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It looks like the first part of my faulty argument still holds, namely that diamonds 4-2 (4 with west) is exactly equally likely as diamonds 3-3
We could deal every combo of 3-3 diamonds in 15 ways (along with 2 out of 6 clubs) and every combo of 4-2 diamonds in 20 ways (along with 3 out 6 clubs).
So assuming all clubs are equal (which is not exactly correct because as noted E wouldn't discard a club from AK or Ax) there are :
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The cases we can be in are when LHO gets:
Jxx / xx - 5C2 * 6C2 = 150 cases
xx / xxx - 5C2 * 6C3 = 200 cases
ways to deal significant diamond layouts. 200/350 = 57.14%
Isn't it correct ?
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By the way, LOVE the title.
The math is simple... once you know what math applies for the situation
EDIT: I see now, while every combination of 4-2 is more probable that every combination of 3-3 there are 20 combinations of 3-3 and 15 of 4-2 so it's evens out. So yeah, first quote is true