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What are the fraction of hands that can be classified as "indisputable winners" (aka "the nuts") after the river is revealed in Texas Holdem? By "hand" I mean the 2 hole cards you have that no one else can see plus the 5 cards on the table.

An indisputable winner is a hand that cannot lose. A clear example would be: you are holding the A,K of spades, and the Q,J,10 of spades are on the table. No one can beat you, no matter what they are holding. For clarity, there are ${47 \choose 2}$ hands that meet this exact criteria in each suit.

A hand that could be tied would be: you are holding A,K, the table has Q,J,10,7,2, with all four suits represented. No one can get better than a straight, and you have the best possible straight. Others could tie you, but you cannot lose. These are also considered "indisputable winners" as there is no risk to betting.

Bonus Question: Of the hands that qualify as indisputable, what are the distributions of types of winners? Straight-flush vs. 4-of-a-kind vs. full-house vs. flush vs. straight vs. 3-of-a-kind.

I'll take estimates if the exact calculations are too complex.

user3294068
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    if you watch carefully the card playing movie with Matt Damon, you get a brief view of his hidden cards. Turns out what he had was unbeatable. http://www.imdb.com/title/tt0128442/ – Will Jagy Aug 31 '15 at 19:58
  • Another way to word this question: Draw 5 cards at random from a deck. Select 3 of them, and choose any 2 other cards from the deck in such a way to make the best poker hand possible. What is the resulting distribution of straight/flush/three of a kind/etc. – Eric Naslund Aug 31 '15 at 20:01
  • In your question, how are you weighting the probabilities when asking about the distribution? For a given 5 card board there may be different hands that are best possible depending on the whole cards. How do we count these? For example, consider the board AsAh5d5c9h. There are 7 indisputable hands, AcAd, Ac9s, Ac9d, Ac9c, Ad9s, Ad9d, Ad9c. Does this count as 6 full houses and 1 four of a kind, or 6/7 of a full house and 1/7 of a four of a kind - are boards that have many best possible hands being counted with multiplicity? Does the board As2s3s4s5s get counted as a straight flush 1081 times? – Eric Naslund Aug 31 '15 at 20:12
  • @Eric, I think you are misunderstanding. You are dealt 2 cards. 5 cards are on the table. Each other player also gets 2 cards that you don't know about. I'm asking about which combinations of cards you do see tell you that no matter what your opponents have, they can't beat you. – user3294068 Aug 31 '15 at 20:25
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    I think the unclarity results from different uses of the word hand. You seem to be using "hand" to refer to the entire $5$+$2$ deal, whereas @Eric seems to be using "hand" once to refer to the $3$+$2$ cards that determine the value and once to just the $2$ cards on the player's hand. If I'm right in thinking that by "hand" you mean the entire $5$+$2$ deal, then the answer to Eric's question would be that his example counts as $6$ full houses and $1$ four of a kind? – joriki Aug 31 '15 at 21:38
  • @Eric: As2s3s4s5s only gets counted $46$ times; indisputable completions must contain 6s. – joriki Aug 31 '15 at 21:42
  • Each individual collection of cards, 2 hole cards you can see + 5 cards on the table, counts once. Added a comment to the question to clarify this. – user3294068 Sep 01 '15 at 13:03
  • This type of hand is called "the nuts". – DanielV Sep 01 '15 at 13:28
  • I have an algorithm than can calculate an exact answer in estimately 5 days. Unfortunately I need my computer for other things :-(. – Albert Hendriks Jan 01 '16 at 04:18
  • It seems like the biggest challenge to this problem is to describe all the community hands that allow an indisputable hand of a particular type. For example, for an indisputable hand of three of a kind, the community cards must not allow a straight, which means none with: 4 consecutive ranks (an outside straight draw), 3 consecutive ranks (double outside draw), 2 sets of 2 consecutive ranks separated by one (inside draw) or two ranks (consecutive inside), 1 set of 2 separated by 1 (inside plus outside draws) or two ranks from another card (consecutive inside), etc. – Χpẘ May 08 '17 at 23:13
  • @EricNaslund Not exactly, there are a lot of interresting cases here. You don't need to use your both hole cards for example: if there's a royal flush on the table you've got the nuts (no matter what you're holding). Another case is where a card you're holding don't contribute to your holding, but blocks other players from beating you. For example if the table is 4,7,8,Q,K of hearts and you're holding 5 and A of hearts you've got the nuts (A high flush) where your 5 doesn't count, BUT that card is important because it hinders somebody from holding 5, 6 of hearts making a straight flush. – skyking Jun 16 '17 at 11:11

2 Answers2

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Any Ax can be the undisputed winner. It could be a flush, straight flush, trips, quads, or even just a pair. There 4 X 48 or those.

Any connector 45 or higher can to make the top of a straight there are 16 each and 9 combination for for 9 x 16

1 gapper same thing

2 gapper same thing

Any pair 4 or higher could be the nuts 13 X 6

paparazzo
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An approximation can be reached by consider the most common setups of table cards.

For example if the table is paired the nuts is most often the quads which is unique. This means you have about $1/1081$ of getting that.

If the table is three-suited the nuts is most often an ace-high suite which can be achieved in nine ways. This means you have about $9/1081$ chance then.

If the table is otherwise unpaired the nuts is the top trip which you have three possibilities for. This is $3/1081$ chance.

Now a paired table is about $50\%$ chance, a 3-suite is about $21\%$ and the rest is $29\%$. Putting this together we giving about $0.3\%$ chance overall.

If you want a more accurate number you can do a more detailed breakdown of the table possibilities. I think the next most important case is top pair on the table which means that you have additional ways of getting the nuts (ie top houses).

skyking
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