Re: Simple combinatoric problem
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Thu, 28 Feb 2008 02:32:43 GMT
In article <rbisrael.20080227232814$379a@xxxxxxxxxxxxxxxx>,
Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
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...
Alternatively, use inclusion-exclusion.
The event ace-of-spades adjacent king-of-hearts can happen in
51 x 2 x 50-factorial ways.
Similarly for each of the other 15 adjacency events, so we have
16 x 51 x 2 x 50-factorial.
Now you have to subtract all the ways that 2 of the 16 adjacency
events can take place. E.g., (AS adj KH) and (AS adj KD) happens
in 50 x 2 x 49-factorial ways; (AS adj KH) and (AD adj KC)
happens in 52-choose-4 x 8 x 48-factorial ways, etc.
Then add back in all the ways 3 adjacency events can take place,
etc. You can't have more than 7 adjacency events taking place, so
you'll know when to stop.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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- From: Richard Tobin
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