Re: Simple combinatoric problem
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 27 Feb 2008 17:54:30 -0600
richard@xxxxxxxxxxxxxxx (Richard Tobin) writes:
In a shuffled pack of cards, what is the probability that an aceYou want the probability that at least one ace is adjacent to at least
is adjacent to a king?
The answer (found by enumerating all the positions of the aces) is
284622747/585307450 or 48.63%, but can anyone find a simple way to
show this?
one king. Generalizing a bit, let P(n, a, k) be the probability that at
least one ace is adjacent to at least one king in a deck of n cards
containing a aces and k kings, where n >= a + k. Of course P(n,a,k) = 0
if a = 0 or k = 0. Let Q(n, a, k) be the probability given that the top
card is an ace, and R(n, a, k) the probability given that the top card is a
king. Then by conditioning on the top card we have
P(n,a,k) = a/n Q(n,a,k) + k/n R(n,a,k) + (n-a-k)/n P(n-1,a,k)
and by conditioning on the second card
Q(n,a,k) = (a-1)/(n-1) Q(n-1,a-1,k) + k/(n-1) + (n-a-k)/(n-1) P(n-2,a-1,k)
R(n,a,k) = (k-1)/(n-1) R(n-1,a,k-1) + a/(n-1) + (n-a-k)/(n-1) P(n-2,a,k-1)
Now calculate...
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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