The table shows the probability of another player having a pair of aces, when you have a pair of kings, in a player game, to be 4. Let's call the players A, B, and C. However there are 3! Thanks for the kind words. However that is a big if. So the probability of getting pocket aces and then losing is 0.

I tend to agree with your strategy of calling, which will keep more players in the hand, and increase your chance of losing. However, in a ten-player game there are 10 possible players who could get the aces, and 9 possible players for the kings. This answer is slightly too high, because it double counts the situation where two players have aces, or two have kings, or both.

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The math of this gets very messy due to the possibility of more than one playing having a higher pair, including the same type of pair. For example if you have pocket kings two players could have pocket aces. However it is easy to show the expected number of players who will beat you. The following table shows the average number of players who will have a higher pocket pair according to your pocket pair left column by the number of opponents top row.

To get the probability that at least one player will beat you I will make the not entirely correct assumption that the number of players with a higher pocket pair is a Poisson random variable with a mean in the above table. For examle if you have pocket queens and there are 9 other players the expected number of players with a higher pocket pair is 0. The table below shows those probabilities.

After this column appeared one of my fans, Larry B. Here are his results. Later Stephen Z. Take the number of higher pairs, multiply by the number of other players, and divide by 2. That is the percentage probability that there will be at least one higher pair. Using that formula you get the following for all situations. For those not familiar with the hold 'em terminology, you are asking for the probability of at least a pair in six cards, given that the first two are the hole cards of different ranks.

I hope you'll forgive me if I just do the probability of getting exactly a pair, including hands that also form a straight or flush. The other three cards must all be of different ranks from the 11 left. For each of these there are four suits to choose from. Now let's look at the number of ways to get a pair outside of the two hole cards. There are 11 ranks to choose from for the pair. For both of those ranks there are 4 possible suits. Take the product of all this and the probability is 0.

For readers who may not know, a "set" is a three of a kind after the flop, including a pocket pair. So the probability of making a set is In 2, pairs you should have made a set So you are The standard deviation is the square root of that, or The probability of luck this bad or worse can be found in any Standard Normal table, or in Excel as norsdist The probability of being on the losing end of KK vs. That is once every 45, hands, so your math was right. The following table shows the probability of 3 or more instances of having KK against AA in hands, by the number of opponents.

So, yes, I would say this looks fishy. The fewer the players, the more fishy it looks.

I would be interested to know where this game was. There are 50 cards left in the deck, and 42 of them are not aces or kings.

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Given 9 other players, the probability is 9 times that, or 1 in This might seem like an abuse of taking the sum of probabilities. However, it is okay if only one player can get the two aces. Enter your email address to receive our newsletter and other special announcements. Wizard of Odds uses cookies, this enables us to provide you with a personalised experience.

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Rick from Gardnerville, USA. Thank goodness I just discovered your great site. I have been trying to solve the following, and keep getting different answers. If I am dealt a pocket pair in Hold'em what are my chances of getting either three of a kind, or four of a kind on the flop next three cards. Elliot from Harwich, Massachusetts.

In hold em poker, what are the odds of being dealt pocket aces?

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And what are the odds of being dealt pocket aces twice in a row? Adam from Redding, USA. If ten people are each dealt two cards from a single deck what is the probability that two players will get a pair of aces? In a handed game of Texas Hold 'em, and the flop is three different ranks, what is the probability that three players have a set? Hi - Thank you for your web site. Thank you! In Texas hold em, if two players are dealt a pocket pair pre-flop, what are the odds of each of these players flopping a set three of a kind? I have read this book and I highly recommend it.

It is basically a book on exploitation.

- Sozialreportage: Einführung in eine Handlungs- und Forschungsmethode der Sozialen Arbeit (German Edition).
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Specifically, it teaches you exactly who to target in live low limit games and and sometimes It also teaches you how exactly to exploit those who are most profitable to target. The author does a good job of ranking these targets in terms of which one you should be prioritizing. There are targets that you exploit with "trapping" and there are targets that you can exploit with "milking". Basically, when you're trapping you're basically giving bettors and raisers the rope to hang themselves.

Milking is done to exploit donks that tend to bet or raise less, calling stations. It is the texture of the board that tells you how much to bet. Ummmm, what else. As to the quality of the writing, this book gave me eye strains.

The paragraphs are very long and I sometimes found myself lost mid-paragraph. Even though it is a short book 36 pages it is a brutal read but very focused in its subject matter, namely: Who are the most profitable players to exploit in and How exactly to exploit them. As to or higher, I think that this book is useful mostly insofar as there are or type players there.