without flopping a Straight, and without flopping a 4-Flush AND a Straight draw :

There are (13 hearts in all - 2 hearts in our hand)=11 hearts remaining in the deck. Since C(11,2)=55, there can be 55 2-card combos made from 11 hearts. There are (52 -2 cards in our hand -11 other hearts) = 39 other cards which can act as a 3rd card to complete the flop. So there are 55*39=2145 possible 2-heart flops in all.

We must sanitize the above number of 2145 combos by taking out all the unwanted flops. When we sanitize it, we will divide the sanitized number by the number of all possible 3-card flops and we will obtain the desired probability.

The sanitizing process involves taking out
- the 4-Flush and Straight-making flop combinations,
- the 4-Flush and open-ended Straight-draw flop combinations,
- the 4-Flush and Double-Belly Buster Straight-draw flop combinations, and
- the 4-Flush and Inside Straight-draw flop combinations.

The first act is the toughest. We got to identify the NON-Straight-making flops and deduct them from the total number of the total 2-heart flops in order to get the 4-Flush and Straight-making flop combinations.

We take out the combos whereby the 3rd card makes a Straight. Each flop's non-suited card is inside the parenthesis: 7h8h(9); 7h9h(8); 8h9h(7); 8h9h(Q); 8hQh(9); 9hQh(8); 9hQh(K); 9hKh(Q); QhKh(9); QhKh(A); AhQh(K); AhKh(Q).

The number of available cards from which we will be making up every time the combinations of flops is (52 cards - 2 cards in our hand - 2 cards outside the parenthesis every time)= 48. If we take out the hearts, the remaining hearts are (13 hearts in all - 2 hearts in our hand - 2 hearts in each of the above combos)= 9 hearts remaining every time.

There are (48 - 3 Nines - 9 remaining hearts)=36 cards whereby each can complete the [78X] flop to produce a 3-card flop. This makes for 36 flops. Each of these flops does not flop us a Straight. So, the number of 3-card flops, which every 2-card combo from the above detailed combos, can produce without making a Straight, is

For 78 : 48 - 3 Nines - 9 remaining hearts = 36
For 79 : 48 - 3 Eights - 9 remaining hearts = 36
For 89 : 48 - 3 Sevens - 3 Queens - 9 remaining hearts = 33
For 8Q : 48 - 3 Nines - 9 remaining hearts = 36
For 9Q : 48 - 3 Eights - 3 Kings - 9 remaining hearts = 33
For 9K : 48 - 3 Queens - 9 remaining hearts = 36
For QK : 48 - 3 Nines - 3 Aces - 9 remaining hearts = 33
For AQ : 48 - 3 Kings - 9 remaining hearts = 36
For AK : 48 - 3 Queens - 9 remaining hearts = 36

36+36+33+36+33+36+33+36+36= 315

The other 2-heart, 2-card combinations that are in no danger of producing a Straight are 55combos - 9 combos of 78/89/etc = 46combos. Each of those safe combos can combine with the (52cards - 2hearts in our hand - 2hearts in the flop - 9hearts remaining) = 39 non-heart cards, to give 46*39= 1794 combinations which don not produce a Straight.

The total flop combos that do not produce a Straight are 315+1794= 2109. Therefore, the total flops that do produce a Straight (along with a 4-Flush draw) are 2145-2109= 36.

Amongst the 2109 combos that do not produce a Straight right away (see previous), there are some combos that produce a 4-Flush and an open-ended Straight-draw flop. Those possible combos are
89X (where X is not 7 or Q)
9QX (where X is not 8 or K)
QKX (where X is not 9 or A)

For the X89, we can have three cases :
(a) 8h9h and Xd or Xc or Xs,
where X is d,c,s of 2,3,4,5,6,8,9,T,J,K,A=11cards
So 3*11=33cards which can be X, making 33 combos for (a)
(b) 8hXh and 9d or 9c or 9s,
Now X must be a heart from 2,3,4,5,6,K,A=7cards
So, the (1 card for 8h)*(7cards for X)*(3cards for 9 d/c/s)=21cards, making 21 combos for (b)
(c) 9hXh and 8d or 8c or 8s,
Again X must be a heart from 2,3,4,5,6,K,A=7cards
So, the (1 card for 9h)*(7cards for X)*(3cards for 8 d/c/s)=21cards, making 21 combos for (b)

The total of (a)+(b)+(c)= 33+21+21 = 75 flop combos that produce a 4-Flush and an open-ended 89X Straight-draw. The same calculation is done for 9QX and QKX so, there are 3*75= 225 flop combos that produce a 4-Flush and an open-ended Straight-draw.

Next, the Double-Belly-Buster Straight-draws : The possible DBB flops are
8QA and 79K. The first possible flop can be made C(3,2)= 3 ways cause there are 3 2-heart combinations made of 3 hearts. Each of those 2-heart combinations combines with the non-heart suits of the other, the third rank in the flop and gives 3*3 =9 DBB combinations with 2 hearts. Same is true for the second possible DBB flop, so 2*9 = 18 total 4-Flush and DBB Straight-draw flop combinations.

