Aha! Finally a question I can answer; I think.
Let's say you have unsuited A, K, Q and two other cards. The strategy is
is to hold only two picture cards. The combinations are AK, KQ, or AQ.
If you hold KQ, the possibilities for a straight are: A-KQ-JT, KQ-JT9,
whereas if you hold AK, the only chance is for an AK-QJT. Thus you have
twice the chance for a straight when you discard the Ace.
If the three high cards initially are AKJ, it's the same logic. With KJ
held, the two possibilities for the straight are again A-K-Q-J-T,
K-Q-J-T9
It is even better with an initial AQJ. When you hold QJ, you have the
chance for a straight three ways: AK-QJ-T, K-QJ-T9, and QJ-T98
The logic doesn't apply for the game where there is a bonus for 4-Aces,
as opposed to four Kings, Queens, or Jacks. So the above rule is for
the plain vanilla Jacks or Better.
Corrections welcome.
We mere mortals play VP for fun; unlike Bob who makes the cards dance to
his tune. - A Myth
I know the math says to discard an unsuited ace when you hold 2 other
picture cards but can anyone give me the logic behind it (other than the
math). The wife asked me and I cannot give a clear explanation
Chet K
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--- In vpFREE@yahoogroups.com, chet Klyn <chetkl@...> wrote: