# Tag Info

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It turns out that each shuffled deck is in the order that may have never before existed in the history of the universe! :) The odds of you getting two 52 card decks arranged in the exact same order are 52! ~= 8 x 10^67, which is waaay more than the number of atoms on Earth (~ 10^50). For a detailed explanation, please check out a great video answer on TED....

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This depends on number of things you have to consider, not only the direct odds. What I mean: The pure odds you calculate should be used if you expect your opponent to check the turn and you see free river. Always calculate implied odds! People usually bet on turn and river! If you expect your opponent to bet on the turn, you have to calculate this too - ...

3

That depends on what you're contemplating, and what you think future action will be. For example, if you're contemplating a bet that will put you or your opponent all in, then the odds of the next two cards are what matters, because you're buying the right to see both of them. But if you're contemplating calling for a draw, and you both have stacks, well ...

3

Frisbee's answer is correct. But it only answers for eight players (although the other numbers are hidden in that table). Since the OP asked for 8-10, I will add the answers for nine and ten players. The odds of an ace not occurring in the next x cards dealt (after you're dealt two aces) is ( 48!/(48-x)! )/( 50!/(50-x)! ). That's the probability that no one ...

2

Your question is confusing, and it appears that you may have misunderstood the terminology. "So I can calculate what the maximum bet size is with a positive EV. However, that would be too high, since we surely don't want a 0 EV." Do you mean if a bet size is too high, it would not be called, therefore the final value would be zero? This is not the standard ...

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Not really, I'd have played it exactly the same, especially given this is a multi-way pot, you want to isolate here against baby aces and small pairs hitting a set on subsequent streets. A King is well within any of the other players' ranges too. It's the right play.

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Pick is it 8 or 10 At 8 I think it is 0.4114 Or 1.4306 : 1 This is for exactly one ace out - (not two) Using combination (2/1) * (48/13) / (50/14) (2 aces need 1) (48 non ace need 13) / (50 cards need 14) Both other aces out would be 0.0743 Add them for 1 or 2 aces out = 0.4857 You can get that same number with 1 - 48 / 50 * = 0....

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They both have aces full of kings. The pot would be split. Aces are always higher than kings in holdem, except when played low as part of a A2345 straight.

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I have seen it on TV a few times Based on hand independent it is 0.000256 (quad) * 0.0017 (boat) = 0.00000044 From the boat there are 45 cards out and need to match the pair on the board So the odds of getting quadded are: (2/2) / (45/2) = 0.00101 = 0.1% Which also = 2/45 * 1/44 Highest or lower boat is immaterial. Quads beat a full house.

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Those number are wrong 19.15% turn 19.57% river 34.97% turn or river (I think you are missing that 5 cards are out) At the flop if either are all in then you can count on no more bets on the turn. You have to base it on what you think your opponent will do on the turn If you don't hit If they bet in to you on the flop then highly likely they are going ...

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When playing heads up I find you almost have to open with anything, if you are normally a conservative player, heads up can be hard, especially if short stacked. Try and adopt a really aggressive strategy, as if you continue to play conservatively and your opponent is marginally more aggressive, they will eventually win due to the swallowing of the blinds/...

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As you play and will gather experience you might start feeling things when you play and it will affect your decisions. (At least that's what happens with me) Sometimes when you have a really good hand(2 pairs or better) and your oponent starts to be aggressive when he normally never is. That's when you know it's time to fold. Even if the situation says it's ...

1

It can depend on the casino or cardroom you are playing at, some have different house rules. With that said from my experience it doesn't matter which seat he picks, he will be big blind and the game will continue as normal. This is by far the normal in most places I have played and dealt in. I have dealt in places that did it a little different. I have ...

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There are far too many situations you'd need to work out to calculate by hand. 1a. Just for AQo vs P1 & P2 range, the permutations of hand matchups is already huge. P1 range: 40% x 1326 = 530 hands P2 range: 10% x 1326 = 133 hands 530x133 = 70,490 match ups 1b. And each one of those would still be too difficult to calculate for the ...

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Just outs to improve on the next card is fairly straight forward. Probability to win is much more complex. If it is just you against a single player on the river then you can calculate. A starting hand against 3 random hands is very complex. I assume you mean chances (not change). You have poker calculators but there is no formula A straight can get ...

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You can use this this calculator, but basically you would need to know your opponents hand in order to actually calculate the odds. The probability you have of hitting your outs however, can be calculated. Take a look at my other answer to learn how. If you have a solid read you could try to include the probability of him hitting his hand after you have ...

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Everyone check to the button, no one is showing strength. Someone wants a free/cheap draw The button's job is to show strength, he might have anything, even after raising pre-flop. You know he has something, probably a suited-connected hand or a small pair, it's unlikely that he has a high pair in this action, although it's always possible. We later come ...

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You got your money in with the best hand. And he was not getting pot odds to call. Even if you knew what was in his hand you make the right play. At the flop you were 68%. Even if he filled up the straight you still had 4 outs yourself for a full house. On that board you had a very good hand. I would not have put a player on 24. And I doubt they put ...

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I do not know if the callers are big/small blind, so I am going to be pessimistic and say the pot is 4*6= 24\$ after the flop. On the flop, before you shove the pot is 24+15+15=52\$. You shove with 48\$, hence he has to call with 33\$ the pot odds are 33/100 or 33%. He has 8 outs, hence the probability of hitting is aproximatley 8*4= 32%, or more precise 30% (...

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In general, equity is your expected value when you would run the same hand infinite times. In probability theory, we take the calculated chances as truth in a scenario where the event is repeated infinitely. In poker, your equity or expected value can be calculated by taking your chance of winning and multiplying that with the value of the pot. We have in ...

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A poker hand is only five cards, so it's a split pot.

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Well it seems like you are talking about value betting. The EV is basically if it is worth calling or betting that much. But when you are value betting you should be sure you have the best hand in play. Then the size of the bet is the maximum amount you think your opponent will call. Say you have the nuts if you are putting your opponent on a strong hand ...

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The answer is that you need immediate pot odds. There is absolutely NO basis to assume they are going to let up on the turn if a blank hits. According to Skansky every time you make an opponent make a bad decision you benefit. Opponent gave you 3:1 and you called thinking you would get a free card on the river. Any good opponent is going to turn up the ...

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I have been struggling over the same example for the same reason. Let P(A) = Prob A has nuts = 0.2, P(A') = Prob A has dead hand = 0.8, P(B) = Prob A bets, P(A'|B) = Prob A has dead hand given A bets P(A|B) = Prob A has nuts given A bets Then P(A|B) = [P(A)P(B|A)]/[P(A)P(B|A)+P(A')P(B|A')] = (0.2*1)/(0.2*1 + 0.8x) and P(A'|B) = [P(A')P(B|A')]/[P(A)...

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