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Originally Posted by Fossil Rock
So ALL combinations of J,Q,K,A (with your math), not considering suit or connectors, are considered to be equal in value.... I don't think so.
Next please....
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Gder is on the right track but he forgot to divide by Pi.
The actual equation is:
If A = hole card # 1 and B = hole card # 2 then:
(A + B) / Pi x 1.2 (if suited) + 0.1 (if connected.)