[Paul2432]

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A [img]/images/graemlins/heart.gif[/img] K [img]/images/graemlins/heart.gif[/img]

A [img]/images/graemlins/club.gif[/img] Q [img]/images/graemlins/heart.gif[/img] J [img]/images/graemlins/heart.gif[/img] 6 [img]/images/graemlins/spade.gif[/img]


T-4 ways
[img]/images/graemlins/heart.gif[/img]-8 ways

so 12 ways to the nuts and 2 ways to 3 of a kind

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Actually a 9[img]/images/graemlins/heart.gif[/img] or 8 [img]/images/graemlins/heart.gif[/img] could make someone with T [img]/images/graemlins/heart.gif[/img] 8 [img]/images/graemlins/heart.gif[/img] or T [img]/images/graemlins/heart.gif[/img] 9 [img]/images/graemlins/heart.gif[/img] a straight flush.

Therefore you can't count those two cards as nut outs [img]/images/graemlins/smile.gif[/img]

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Don't forget that the 6h could make someone quads. This example only has 9 nut outs.

If your hand must improve, ten is the most nut outs possible. For example, you hold As Ts, board is Ks Js 3c 2d. Nut outs are the four queens plus six non-pairing spades.

Technically, I suppose you could argue the answer is still 21. You hold Kd Kc, board is Ks 7h 3c 2s. Any K,Q,J,T,9,8 leaves you with the nuts and improves your hand from KKK73 to KKK with a higher kicker.

Paul