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May 1st, 2018, 02:47 AM   #1
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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?
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May 1st, 2018, 10:37 AM   #2
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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.
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May 1st, 2018, 11:29 AM   #3
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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.
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May 1st, 2018, 03:48 PM   #4
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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
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May 1st, 2018, 09:48 PM   #5
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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?
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