Probability of an Indisputable winner at Texas Holdem 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.
 A: 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
A: 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).
