Texas holdem, KKK22 full house showing on table.   Player 2 has the K2 hole cards to make the same full house, is the pot split among all players or is it player 2 pot?
In Texas Hold'em poker, there are some cases where a particular card, often called kicker, acts as a tiebreaker between players to determine who wins the pot, or if the pot has to be shared.  For some combinations such as three or four of a kind, there is no doubt: the kicker is the 5th card of the chosen combination, and determines the winner.  Example     Alice has 7 and 10   Bob has 7 and K   On the board are 7, 7, 2, 3, 4   In that case, both have a three of a kind. But because Bob has a king and Alice only a 10, he wins.  Most of poker websites explaining the rules clearly mention the role of the kicker applying to three of a kind and double pairs. But much fewer say something about colors, straights and full houses, and I found many contradicting answers. I also asked the question to some used online poker players and they also gave me contradicting answers.  I have tried to search for official tournament rules, but most of them only explain what happens with bad behaviors, bad deals, incorrect or confusing betting, showing cards when you shouldn't, acting when it's not your turn, etc. without mentioning subtleties about combinations at all.  To simplify my question, I will take three examples; I think it's better to start with examples before getting to the general answer if one exists. So, what's happening in the following 3 examples? Could you give a more generalized answer?  Example 1 - Flushes     Alice has 2&clubs; and 3&clubs;   Bob has 4&clubs; and 5&clubs;   On the board are 6&clubs;, 8&clubs;, 10&clubs;, 2&hearts;, 5&diams;   Contradicting answer 1: it's a tie, because the highest card included in the flush is the 10, which everybody chooses to include in their 5 showdown cards. Contradicting answer 2: Bob wins, because he has the greatest private card that is part of the flush  Example 2 - Straight     Alice has 6 and K   Bob has 6 and J   On the board are 4, 5, 7, 8, 10   Contradicting answer 1: it's a tie, because the greatest card in the straight is the 8 for both Alice and Bob Contradicting answer 2: Alice wins, because she owns an extra king, compared to the jack of Bob  Example 3 - Full house     Alice has 3 and 7   Bob has 3 and 6   On the board are 3, 3, 2, 2, 5   Contradicting answer 1: it's a tie, because one is supposed to choose only five cards to make a combination, and a full house is already five cards. There couldn't be any kicker, and thus their showdowns are strictly equals. Contradicting answer 2: Alice wins because of her extra 7, compared to the 6 of Bob  Thank you very much for your answers.  Note: I'm unable to post next to you, so I edit my own post; strange not be able to answer to an answer.  Ok, So if I summarize what you are saying :   In the flush case, Bob wins because at some point, their hand differs. Technically, I can continue comparing the highest private card to decide who wins. I had it correct. In the straight case, if the highest card of the straight is public, then it's technically always a tie, no matter what the players had as second private card (asuming that only one of the two cards was part of the straight). In a full house if both players have the same triplet and the same pair, it's always a tie, no matter what the players had as second private card (assuming again that only one of the two was part of the full house).   Please confirm me that what I summarized is correct.  Thank you for your answer.
Meta duplicates don't require an answer. Don't suggest there is one when that's not true This request was sparked . The vote to close as a duplicate on meta, in contrast to the main site, does not require for the duplicate to have an answer. Yet, the closure message still says:     This question has been asked before and already has an answer.    ...even when this is not the case. Perhaps we should either only show this particular sentence if there is indeed an answer, or remove this altogether for meta.   Note: I fear this might be a duplicate request, but I was unable to find one. (Not even one without an answer :) ) Why won't my villagers breed I have created the following structure. The intention is to create fairly compact houses for villager breeding.   This is a section of the houses.   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I dont want to use ANT or special software.  SF must have a native way to achieve this.  And if not...  then I am very disappointed in humanity. A slick proof that a field which is finitely generated as a ring is finite It is a known fact that if $k$ is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism $f:\mathbb{Z}[x_1,\dots,x_n]\to k$ for some $n\in \mathbb{N}$, then $k$ must be finite. Since finite generation as a ring implies finite generation over the prime field ($\mathbb Q$ or $\mathbb F_p$), by Noether normalization it follows that $k$ must be a finite extension of its prime field. In positive characteristic this finishes the job and in zero characteristic, should lead quickly to a contradiction, though I don't see immediately how. Is there an elementary/slick proof of this fact? Why didn't mastery of the Elder Wand move on from Harry? 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