Re: Ranges of size 5 of samples from [1, ..., 52]
- From: j.howroyd@xxxxxxxxxx
- Date: 10 Dec 2005 16:54:43 -0800
Let's do a single case, say P(W=7) (with the general case P(W=w) in
brackets).
First we need to consider the number of sets of 8 (= w+1) consecutive
numbers. These are {1,2,...,8} through to {52-7,...,52}, so I make
that 52-7 (= 52-w) such sets. So the sets of 5 elements with range 7
can be generated from these by taking the first and last element and
selecting the remainder (3) from the middle 6 (= w+1-2 = w-1) elements;
giving C(6,3) (= C(w-1,3)). So the total number of sets of 5 elements
with range 7 is (52-7)C(6,3) (= (52-w)C(w-1,3)). This is out of a
total of C(52,5) possible sets of 5 elements from 52. So the
probability
P(W=7) = (52-7)C(6,3)/C(52,5) (in general, P(W=w) =
(52-w)C(w-1,3)/C(52,5) as stated).
The general case is entirely similar.
Regards,
John.
.
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- Ranges of size 5 of samples from [1, …, 52]
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