Posted by Jay on Friday, July 20, 2012 at 1:32am.
Now, each of those can be arranged in 7!/(4!3!) or 35 ways
let' calculate the prob of one of these, HHHHCCC
But the same prob is true for all of the 35 cases,
so multiply that by 35 to get
prob(4from one suit, 3 from another) = 1573/1029112
so the prob of NOT any of those = 1 - 1573/1029112
So the odds in favour of 4 from one suit, 3 from another
= 1573 : 1027539
check my arithmetic, I did not get the same answer you gave.
oops, for some reason I read it as
4 Hearts, 3 from another
so it could be
So there are 12 of these and each can be arranged in 35 ways
so to get the prob of one of these 420 cases, multiply
(13/52)(12/51)(11/50)(10/49)(13/48)(12/47)(11/46) by 420
(my calculator is overheating)
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