Hot answers tagged

10

K7 steals one winner from A2 K7 wins 3456 and K8 does not K8 does not steal 4567 as A2 does not have a piece of it


8

You should break it into disjoint (non-overlapping) cases, and find the probability that you win with each case, and then add them up: Case 1. Heart on turn Occurs 8/45 times and your probability of winning is 0 Case 2. 8 (not heart) on turn Occurs 3/45 times and your probability of winning is 6/44 (three 3's and three 5's) Case 3. 5 on turn Occurs 3/...


7

The odds of getting aces do not at all depend on the number of cards remaining in the deck. They depend solely on the number of cards in the deck (52), how many aces are in the deck (4), and how many cards you receive from that deck (2 in holdem). You have a 4 in 52 (or 1 in 13) chance to get an initial ace. If you get that first ace, you then have a 3 in ...


6

It depends on what kinds of hands your opponent would go all-in with. If it's a tournament and he's short stacked in late position, he might have many types of hands. If it's a cash game or early in a tournament and he's got a relatively large stack, his range of hands is likely a lot tighter. Of course, if you know this opponent and his tendencies, you ...


5

My short answer is this: as long as your opponents have 100BBs, it doesn't matter how many BBs you have (as long as you have them covered). You will only play for their 100 BBs. Shoving every hand when you have 10000 BBs is as bad as shoving every hand when you have 100 BBs. The difference is that in former case you don't have to re-buy every time after you ...


5

On the flop in hold-em, it is not possible for one player to have a straight flush (or even just a straight or just a flush) and another to have a full house. A full house requires at least two cards on the board to have the same rank. A straight (or flush, or straight flush) on the flop requires that no cards on the flop be of the same rank. These are ...


4

The only way to lose with a King high straight flush is to a royal flush of the same suit. That means that the KQJT must all be community cards. The last community card is either the 9 or you have it as a pocket card. Case 1: 9 is on the board. The odds of this are the same as getting dealt a royal flush in 5 card stud: 20/52 * 4/51 * 3/50 * 2/49 * 1/48 =...


4

When you are dealt a card, it has 13 possible ranks. When you are dealt a second card, for it not to be a pocket pair, that can be of any rank different to your first hole card - i.e. 12 different card ranks. For the division by 2, the calculations are for the number of combinations of different types of hands and a combination disregards order. If you take ...


4

This is actually an interesting question and not because it might relate to good strategy when considering hand equity match-ups or Expected Value, (let's face it, random All-In shoves every hand don't fair well in terms of overall strategies) but because it's a good example of bankroll management, and how people misapply it. The key concept here is how ...


4

You need to calculate the odds of getting the exact flop that you need. Since the order doesn't matter, the first card dealt would have three possibilities, and then if you got one of those you would have two possibilities on the second card, etc. It would look like this: 3/50 * 2/49 * 1/48 = 1/19,600 = 0.005% EDIT Updated based on your comment/...


3

This answer assumes you have no additional knowledge of the opponent's cards. You have five cards in your hand. Of those, three are spades and two are not spades, so you need two more spades to complete your flush. There are 47 cards that you haven't seen, and 10 of those 47 cards are spades. You discard the two non-spades. The odds of drawing two more cards ...


3

There are five board cards in hold'em. Since you start with two known cards, there are 50 unknown. That means there are 50x49x48x47x46 ways the board can come. Since the order of the cards on the board doesn't matter, divide that by the number of ways 5 cards can be arranged (120), that's 2118760 total distinct boards. There are 47x46/2 of those boards that ...


3

The probability would be calculated by taking the odds of having 1 card pair and then the other two cards NOT pairing or forming a set. Calculate this for the three combinations (1st card pairs, 2nd card pairs, and third card pairs) and add them together: (6/50) * (44/49) * (43/48) = 0.096530612 (44/50) * (6/49) * (43/48) = 0.096530612 (44/50) * (43/49) * (...


3

First off, I believe it is easier to understand the probability of getting dealt a pocket pair this way: Your first card can be any of 52 cards, so it is not relevant in the calculation. The 2nd card must be one of the 3 cards that match the value of your other card. There are 51 cards left in the deck now, so the probability of being dealt a pocket pair is ...


3

There are 10 players at the table, any one of them has a 6% chance of having a pocket pair. So the chance that no-one is dealt a pocket pair is 0.94^10 ≈ 0.54, ie chance of someone having a pocket pair at a FR table ≈ 0.46. Excluding card removal for ease: You have a pocket pair, what is the chance that out of 9 players none of them has a pocket pair ? 0....


