two pairs

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3 total messages Started by Peter Pui Tak Chiu Tue, 15 Sep 1992 02:49

#338

Author: Peter Pui Tak Chiu
Date: Tue, 15 Sep 1992 02:49

16 lines
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hi,

can anyone point out what is wrong with my calculation of getting a hand
of two pairs from a shuffled deck of cards?

    the probability i have is:

	(13 * 4C2) (12 * 4C2) (4 * 11)
	------------------------------ = 0.095078
		      52C5

which is exactly two times the correct value (0.047539).

pete

Re: two pairs

#341

Author: Robert Strawderman
Date: Tue, 15 Sep 1992 12:19

14 lines
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You can think of the problem as follows:

You have 13 sets of 4 identical cards, and you first choose two of those sets.  Then,
to get the two pair, you must choose 2 from the 4 in each set.  The remaining card
may be chosen from the 11 remaining sets of 4 cards.  Thus, the numerator is

13C2 * (4C2)^2 * (11 * 4C1)

and the denominator is 52C5.

Rob Strawderman
Dept. of Biostatistics
University of Michigan

Re: two pairs

#347

Author: Arthur Rubin
Date: Tue, 15 Sep 1992 21:26

25 lines
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In <1992Sep15.024945.17690@wuecl.wustl.edu> ppc1@cec1.wustl.edu (Peter Pui Tak Chiu) writes:



>hi,

>can anyone point out what is wrong with my calculation of getting a hand
>of two pairs from a shuffled deck of cards?

>    the probability i have is:

>	(13 * 4C2) (12 * 4C2) (4 * 11)
>	------------------------------ = 0.095078
>		      52C5

>which is exactly two times the correct value (0.047539).

>pete

You can't distinguish between the pairs, so you have to divide by 2.
--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; please mail anything important.

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