Thread: Was the WPT correct?
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Old 03-12-2005, 08:43 PM
uuDevil uuDevil is offline
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Default Re: Was the WPT correct?
If I followed BruceZ's method from this thread correctly, the probability is

5*6/C(50,2) - C(5,2)/C(50,4)= .024446

which is odds of about 39.9 to 1.

So yeah, about 10% too high. Good catch.
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