Finally, the Inside Straight draws, along with the 4-Flush. The possible Inside Straight-drawing 2-heart flops are found in 6 possible flop structures:
AKX
AQX
9KX
8QX
79X
78X

We have to take out the flops that produce a Straight ...straight away.

For the AKX case, the possible 2-heart flops are
(a) AhKh and Xd or Xs or Xc
(b) AhXh and Kd or Ks or Kc
(c) KhXh and Ad or As or Ac
For (a), X can be d,s,c of 2,3,4,5,6,7,8,9,T,J,K,A= 12 cards (the Q-rank is out)
so 3*12 = 36cards which can be X, making 36 combos for (a).
For (b), X must be a heart and K is d,s,c. X can be a heart from 2,3,4,5,6,7,8,9,K= 9cards (the Jh & Th we already hold in our hand, Ah is already in the flop and Qh flops a Straight) so
(1 card for Ah)*(9 cards for X)*(3 cards for K d/s/c) = 27 combos for (c).
For (c), X must be a heart and A is d,s,c. X can be a heart from 2,3,4,5,6,7,8,9,A= 9cards (the Jh & Th we already hold in our hand, Kh is already in the flop and Qh flops a Straight) so
(1 card for Kh)*(9 cards for X)*(3 cards for A d/s/c) = 27 combos for (c).
Therefore, there are 36+27+27 = 90 2-heart flops that give a 4-Flush draw and an Inside Straight draw for AKX. The same results we get for the other five structures, so we have 5*90=540 total 2-heart flop combinations that give a 4-Flush draw and an Inside Straight draw.

Summing up, we have
36 flop combinations that produce a 4-Flush draw and a Straight,
225 that produce a 4-Flush and open-ended Straight-draw
18 that produce 4-Flush and Double-Belly Buster Straight-draw flop, and finally
540 that produce 4-Flush and Inside Straight-draw flop combinations

The total is 36+225+18+540= 819, which we deduct from the (55*39)=2145 total possible 2-heart flops, to get 2145 - 819 = 1326 sanitized flops that produce ONLY a 4-Flush draw.

Therefore, the incredibly valuable /forums/images/icons/grin.gif probability of flopping only a 4-Flush draw, when holding JhTh, is 1326 divided by the total number of 2-card flops, ie C(50,2)=19600, so 1326/19600= <font color="red"> 7%</font color> or 6.765306%.

I would welcome any corrections or suggestions for short-cuts to the above process.

--Cyrus

[Cyrus]

...Yes, the above was a long one! As I wrote, I would welcome any corrections, and, especially, any suggestions for short cuts to the process. Some remarks :

1. The post did not happen out of the blue. I was asked by BruceZ to provide justification for the relevant probability figure which I provided in a post titled "Suited Connectors, max stretch" (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=probability&amp;Number=257 185&amp;page=1&amp;view=expanded&amp;sb=6&amp;o=14&amp;fpart=1) ,in this forum. BruceZ had calculated the probability to be 9.5% if I recall. So, although, my calculation had been done in the past, I re-did it and put it up. Perhaps BruceZ will suggest improvements or point out my errors.

2. In my original post, I had the probability as 6.94897959183673469387755102040816%, which however was not accurate. In revisiting the calculations, I found out that I had erroneously excluded the possibility of pairing the board in the Inside Straight-draw step. (In other words, we can flop AKA.) This makes for the correct figure of 6.76530612244897959183673469387755%. Please note that both figures round to <font color="red">7%</font color>.

3. I commented that the process is 'extremely tedious' and I stand by that characterization. Multiple and repetitious calculations I do not enjoy. So I'm glad that someone else is out there to do them for me. (And someone else to double-check!) This doesn't mean that I cannot do them. Accounting entries are a cinch to learn but a pain to do every day, unless you are a creative guy...

4. I hasten to add that I fully respect the opinion of BruceZ, who does not find such calculations to be tedious at all. To each his own.

5. I also commented that I find that specific figure (the probability of flopping only a 4-Flush when holding max suited connectors) to be more or less useless. I may have been exaggerating, though not by much. Experienced poker players might have a different opinion, as BruceZ has, for example, who finds that figure helpful to know.

Well, that's it. /forums/images/icons/cool.gif

--Cyrus

[BruceZ]

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. My calculation (http://www.twoplustwo.com/forums/showflat.php?Cat=&amp;Board=probability&amp;Number=255744&amp; page=1&amp;view=expanded&amp;sb=5&amp;o=14&amp;fpart=)

Also, the above calculation does not count the number of gut shot draws correctly. There are not 540. While it is true that there are 6 pairs which can flop to give a gut shot draw, and there are 90 flops for each of these pairs, we can not simply multiply 90*6 to get the total number of gut shot flops because these are not mutually exclusive. For example, the flop AK7 has both an AK pair and an A7 pair, and this flop would be counted twice. In fact, for each of the 6 pairs, there are exactly 2 flops which get counted twice by this method ignoring suit. For example, for AK the two flops are AK7 and AK9. For each of these flops that are counted twice, there are 3 ways to choose the unsuited card, and 3 suits that the unsuited card can be, for a total of 9 flops. The total number of flops that are double counted then is 6*2*9 = 108. Additionally, the 18 double gut shot straights should not be counted at all because they are already being counted separately. Thus there are 108+18 = 126 flops which need to be added back in, and the correct probability should be (1326+126)/19,600 = 1,452/19,600= 7.4%.

