11

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 ...


11

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

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 showdowns,...


7

This is an interesting question. The key consideration to me is "Player A cannot win the pot via a split pot", meaning that there should be no scenario whereby the board contains the nuts. This is essentially the error with the 2 answers above. IF the board comes A♠2♠3♠4♠5♠, then someone other than player A needs to be holding the 6♠, otherwise it is a ...


6

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 ...


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 ...


6

I exhaustively simulated (every possibility is run) AA vs any pocket pair and AA vs any suited connector. Firstly let's cover the pocket pairs: Obviously it is best for the under-pair to have different suits, in order to win with flushes. The best pocket pair to have is pocket eights. This is because all straight possibilities will result in a win. This is ...


6

Probability is a measure of information. It's not about where the cards are--it's about what you know about them. Let's say you have a flush draw on the turn--9 outs. What odds should you use for betting? You should assume a 9/46 chance you'll hit that spade. Now take 10 cards off the bottom of the deck stub and throw them out the window. What are your odds ...


5

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 = ...


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 ...


5

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 ...


5

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/updated ...


5

In general, if you remove luck from the game, it becomes much more difficult to win money. This is because the average poker player thinks they are better then average. Mathematically this cannot possibly be true, so therefore many players think they are better than they actually are, and the reason they can continue to believe that is sometimes they get ...


4

A poker math geek gave me the following formula to use as an estimation. (Edit: The credit for this shortcut goes to Phil Gordon and is known as the Gordon Pair Principle.) Let's say you have 55: Work out how many pairs are higher than your pair - 66, 77, 88, 99, 10, JJ, QQ, KK, AA = 9 pairs left Count how many players are left = 8 Multiply higher pairs ...


4

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) * (...


4

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) / ...


4

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)


4

At first, I thought, OK, lets quickly answer this, returning the favor. But it turned out to be a rather tricky one! One online calculator gives 48.18%, the other 48.28%, off by 0.1%, already very strange. My first result was 49.09%, off by exactly the runner-runner 0.909%. You also confused me with your cases and some calculations. The tricky part is - of ...


4

As I mentioned in a comment above, this was not my answer, found it here. Although I did find you could also substitute the 7s with 8s, and then the Queens with Jacks, 10s or 9s. Suits matter in the solution. K♣K♠ A♣A♠ A♦K♥ A♥K♦ Q♣Q♠ 7♦7♥ 7♣7♠ Can check it out here.


4

Rich? It's extremely hard to be "rich" just by playing poker, but it's simple enough to make a living off it. The reason being that you have to be vastly superior to the field to make steady money off poker with a rake (like at Casinos). For example, in order for you to "break even" with a 7% rake, you need to be a 53.5% winning player. If you eat fancy ...


3

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 ...


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

The hands that can't happen with seven cards are exactly the five card hands that improve no matter which two cards are added to them. This basically means hands with low kickers. A high card hand (think of very card as a kicker) like 7 5 4 3 2 is impossible because any cards added to it will improve it. Some classes are easy to show that there are always ...


3

Each players' hand is comprised of their best 5 card hand... Ruane has Two Pair: Queens and Tens with a King kicker Hallaert has Two Pair: Queens and Tens with a 6 Kicker Ruane wins with the higher kicker.


3

Assuming heads-up. Since you hold AQ and villain has AJ, there are 52-4=48 cards left on the deck, with 2 Aces remaining. So, Prob[at least an Ace on the Flop] = 1-Prob[no Ace on Flop]=1-(46/48)(45/47)(44/46) = 1-0.878 = 12.2%


3

The standard line would be to re-raise to what ever your defend re-raise is. Depending on how often the BU raises you should be defending with a re-raise a decent range. If you defend with 88+ and sometimes just suited connector then AA will not stick out. Just call with some of your re-raise range sometimes but I would always re-raise AA here. Give ...


3

I think that calling here would be the worst play and I agree with your reasoning. Additionally, one of your outs (9 of diamonds) either helps another player out even more or else kills your potential action. I would say however that although folding is fine here, raising would be better than calling if you were to play on (although I'd like it much more ...


3

This is quite difficult to answer without making some assumptions. You could just calculate the chances of your opponent having a hand with a maximum of one heart based on the known and unknown cards, which comes to about 95%*. However, this is a flawed approach because it assumes that your opponent is equally likely to get to the river with any two cards ...


3

To elaborate on Paparazzi's answer, you can calculate the odds of a card being dealt next by dividing the number of cards you wish to know the chance of coming by all remaining unknown cards (including those in the deck and those in other players' hands). So if you have a suited hand and one card of that suit has come on the flop, the chance of getting ...


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