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 01-23-2017, 09:26 AM #1 dan3 stranger   Join Date: Jul 2007 Posts: 4 Specific Flop Odds with Pocket Pairs I'm trying to figure out the math to determine all the possible hands a player might make on the flop when holding pocket pairs 22-AA. In this scenario, a paired board that doesn't hit our pair will be calculated separately as "two pair". The odds of flopping quads, a full house, a set, or "two pair" are static meaning that these odds do not change whether we're holding 22 or AA or any pair in between Odds to Flop with PP ============= Quads = ((2/50)*(1/49)*(48/48))*3 = 0.24% Full House = ((2/50)*(48/49)*(3/48))*3 = .73% Set = ((2/50)*(48/49)*(44/48))*3 = 10.78% Two Pair = ((48/50)*(3/49)*(44/48))*3 = 16.16% This gives us a total of 27.92% for every pocket pair. The odds of flopping an over pair or otherwise varies with our pocket pair rank. However, we know the total odds must equal 100-27.92 = 72.08% This is where I run into problems. Let me share my calculations which are certainly incorrect. Odds to Flop Over Pair with PP =================== Example Hand 77: (20/50)*(16/49)*(12/48) = 3.27% How I came up with this: (20/50) - all cards lower than 7 (16/49) - all cards lower than 7 that doesn't pair the board (12/48) - all cards lower than 7 that doesn't pair the board Odds Flop has at least 1 Over card and does not pair ================================ Example Hand 77: (28/50)*(44/49)*(40/48) = 41.9% How I came up with this: (28/50) - all cards higher than a 7 (44/49) - all cards that don't give us a set or pair the board (40/48) - all cards that don't give us a set or pair the board The total odds for the last two calculations should equal 72.08% so I am certain that my math is flawed. I thought I might need to multiply the result by 3 as in the calculations provided for quads down to two pair but doing so results in some hands coming out to over 100% which really confuses me. On a side note I want to thank forum members heehaww and whosnext because I've learned so much from your posts in the last few days.
 01-23-2017, 10:34 AM #2 md46135 centurion   Join Date: Mar 2015 Location: where they don't respect my raises Posts: 160 Re: Specific Flop Odds with Pocket Pairs your full house probability only contains "strong" full houses where you have xx and the board is xyy. there's also a possibility of the board coming yyy, which is still a boat if you have a pair. probability for that is 12C1 * 4C3 / 50C3 = 0.24%, so that adds to your full house probability. quads, set and two pair is good. overpair with 77 is correct, that's 5C3 * 4C1^3 / 50C3 = 3.27%. at least 1 overcard with no pair: (7C1*4C1*5C2*4C1^2 + 7C2*4C1^2*5C1*4 + 7C3*4C1^3 ) / 50C3 = 13440 / 50C3 = 68.57%. btw with poker (card) problems combinations are usually an easier way to solve these problems, so maybe try to learn them https://en.wikipedia.org/wiki/Combination Last edited by md46135; 01-23-2017 at 10:39 AM.
 01-23-2017, 05:37 PM #3 dan3 stranger   Join Date: Jul 2007 Posts: 4 Re: Specific Flop Odds with Pocket Pairs Is it possible to solve this problem using the type of math that I'm using? I have a hard time understanding the combination math that you posted without an explanation attached.
