Calculating odds

[PygmyHero]

Hi there, thanks for taking the time to read this. I apologize in advance for the length. I’m very new to poker so I will appreciate any help. Basically I am trying to refine my starting hand requirements (I am playing low limit HE) by being aware of the odds of certain types of hands improving on the flop. I feel I already have a good grasp of the odds on the turn and river (those calculations are fairly simplistic since they only involve one card), so I am ONLY concerned with the flop.

I have been using the method set out in Lee Jones' WLLH. Briefly: you have a PP, to calculate the number of times you will flop a set calculate the number of times all three community cards will not help you (48*47*46), and subtract that from the number of possible flops (50*49*48). This gives you 13,824, the number of times you will help. Divide this by the total number of flops (117,600), and you get 11.7% - this is the percentage of times you can expect to flop trips. Converted to odds this is 7.5:1 against.

The problem is that this technique only works if you are calculating the odds of catching ONE card. I don’t know what to do it if I need to catch TWO cards.

I would really appreciate if someone could walk me through the steps I would need to take to solve this problem. If you would use an example, that would be extremely helpful. I’m mostly concerned with situation where I would have a suited ace and am trying to flop a four flush, and situations where I have connectors and am hoping to flop an open ended straight draw (both of which would require catching two cards).

Last question: whatever process you reply with, I assume I can use the same method to calculate my odds of more complex situations (e.g. catching three card to flop a complete hand, playing 76s and flopping an open ended straight draw AND a four flush, etc.).

Thank you again for your assistance. I appreciate it.

[Stephen H]

I prefer to use combinatrics on these problems, since the order the cards appear in the flop doesn't matter.
To illustrate how this works on the pair looking for a set:
50 choose 3 total flops (denoted 50C3 from here on out) = 19600.
For a set, we need to count all the ways we could get exactly 1 of our rank, and then add all the ways to get exactly 2 of our rank.
2 cards left in our rank, 48 cards left not in our rank.
Flop trips: 2C1 * 48C2 = 2256
Flop quads: 2C2 * 48C1 = 96
(2256+96) / 19600 = 13824/117600 = about 11.5%

Describing any number of cards on the flop is the same.
For example, suited ace catching a four flush:
11C2 * 39C1 = 2145
2145/19600 = about 10.9%
Add in flopping the flush:
11C3 = 165
(2145+165)/19600 = about 11.8%
More complex is connectors catching the open-ender.
Here you have multiple cases to add up. Let's look at having the 8-9.
You either want:
6-7-X (X is not 5 or 10)
7-10-X (X is not 6 or J)
10-J-X (X is not 7 or Q)
There are the same number of flops for all 3, so for all 3 you get:
3* ( 4C1 * 4C1 * 40C1 ) = 1920 flops
1920 / 19600 = about 9.8%
Plus 4 ways to actually flop the straight; 567,67T,7TJ,TJQ
4* (4C1 * 4C1 * 4C1 ) = 256 flops
(1920 + 256) / 19600 = about 11.1%
Of course, if you want to devise strategy around this, you might want to factor in the fact that you're less confident when TJQ flops as you are dead to AK; similarly the TJ draw isn't so exciting. Plus, these numbers didn't factor in the possibility of a flush/flush draw out there against you.

The straight + flush situation is extrememly complicated, but can be done using the above methods. You must be careful to define your cases where there are no overlaps, or you will double-count the overlap and must subtract out the overlap once to make your flop count correct.

Hope this helps!

[PygmyHero]

Brilliant. Thanks for your response Stephen. You laid everything out clearly and concisely. I'd like to clarify one point.

Regarding flopping an open ended straight draw, from your response:

Here you have multiple cases to add up. Let's look at having the 8-9.
You either want:
6-7-X (X is not 5 or 10)
7-10-X (X is not 6 or J)
10-J-X (X is not 7 or Q)
There are the same number of flops for all 3, so for all 3 you get:
3* ( 4C1 * 4C1 * 40C1 ) = 1920 flops

I am a little confused as to the 40C1 part. I think the 40 represents the universe of remaining unknown cards given 52 to start, minus my two hole cards, minus the first two cards of the flop, minus the eight remaining cards that would give me a straight (e.g. if I hold 98 and the first two cards on the flop are 6 and 7, then 40 represents the universe of cards remaining minus all the fives and all the tens, likewise for the other flops). Is this correct?

One last thing. From your trips example flopping quads:

Flop quads: 2C2 * 48C1 = 96
(2256+96) / 19600 = 13824/117600 = about 11.5%

2C2 is 1 so the first line is equal to 48, which should also be in the second line. 13,824/117,600 reduces (divided by six) to 2304/19,600, giving us the expected result, about 11.7%. Sorry if that seems like nit-picking.

Thanks again for your assistance.

[gamble4pro]

Your calculation starts with an error: the number of possible flops is C(50,3)=19600 (this is 50*49*48/6). When you multiply 50*49*48 you are doble-counting a lot of combinations. When calculating odds for a gaming event, this event must be described by unique combinations (ABC is the same with ACB or BAC).

