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 May 1st, 2018, 02:47 AM #1 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 Probability The question is “a 5-card poker hand is selected from a deck of 52 playing cards. How many different poker hands contain exactly one pair only?” My answer is (52C1 x 48C1 x 44C1 x 40C1 x 3C1) / 5! = 109 824 However, the suggesting answer is 10 times more, which is 1 098 240 Is there any mistakes for my calculation? May 1st, 2018, 10:37 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 Let the deck contain $v$ values of $s$ suits and we deal an $h$ card hand. Our values for this problem are $v=13,~s=4,~h=5$ I'm seeing that the formula for the number of hands with only a single pair is $n = \dfrac{s^{h-2}\dbinom{s}{2} \prod \limits_{k=0}^{h-2}(v-k)}{(h-2)!}$ for our current problem this does lead to $n=1098240$ You're neglecting for the variety of suits the pair can take. Last edited by romsek; May 1st, 2018 at 10:41 AM. May 1st, 2018, 11:29 AM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 Ok I have a good way to look at this now. Pick a pair. There are $v \dbinom{s}{2}$ different pairs. Now place them in the hand. There are $h\cdot (h-1)$ ways of doing this as the two cards are distinguishable. Now pick the remaining $h-2$ cards. There are $\prod \limits_{k=1}^{h-2} s(v-k)$ ways of doing this So all told the number of ordered hands with 1 pair is given by $v \dbinom{s}{2} h(h-1) \prod \limits_{k=1}^{h-2} s(v-k) = \dbinom{s}{2}h(h-1)s^{h-2}\prod \limits_{k=0}^{h-2} (v-k)$ Now as we don't care about the order of the hand at all we divide all this by $h!$ to obtain $n1p = \dfrac{\dbinom{s}{2}h(h-1)s^{h-2}\prod \limits_{k=0}^{h-2} s(v-k)}{h!} = \dfrac{\dbinom{s}{2}s^{h-2}\prod \limits_{k=0}^{h-2} s(v-k)}{(h-2)!}$ Last edited by romsek; May 1st, 2018 at 11:32 AM. May 1st, 2018, 03:48 PM #4 Newbie   Joined: Sep 2013 From: Reno,NV Posts: 14 Thanks: 1 Math Focus: Number Theory $\dbinom{13}{1} \dbinom{4}{2} \dbinom{12}{3} \dbinom{4}{1}^3$ This is from 'The theory of Gambling and Statistical Logic" by Richard A. Epstein The explanation is: The rank can occur in $\dbinom{13}{1}$ ways The cards forming the pair can be selected in $\dbinom{4}{2}$ ways The three other ranks can occur in $\dbinom{12}{3}$ ways A card of each rank can be selected in $\dbinom{4}{1}$ ways May 1st, 2018, 09:48 PM #5 Member   Joined: Nov 2012 Posts: 80 Thanks: 1 I think I know what’s wrong with my calculation. It seems that I cannot assume different suits of the same rank could be eliminated from the selection after each pick. That’s why it is incorrect to write 52C1 x 48C1.... What about this question: there are five classes. A committee consists of 15 members, with 3 representatives from each class. In how many ways can 3 members be selected from the committee if the 3 selected members are from different classes? There are two ways to approach the answer for me Approach 1: 5C3 x (3C1)^3 = 270 Approach 2: (15C1 x 12C1 x 9C1)/3! = 270 Although both approaches give the same result, is it the second approach not a correct one to do so? Tags probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rivaaa Advanced Statistics 5 November 5th, 2015 01:51 PM hbonstrom Applied Math 0 November 17th, 2012 07:11 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM

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