This is a discussion on Pocket pair vs two overcards odds within the online poker forums, in the Learning Poker section; Hi everyone, It might just being missing something, so please forgive me if I seem stupid asking this question. Everywhere you go you hear that eg. 66 


#1




Pocket pair vs two overcards odds
Hi everyone,
It might just being missing something, so please forgive me if I seem stupid asking this question. Everywhere you go you hear that eg. 66 vs AK is a coinflip situation, or close. Like 55% vs 45%. But when holding two cards you have 7 outs  roughly that give you a 32% chance of atleast 1 of your cards making a pair. That's what i don't get. 66 is beating AK unless it hits, so why is it a coinflip situation? The straight possibilities that AK has can't make up for those 13%. Ofcourse it weighs a little up, but then the probabillity of 66 making a set must drag something down again since you have 12.5% chance to hit your set which would kill AK unless it hits the straight or hits a full house which is extremely unlikely. Any kind of explanation is appreciated Thank you 
#2




Alex, I asked myself that question too when I started playing.
A very helpful site for odds is Wizard of Odds: http://wizardofodds.com/games/texasholdem/ If you look at the headsup odds of hands, you'll see the coin flip again: http://wizardofodds.com/games/texasholdem/2playergame/ These types of tables are created with the math explained on the wikipedia page and the numbers are confirmed with software that runs millions of hands and comes up with the same percentages for these types of tables. I've used Poker Academy (http://www.pokeracademy.com/) to look at hand combos. It runs 50,000 hands to several million and confirms the results of the Wizard of Odds (http://wizardofodds.com/games/texasholdem/) tables or hand calculators (http://wizardofodds.com/games/texasholdem/calculator/). Hope that helps a bit. 
#3




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If the overcards don't make a pair on the flop and have no draws, now they are significantly behind (~27% to win) and it gets worse if the turn is a blank. Another scenario is when the overcards flop a flush draw, then they are actually slightly ahead of the pocket pair postflop with two cards to come. 
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#10




66 vs AK is better to be played only coinflip if you miss the flop 66 is awful hand even when flop is 9 10 2 so it is easy fold, so 66 is not good hand it is good when you hit set on flop, all other ways you are behind against every hand only coinflip 6's pair is little favourite vs 2 over cards.

#11




I think because if it were heads up your chance of hitting your outs would be about (6/48) *5 because of the 5 community cards. That would equal to a 0.625 probability.
This is obviously very oversimplified because there is also the possibility that your opponent makes a set and also you have to take into account the other cards folded if you have several players. So that would decrease the probability below 0.625. Figuring out the probability is not as simple as counting the outs, which is why they usually get the percentages by running the hands millions of times. 
#13




Here are some interesting stats. There are 1716 unique pair vs. overcard matchups. That's after eliminating equivalent hands, i.e. 6c6d vs AcKc is has the same equity as 6c6d vs AdKd, which is the same as 6h6s vs AsKs, but not the same as 6c6d vs AhKh. Pairs have an advantage in 1435 cases, and a disadvantage in 281 cases. The maximum equity for a pair is 57.5% for 8c8d vs Ah9s. The minimum equity for pairs is 46% for 2c2d vs JhTh. The most even match is 3c3d vs AcTc with an equity of 50.007%, a mere .007% from equal.
I suppose it's a bit late, but as far as the OP question, I don't understand where 32% comes from. Here's a rough calculation. Preflop there are 4 cards missing from the deck, leaving 48. From those 48 we have to choose 5 for the board (flop+turn+river). As you say, 6 of those will pair the overcards. Suppose they don't pair? Then all 5 board cards must have come from the 42 cards which don't pair the overcards. Thus, the odds of NOT making a pair are (42 choose 5) / (48 choose 5) = 850668/1712304 = 49.7%. So the odds of making a pair are 50.3%. Of course this completely ignores straights and flushes and the fact that the pair might also improve. Which is why the exact numbers vary as above. 
#14




Might be a dumb question here? But I never have gotten the math in this. No one seems to factor in the cards that were folded before the flop. Seems to me the odds of an Ace or King being in the muck must be pretty high therefore removing them from the possibility of hitting. Which, and I could be convinced otherwise, I limp AK unless I have chips to burn.

#15




Well if you knew exactly what cards were mucked you could factor them it. But if not, then you must consider them random. You could say the same thing about the bottom of the deck that never gets dealt. The odds of an A or K being there must be pretty high. But that doesn't mean you can count on it. That's what the whole notion of "odds" is about  the fact that you don't know for sure. The fact that an A or K could be in the muck is already factored in by considering those cards random, which includes A & K, as well as other cards.
OP: Thinking about it more, the 32% you mentioned looks like the odds of hitting a pair on the flop. The "coinflip" situation you asked about is for the showdown, which includes the turn and river. 
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Im wondering because I don't know myself Quote:

#18




re: Poker & Pocket pair vs two overcards odds
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There is also a small chance the board hits four of a kind to deny the pocket pair 
#19




I know I'm learning my lesson with AK.
AK does dominate many players all in range, but AK is a drawing hand and needs to improve post flop. I think lots of people, me included, have been overvaluing the strength of AK Preflop. But it does seem to work fine for some. Straight up probabilities would suggest AK be played with caution PF. That said, looking at hand histories of many MTT's A or K does seem to hit the board at a greater frequency than probabilities would suggest. That in turn influences future decisions regarding PF play for AK. Factoring in the many players PF ranges and their odd decisions PF, across all skill levels, AK would dominate many situations against many players. Bottom line, IMHO, Stack, position, PF action, AK should be played completely dependent on the present situation and not in any way an automatic shove. 
#20




@alex if you want to estimate EV pre, you need to multiply by 10 (2x per streets ), giving 10*6 = 60%, now the correct question is why we don’t have 60% but less.
To partly answer to this question, we reduce set outs for villain who has 2 outs for set, 10x2=20% The final answer in this topic is that our equity AK vs 66 is 60%20% = 40% Rest of the 5% difference comes probably from the fact that 10 is not precise number, and that A or K hitting set is more likely than 66 hitting quads, dominated flushes, board hitting funny 88992, draws and so on. Just a warning though that above is rude black box model, and totally made by me, so it will contain quite large flaws. 
#22




How do you get 32% for the odds of hitting the flop with unpaired hole cards? You have 2 cards in your hand, leaving 50 in the deck. 6 of those cards left in the deck will pair your hand (or better), which means 44 won't. So from the (50 choose 3) total possible flops (44 choose 3) of them don't pair your hand. That means the odds of NOT hitting the flop are 44x43x42/50x49x48=67.6%. Therefore the odds of hitting the flop are 1 67.6%=32.4%. This counts all hands that include a pair which hits your hand, i.e. one pair, two pair, trips, full house, and quads, but not any straights or flushes. It also does not count the chance of the board making a pair or trips by itself. But you probably don't want to count those anyway because then everyone has that hand and you didn't really improve.
If you want the odds of flopping one pair exactly, and nothing better, look at the combinations that give that result. There are 3 cards that will pair your first card and 3 that pair your second. For each of those you have 44 cards left that will not pair your hole cards. But you can't select just any two of those because you don't want another pair on the board, giving you two pair, which is the way I have seen it posted several times. Instead, you select 2 of the 11 remaining ranks, then give each one of the four suits. So the calculation looks like this: 3x2x(11 choose 2)x4x4 = 5280. Again the total number of flops is (50 choose 3). So the odds of flopping exactly one pair, using one of the unpaired cards in your hand, is 5280/19600 = 26.94%. (The mistaken calculation, which includes two pair, gives 28.96%) 