On 11/20/2016 01:33 AM, Dominique Laurain wrote:
How do I get a solution similar to the one for flushes?
I guess you don't hope for the exact same solution, for exercice "hand
of four" than for "flush" :-)
If your question "How do I get" means "How I can code " and is related
to the probability to get hand of four (end of exercice),.., why not :
Hands = Subsets(Cards, 5)
print len(FourOfKinds) / Hands.cardinality()
len(FourOfKinds) is 624 because 13 choices for the "kind", 12 choice
for the kind of the "odd card" when kind choosen, 4 suits for that card.
624 = 13 x 12 x 4
I managed to get that far. I am looking for a solution that uses a set
for FourOfKinds and then
print FourOfKinds.cardinality() / Hands.cardinality()
Tom Dean
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