[Aisthesis]

Ok, here goes. First, I'm going to skirt the ranking issue for the moment and just assume that we have one (Karlsson-Sklansky might be a good start, but it depends on what we're doing with the results).

I'm going to define "median best starting hand" in the following way: GIVEN a particular hand, it is "median best" if the probability at a table of n players is 50% that yours is the best hand among the random hands dealt.

So, assume we have a way of ranking starting hands, of which there are 1326. There are obviously lots of equivalencies, but it won't hurt the solution if we just assume that we can rank them all 1, 2, 3, 4, ...

You are dealt the hand ranked r at a table of n players. Then the probability that no other hand beats you is: [(1326-r)/1325]~(n-1). (Sorry, but "~" is the best I can find on my keyboard to represent an exponent).

So, at a table of 9 players we want .50 = [(1326-r)/1325]~8 or 2 = [1325/(1326-r)]~8. Solving this, my computer calculator puts r at 111.

Hence, we want the hand ranked 111.

Now, as to particulars (will depend on how to rank these hands, obviously): Pairs all have 6 instances. Unsuited non-pairs 12 instances; suited non-pairs 4 instances.

Pairs 88-AA cover 48 cases. ATs-AKs cover 16 instances. ATo-AKo cover 48 cases. That puts the total at 112.

So, assuming ATo is "better" than any hand other than the aforementioned and the worst of those, ATo should be the median best starting hand at a 9-player table.

Side note to PairTheBoard: This means my initial attempt on UPF, where I came up with AJo on the basis of a slightly faulty attempt at assigning probabilties, was indeed a little off.

As to the practical relevance: One fairly obvious case is shortstacked in a tournament looking for an opportunity to double up. What do you need in EP to go all-in? For me, anyway, it would depend also on just how short my stack is relative to the blinds. But it at least provides some orientation in making this decision, I think.