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joeu2004[_2_] joeu2004[_2_] is offline
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Default List all combinations of 6/36 with unique 4 numbers

Errata.... I wrote:
Given the condition for full wheels ("all 6 of the drawn numbers are in
your subset"), the probability that the condition is met is COMBIN(18,6) /
COMBIN(36,6) for a subset is 18 numbers.

Similarly, for abbreviated wheels, the probability that the condition ("at
least 4 of the drawn numbers are in your subset") is met is COMBIN(18,4) /
COMBIN(36/6).


While I believe the full-wheel conditional probability is correct, I have my
doubts about the formula for the abbreviated-wheel conditional probability.
I need to give that more thought.