 01-23-2017, 06:24 PM #4 md46135 centurion   Join Date: Mar 2015 Location: where they don't respect my raises Posts: 160 Re: Specific Flop Odds with Pocket Pairs yeah, it's possible. if you want an explanation for combinations, maybe the board coming yyy while you have xx is the easiest starting point. you have 2 cards in 1 rank, meaning that there are 12 other ranks from which 3 cards can come (they can't come from yours). 12C1 - choose one rank from 12 possible. clearly, there are 12 ways to do this. 4C3 - you have already chosen a rank, now you need 3 cards (different suits) from that rank out of 4 possible suits. you can choose 123, 124, 234, 341 - total 4 combinations. multiply these to get the number of combinations that fit the initial condition. now, you need to divide that by the total number of possible selections of 3 cards from 50 cards remaining in the deck, that is 19600. if the notation confuses you, the first number before C would be "n" from the link i gave you, the second would be "k", so it would be written nCk (that's the formula on the link). also, here's a calculator for combinations http://www.ohrt.com/odds/binomial.php for the 3rd example with 77 with at least 1 overcard, you need to divide ranks into two groups: lower ranks (2,3,4,5,6 - total 5 of them) and higher (8,9,T,J,Q,K,A - total 7 ranks). there are 3 cases separated by "+", each of them fits your selection criteria so you add them together. if you want exactly 1 higher card, first choose 1 high rank out of 7 possible, then 1 card of 4 possible suits. then choose 2 ranks from the lower 5 possible ranks and 4 cards for each one of those. you do the similar for other 2 options. basically, combinations come down to putting things in groups and then subgroups and so on (you can think of lower ranks as 5 different groups, each of those with 4 different members i.e. suits). if the order of things doesn't matter like in this case (flop 224 is the same as 422), combinations should be the preferred way of solving things. Last edited by md46135; 01-23-2017 at 06:45 PM.
 01-23-2017, 06:35 PM #5 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,787 Re: Specific Flop Odds with Pocket Pairs Just to echo, many of us do virtually all poker calculations via combinatorics (the "Choose" function). I find them much easier to use and understand. Note there exists a vocal minority who disagree.
 01-24-2017, 06:33 AM #6 dan3 stranger   Join Date: Jul 2007 Posts: 4 Re: Specific Flop Odds with Pocket Pairs Thanks md46135. I think I'm starting to understand these combination calculations much better now. I would like to share what I believe is a pretty full explanation of your math for counting the # of combinations where the player will flop at least 1 overcard when holding a pocket pair when that flop will not contain a pair itself. This is just to provide a clear explanation for others trying to understand these calculations. When holding 77, how often will we flop a AT LEAST 1 OVERCARD when the BOARD DOESN'T PAIR Answer -------- (7C1*4C1*5C2*4C1^2 + 7C2*4C1^2*5C1*4 + 7C3*4C1^3 ) / 50C3 = 13440 / 50C3 = 68.57% Explanation ----------- 7C1*4C1*5C2*4C1^2 = Combinations with 1 overcard 7C1 Choose 1 rank from the 7 ranks higher than us 4C1 Choose 1 suit from the 4 suits for the 1st card 5C2 Choose 2 different ranks from the 5 ranks lower than us 4C1 Choose 1 suit from the 4 suits for the 2nd card 4C1 Choose 1 suit from the 4 suits for the 3rd card 7C2*4C1^2*5C1*4 = Combinations with 2 overcards 7C2 Choose 2 different ranks from the 7 ranks higher than us 4C1 Choose 1 suit from the 4 suits for the 1st card 4C1 Choose 1 suit from the 4 suits for the 2nd card 5C1 Choose 1 rank from the 5 ranks lower than us 4C1 Choose 1 suit from the 4 suits for the 3rd card 7C3*4C1^3 = Combinations with 3 overcards 7C3 Choose 3 different ranks from the 7 ranks higher than us 4C1^3 Choose 1 suit from the 4 suits for each card on the flop (3)