I agree that the odds of making a clean flush draw without a straight draw or a gut shot draw are not very useful. The reason this question came up originally was because someone wanted the odds of making a 4-flush draw or an outside straight draw, so it became necessary to compute the odds of making each of these independently. For that purpose, we do not want to exclude the flushes that are gut shot draws.

I also agree that this calculation is long and tedious the way it is done above. In fact it is a God awful mess, but this needn’t be the case. For example, after much work and counting, we learn from the above calculation that there are 36 straights which are 4-flush draws. This is obvious. There are exactly 4 straights we can make with JT ignoring suit, for each one there are 3 ranks which can be unsuited, and there are 3 unsuited ranks, so 4*3*3 = 36. Done. We can also do this entire calculation using the method I provided earlier, and since we are now excluding gut shots, this method becomes even shorter. The reason is that now we do not have to count any of the 8,Q or 9,K flops, so there are only 3 types of flops to consider. Here is the entire calculation, which replaces the entire above post, and which is almost simple enough to do in your head:

No 8,9,Q,K: C(7,2)*27 = 567
Exactly 1 denomination of 8,9,Q,K but not 87,97,QA,KA: 4*6*27 + C(6,2)*12 = 828
8,K but not 7,8,K or 8,K,A: 2*3*5 + 1*27 = 57
------------------------------------------------
(567 + 828 + 57)/19,600 = 1,452/19,600 = 7.4%. Same as the long method to the flop.

Done!

[BruceZ]

For example, the flop AK7 has both an AK pair and an A7 pair, and this flop would be counted twice. In fact, for each of the 6 pairs, there are exactly 2 flops which get counted twice by this method ignoring suit. For example, for AK the two flops are AK7 and AK9.

Make that AKQ and AK9. AK7 is not a problem. Here are the two double counted flops for each pair. Note that each flop appears on the list exactly twice.

AK: AKQ and AK9
AQ: AKQ and AQ8
K9: AK9 and K97
Q8: AQ8 and Q87
97: K97 and 987
87: Q87 and 987

While we're fixing typos, in my original calculation which included gut shots, My calculation w/gut gots (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=probability&amp;Number=255 744&amp;page=1&amp;view=expanded&amp;sb=5&amp;o=14&amp;fpart=) which I have modified here for the case where we exclude gut shots,

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

This should read 8,9,Q, or K. These were the basis for the calculation.

[BruceZ]

My characterization of the Cyrus error was not completely correct or complete. First of all, we are not double counting 12*9 flops. 4 of the 12 flops on the list are straights (AKQ and 987 each listed twice) so they never got counted at all which is correct. Of the remaining 8, there are only 4 unique flops. They got counted twice because they appear on the list twice. These flops are AK9, AQ8, K97, and Q87. So there were 4*9 = 36 flops that were double counted. Add to this the 18 double gut shots which were already taken into account elsewhere to get 54.

Now in addition, there is an error in the way paired boards are handled in counting the gut shots. Pairs appear to be handled correctly for outside straights, but not for gut shots. For example, AhKhKx can be gotten from both cases a and b. AhKhAx can be gotten from both cases a and c. This contributes 6 extra flops times 6 pairs or 36 extra flops. Add that on to the previous 54 to get 90. To make this agree with my answer, this number needs to be 126, so we are still 36 short.

I've been looking at it for awhile, but I haven't been able to account for the missing 36 flops. It is either an extra 36 in Cyrus' calculation which he subtracts from 1326, or an extra 36 in my calculation. I'm sure there is a good explanation for this number. Since our answers agree exactly when we don't exclude gut shots, the problem must lie in the gut shot portion of the calculation. This makes a difference of 0.2% to the final answer. Until this is found, my answer stands at 1,452/19,600 = 7.4%, and Cyrus' method stands at 1,416/19,600 = 7.2%. If I find the missing 36 flops I'll let you know.

[BruceZ]

So I asked myself, are we counting any gut shot draws that are actually outside straight draws? My answer came back yes! In fact there are 4*9=36. They are 89K, 8QK, 79Q, and 9QA. There are 9 of each of these. These are being counted as both outside straight draws and gut shots.

With these 36 added back on, along with the other 90 we found in the last post, Cyrus' method produces the same answer my short 4 line solution did from the beginning, 1,452/19,600 = 7.4%. My original post gave the correct solution, but it didn't give a completely correct explanation for the discrepancy. Now it is completely explained.