[Stephen H]

[ QUOTE ]

Regarding flopping an open ended straight draw, from your response:

Here you have multiple cases to add up. Let's look at having the 8-9.
You either want:
6-7-X (X is not 5 or 10)
7-10-X (X is not 6 or J)
10-J-X (X is not 7 or Q)
There are the same number of flops for all 3, so for all 3 you get:
3* ( 4C1 * 4C1 * 40C1 ) = 1920 flops

I am a little confused as to the 40C1 part. I think the 40 represents the universe of remaining unknown cards given 52 to start, minus my two hole cards, minus the first two cards of the flop, minus the eight remaining cards that would give me a straight (e.g. if I hold 98 and the first two cards on the flop are 6 and 7, then 40 represents the universe of cards remaining minus all the fives and all the tens, likewise for the other flops). Is this correct?


[/ QUOTE ]

Yes, that's right; 40 is number of cards left in the deck that don't give us the made straight (48 cards in deck, 8 give us the straight). I do it this way to avoid double counting some of the made straights; if you use 48C1 (the number of cards left in the deck), you would get 6-7-X (where X is any card) and 7-10-X, but 6-7-10 would be counted once as 6-7-X and once as 7-10-X. You can actually account for this by then subtracting out the overlap; this is a viable alternative and actually can be easier depending on the situation. The overlap in this case is the 6-7-10 and the 7-10-J; the numbers would be:
3* (4C1 * 4C1 * 48C1) - 2* (4C1 * 4C1 * 4C1) = 2176, which is the same as 1920+256.
48 = cards left in deck; 4C1*4C1*4C1 = ways to get 6-7-10 (or 7-10-J). Notice this time we don't have to specifically count up all the made straights, since they're included in the calculation; we just have to subtract out overlaps.

[ QUOTE ]

One last thing. From your trips example flopping quads:

Flop quads: 2C2 * 48C1 = 96
(2256+96) / 19600 = 13824/117600 = about 11.5%

2C2 is 1 so the first line is equal to 48, which should also be in the second line. 13,824/117,600 reduces (divided by six) to 2304/19,600, giving us the expected result, about 11.7%. Sorry if that seems like nit-picking.

Thanks again for your assistance.

[/ QUOTE ]
No, it's important to get the right numbers; good catch. I actually did it right once on my scratch notes here *points to his paper as proof* but then I redid it wrong when I typed it in. Oops! Either that, or I intentionally put the error in to see if you were paying attention.

[Cobra]

[ QUOTE ]
More complex is connectors catching the open-ender.
Here you have multiple cases to add up. Let's look at having the 8-9.
You either want:
6-7-X (X is not 5 or 10)
7-10-X (X is not 6 or J)
10-J-X (X is not 7 or Q)
There are the same number of flops for all 3, so for all 3 you get:
3* ( 4C1 * 4C1 * 40C1 ) = 1920 flops
1920 / 19600 = about 9.8%


[/ QUOTE ]

by doing it this way your are double counting certain hands. An example is when you are looking at 67x you should exclude the 5 and 10 because they make a straight you should also exclude the 6 and 7 because you are using them in another portion of your equation. The real answer is as follows:

=3*((4c1*4c1*34c1)+(2*4c2*4))=3*(544+48)=1776 flops not the 1920 that you came up with.

Cobra

[Stephen H]

Ah, drat. I'm double-counting 6-6-7 and 6-7-7 and the like, aren't I? Well, that's the real reason I'm posting on these boards...to help get all the kinks out in my math on this stuff.
Thanks!

[PygmyHero]

I don't know if anyone's still following this thread or not but I have to ask a few stupid questions.

The real answer is as follows:

=3*((4c1*4c1*34c1)+(2*4c2*4))=3*(544+48)=1776 flops not the 1920 that you came up with.

The three at the start of the equation is because each card can come in one of three positions on the flop, correct (e.g. A x x, x A x, x x A)?

I understand the first term in parentheses, but not the second. Could you please indulge me and explain what each number in (2*4c2*4) is?

Sorry if these questions are overly stupid. Thank again for all the help - I feel like I'm BEGINNING to have a rudimentary grasp of what's going on here.

[BruceZ]

[ QUOTE ]
I don't know if anyone's still following this thread or not but I have to ask a few stupid questions.

The real answer is as follows:

=3*((4c1*4c1*34c1)+(2*4c2*4))=3*(544+48)=1776 flops not the 1920 that you came up with.

The three at the start of the equation is because each card can come in one of three positions on the flop, correct (e.g. A x x, x A x, x x A)?

[/ QUOTE ]

There are 3 pairs of ranks that can flop to make an open-ended straight draw with 89. They are 67, 7T, TJ.

[ QUOTE ]
I understand the first term in parentheses, but not the second. Could you please indulge me and explain what each number in (2*4c2*4) is?

[/ QUOTE ]

He is separating out the cases where the board pairs. That is why he has 34 in the first term, it is 48 minus 6 cards that pair the board minus 8 cards that make the straight. Either of the 2 ranks can pair (e.g. 6 or 7), and there are 4c2 = 4*3/2 = 6 ways to make each pair, times 4 ways to pick the unpaired card.

Here is my calculation which I posted awhile back. I am assuming suited connectors, so I have separated out the 1 case out of 16 where 2 suited cards hit the board in order to exclude the case where we flop a flush. I also include double gutshots. Otherwise, our calculations are the same, and I happen to remember 1776 as being the number you get if you do not take these things into account.

[ 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.

See this archived post for the separate staight and flush draw calculation computed a couple different ways.

These are among the toughest practical counting problems that we do here, so feel free to ask anything about it that still isn't clear.