 01-24-2017, 06:46 AM #7 dan3 stranger   Join Date: Jul 2007 Posts: 4 Re: Specific Flop Odds with Pocket Pairs With this newfound understanding I was just able to achieve the same result using the type of math I used in my original post. However, one thing confuses the heck out of me and I really hope someone can explain it. So here is what I came up with: Example Hand 77 --------------- ((28/50)*(20/49)*(16/48)*3)+((28/50)*(24/49)*(20/48)*3)+((28/50)*(24/49)*(20/48)) = 68.57% Explanation ------------- (28/50)*(20/49)*(16/48)*3 = % of time we get 1 overcard and no paired board (28/50) - all cards higher than a 7 (20/49) - all cards lower than a 7 (16/48) - all cards lower than a 7 that doesn't pair the previous low card (28/50)*(24/49)*(20/48)*3 = % of time we get 2 overcards and no paired board (28/50) - all cards higher than a 7 (24/49) - all cards higher than a 7 that doesn't pair the previous high card (20/48) - all cards lower than a 5 (28/50)*(24/49)*(20/48) = % of time we get 3 overcards and no paired board (28/50) - all cards higher than a 7 (24/49) - all cards higher than a 7 that doesn't pair the previous card (20/48) - all cards higher than a 7 that doesn't pair the previous cards ((28/50)*(20/49)*(16/48)*3)+((28/50)*(24/49)*(20/48)*3)+((28/50)*(24/49)*(20/48)) I get the exact same result so the math seems good but why is it that in order to achieve this result, I could not multiply the last calculation by 3 as I did the others. Why did I have to do (28/50)*(24/49)*(20/48) for the last calculation and not ((28/50)*(24/49)*(20/48))*3 like the previous two?
 01-24-2017, 09:19 AM #8 md46135 centurion   Join Date: Mar 2015 Location: where they don't respect my raises Posts: 160 Re: Specific Flop Odds with Pocket Pairs first post: correct second post: when you're using this way of counting, the problem is that you need to count all the permutations. that can lead to confusions that don't appear if you use combinations. if you have 1 overcard and no paired board: first you can choose 1 overcard (28), then 2 unpaired undercards (24 and 20). but what if you want to choose the undercards first and then overcard? you could write it like this: 28/50 * 20/49 * 16/48 + 20/50 * 28/49 * 16/48 + 20/50 * 16/49 * 28/48, which is just 3 times the same thing. you can choose the overcard first and then 2 undercards, 1 undercard + 1 overcard + 1 undercard or 2 undercards + 1 overcard. second case (2 overcards) is similar, you can have undercard first, in the middle or in the end. third case, which confuses you, you're choosing 3 overcards from group of overcards, so it needs to be 28*24*20 and that's the only way. maybe now you see why i told you in the first post why combinations are preferred for this though that doesn't mean you can't use this "simple" counting, in fact for some situations it's easier to do it that way. that is usually if the order of things matters. for example, suppose flop is "abc" and you want the board after the river to be unpaired ("abcde"). in this case, simple counting is the first thing that should come to mind for solving this because the cards arrive in sequence one after another. so, recap: in general, when sequencing matters use simple counting and if it doesn't use combinations.
01-24-2017, 01:40 PM   #9
heehaww
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Join Date: Aug 2011
Location: It was an attractive goat in AR
Posts: 4,182
Re: Specific Flop Odds with Pocket Pairs

