Re: Card pairing probabilities
- From: "Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx>
- Date: Tue, 31 Jan 2006 03:56:46 -0500
Phil Freedenberg wrote:
Suppose a standard deck of 52 cards is randomized.
How would you calculate the probability that no adjacent cards have the same value (e.g., pair of
treys, pair of Kings, etc.)
What is the expected number of such pairs in a deck?
The latter question is easy to answer. For i = 1, 2, 3,..., 51, let I_i be the indicator for the ith and the (i+1)st card being the same rank. Then the number N of pairs of adjacent cards cards having the same rank is sum(i=1..51, I_i). Thus, EN = sum(i=1..51, EI_i) (by linearity) = 51 EI1 (by exchangeability) =
51 * 13 C(4,2) / C(52,2) (where C(n,m) denotes the binomial coefficient) = 3.
The probability of no pairs requires more thought. It is not immediately obvious that there is a nice expresssion or an easy way to compute it.
-- Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx Math Tutor on the Internet and in Central New Jersey and Manhattan
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- From: Phil Freedenberg
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