QQQ87
A “set” of Queens with 87 kickers is the worst such hand. The board reveals Q 8 7 3 2 with no attainable flush, and hero holds QQ. Proof that that is one of the best (worst) we will do:
“Prime set or higher” is all the time attainable
If the best card on the board is unpaired, trivially a participant can have at the very least a set of that card. A participant can’t have “blockers to a set” besides by having the set.
If the best card on the board is paired, that solely allows even stronger fingers. If we have now “journeys” with a 3rd card of that rank and an unpaired card, that doesn’t block the opponent from having a full home (with the final card of that rank and a paired card). Subsequently, we all know that pairing the highest card — or another card — would not permit us to decrease the rank of the “worst nut” three-of-a-kind hand.
The Pigeonhole precept constrains ranks
For a set to be the nuts, there should not be any attainable straight on the board. Thus, from any group of 5 consecutive ranks, we will put at most two of these playing cards on the board; if we put a 3rd, then a participant may have the lacking two. (Trivially, if we have now a set, we do not need any blocker to a straight, even disregarding that there are a number of fits to select from.)
Specifically, from the ranks 2 by means of 6 inclusive we will select at most two playing cards, and from 7 by means of J inclusive we will select at most two playing cards. To have 5 whole neighborhood playing cards, subsequently, we should add a card greater than J; and Q is the bottom accessible possibility.
It stays to confirm by inspection that from the 7-through-J vary we could select 8 and seven, after which from 2-through-6 select 2 and three, with out making a straight attainable. The 8 and seven will essentially rely as our kickers.