3

Welcome to the world of probabilities! 1. Questions, Answers, Frequencies Probabilities calculation is about a question requiring an answer. It is about a question involving: An initial state. This is the configuration of the world you want to analyze. This world is just a narrowed view of what you want to analyze. When analyzing poker hands and ...


2

This question is actually very similar to the "double your bet" strategy for baccarat, or any other close to even odds game. Casinos typically will always have a table max bet for these games to prevent some billionaire from coming in and keep doubling their bet until win back all of their loses. The number of times you can lose before you hit the limit is ...


2

TL;DR Given a flush on the board, five players will split 19.7% of the time, or about once every 5 hands. Given a fresh deal, the deal will come up with a flush showing and a five-way split 0.0389% of the time, or once every 2,569 hands. Some general probabilities Overall odds that none of the five players have a given suit (say, hearts): comb(39,10) / ...


2

From the Wikipedia article on Poker Hands: Of the 2,598,960 possible five-card combinations, {formula elided} 1,302,540 do not contain any pairs and are neither straights nor flushes. As such, the probability of being dealt "no pair" in a five-card deal is approximately 50.11% In seven-card poker, the frequency of such "no pair" is 23,294,460; the ...


2

The worst case for a possible straight flush is holding something like A2s, AKs, A5s, etc., where there's only one possible way to flop the straight flush. In that case, the probability is one in 50C3, or 19600. The best case is 45s..TJs, which is 4 in 19600, because there are 4 ways to flop the straight flush. Hands like 58s are 2-ways, hands like 79s are ...


2

Here is the maths for you. Or well my maths anyway. With 7 cards to choose from in hold'em, your hole cards and the board, the odds of making quads is about 1 in 595. (13 * (48 choose 3)) / (52 choose 7) which = 0.00168067227 or 1 in 595. This is over the entire 7 cards. So for another person to have quads in the same hand we figure out how many possible ...


2

Well on two different tables I think you have to take each table as a individual, independent scenario. It is unlikely to happen but it happens. So for a single specific hand the odds are: (2 / 52) * (1 / 51) = (2 / 2652) => (1 / 1326) Which in percentage gives you a probability of 0.0754% for this event to happen once. From here you can multiple this by ...


2

To be brief, In poker a player is never dealt two CONSECUTIVE cards. this in my view changes the odds, No. while also making the number of people at the table as well as the players position a material factor. No. Since the desk is randomly ordered, the order of dealing does not in fact change the likelihood of receiving any two cards. In ...


2

I think that it's definitely possible to win large field tournaments without any big suck outs, but it would be difficult to quantify how often it happens. My hunch is that it's more often than you think. A person can win a couple big pots early, play smart and aggressively, have good hands at the right times, and just keep building their stack as the ...


2

Stop saying winning. The best hand often does not win as it is folded. And can win without the best hand as a better hand was folded. You will see top pros like a durrr win with marginal hands a lot because they just know how to play them. All an odds calculator can do is give you the odds that it will be the best hand at the river. You have to do ...


2

You already account for the dependency of the events since in your second calculation you take into account the probability that the flush did not fall on the turn. You can just add the probabilities to get 0.35. For a quick estimation of the probabilities of hitting the flush: Count your outs Multiply by two on the flop Multiply by four and add 2 on the ...


2

The odds of hitting "one or the other" are dependent. But the odds of "missing both" are independent. So reverse everything: the probability of missing both is (38/47) * (37/46), or 1406/2162. Therefore, the probability of not missing both is 1 - (1406/2162), or 756/2162, or 34.97%.


2

These are the chances (assuming you have no ace): These are valid only preflop assuming that there are 50 cards left in the deck (you are holding 2) and you are one of the players (2 players = 1 opponent with 2 cards; 3 players = 2 opponents with 4 cards and so on)


2

Based on figures from http://www.flopturnriver.com/poker-strategy/pyroxenes-common-flop-odds-19147, the odds of flopping 2 pair or better, not including draws, is just over 5.6%. Including flopping a straight or flush draw, the odds are just a hair over 27%. If your facing a raise and re-raise, you should probably fold, as you're not getting the right odds....


2

Whether or not it is profitable to call depends not only on # of outs but on the format of the game (i.e. ICM may be the biggest consideration in MTT/SnG) A really simple rule I was taught to calculate equity is: # of outs after flop * 4 ~ equity E.g. You are 4 to a flush on the flop. Without knowing any other cards, you have 13-4 = 9 outs. Then 9*4 = 36%,...



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