Quote:
 Originally Posted by dan3 Odds Flop has at least 1 Over card and does not pair ================================ Example Hand 77: (28/50)*(44/49)*(40/48) = 41.9% I thought I might need to multiply the result by 3 as in the calculations provided for quads down to two pair but doing so results in some hands coming out to over 100% which really confuses me.
You do need to multiply by 3 (because of the 3 possible places for the overcard), but doing so over-counts the ways of getting multiple overcards (it double-counts the ways to get two overs, and triple-counts the ways to get three).

You have an arrangement Oxx and you also wish to count xOx and xxO.

But the (44/49) and (40/48) in your expression already allow the x's to be O's. Multiplying by 3 counts the specific arrangement OOU twice because Oxx and xOx each counted it once (because the first x in each was allowed to be an O). Similarly, it double-counts OUO and UOO. It triple-counts OOO.

The over-counting is why you get a result greater than 100% (but over-counting doesn't always break the 100% barrier, and that's when it can dangerously go unnoticed). To fix this, you can use the inclusion-exclusion technique (the principle of inclusion-exclusion, or PIE). The following would have worked:

3*(28/50)(44/49)(40/48) - 3*(28/50)(24/49)(40/48) + (28/50)(24/49)(20/48) = 24/35 =~ 68.57%

1) We take your original expression and multiply by 3 to count the 3 arrangements of Oxx.
2) By doing that, each of the 3 OOx arrangements is counted twice, so we subtract each OOx arrangement once.
3) The first step triple-counted the OOO possibility, because OOO is counted by each Oxx. The 2nd step subtracted it 3 times, because each subtraction of an OOx also subtracted OOO. Thus, by now, OOO isn't counted at all, so we need to add it back in.

PIE can come into play whenever there are overlapping possibilities. The 1st card being an over is not mutually exclusive of the 2nd card being an over, so adding the two possibilities double-counts the possibility of both happening. On the other hand, if you were talking about a specific card like the A, the possibilities would be mutually exclusive, so you could multiply or add without a care in the world.

PIE is a good approach to this problem, and is sometimes a far superior approach, but here, what you did in your latest post (adding non-overlapping cases like md did) was equally good. However, we can improve the execution of both approaches. As md alluded to, your strings of fractions are the same as using permuations. The above is equivalent to writing:

[3(28*44*40) - 3(28*24*40) + (28*24*20)] / (50 P 3)

Another example of fractions vs perms: the probability of an all-diamond flop is (13/52)(12/51)(11/50), but that's the same as (13 P 3) / (52 P 3)

Not only is the latter is easier to type, but calculators have the permute operation. Like you, I began with multiplying fractions. Once I realized they were permutations, the transition to combinations was painless: just replace some P's with C's, and this spares you the extra work of multiplying by 3's and such (when you're not using PIE, that is). When order doesn't matter, you can use perms, but doing so requires you to count all the arrangements in the numerator so that they cancel out with the denominator. With combinations, you don't have to worry about arrangements anywhere. Also notice that nCr = nPr / r!, so combos are just perms with the arrangements "divided out of them".

For a probability problem where order doesn't matter, you'll get the same answer with perms and combos because order either cancels out (perms / perms), or is not counted at all (combos). Normally, it's simpler to use combos in that situation. However, imo this problem is an exception because of how the paired boards aren't allowed.

Here's what the PIE solution looks like with...

permutations: 28[3(44-24)40 + 24*20] / (50 P 3)
combinations: {3*42[28(11C2) - (7C2)40] + (7C3)43} / (50 C 3)

Here's what your latest solution looks like with...

permutations: 28[3(24+16)20 + 24*20] / (50 P 3)
combinations: {42[28(5C2) + (7C2)20] + (7C3)43} / (50 C 3)

Quote:
 On a side note I want to thank forum members heehaww and whosnext because I've learned so much from your posts in the last few days.
What motivates me to write posts like this are eager learners like you. I'm glad to have been helpful!

 01-24-2017, 07:41 PM #10 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,787 Re: Specific Flop Odds with Pocket Pairs Great posts as always. Again, to echo what has been stated above, my mindset when tackling these types of problems is something like the following: 1. Try to solve the problem with combinations 2. If the order of the cards is important, take special care when using combinations (include either additional factorials or permutations in the calculations) 3. If over-counting is an issue, take special care. Either break the problem down into cases in which the issue of over-counting is eliminated or utilize the principle of inclusion-exclusion (PIE) as described by heehaww above (and in many other great posts in this forum). My over-arching advice is to continue practicing solving these types of problems. Get comfortable using Combination methods (they are actually quite easy.) Practice identifying when "order" is important and solving these via permutations (or factorials). Practice identifying when "over-counting" may be an issue and try to break down the problem into cases to get around it and/or learn PIE techniques. If you are so inclined, I would open more past threads on this forum where these types of problems and solution methods were utilized. Following a provided solution is often much easier and more illuminating than deriving the solution from scratch. Most importantly, a belated Welcome to the Probability Forum. There are many here who are able and willing to help out.

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