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A special deck of Old Maid cards consist of 25 pairs and a single old maid card. All 51 cards evenly between you and two other players – 17 cards for each player.

(a) how many different hands can be dealt to you?

(b) what is the probability that your hand has exactly 2 pair (and 13 single cards)?

(a) This is easy: $\binom{51}{17}$

(b) This one I'm having trouble. I thought about doing something like this:

$\cfrac{\binom{25}{2}*\binom{47}{13}}{\binom{51}{17}}$

25-C-2 = Choose 2 of 25 pairs

47-C-13 = There's 46 remaining cards (or 23 pairs) but you add 1 because of old maid card

51-C-17 = Total possibilities.

I know this answer is wrong because its greater than 1. The solution is 0.30282.

Any help is appreciated. Thank you.

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    $\begingroup$ Your answer allows for additional pairs among the 13 cards you choose second. $\endgroup$
    – vadim123
    Commented May 9, 2013 at 2:59
  • $\begingroup$ Are the pairs identical? If so, part a is not so simple as $\binom{51}{17}$. $\endgroup$
    – 2'5 9'2
    Commented May 9, 2013 at 3:06
  • $\begingroup$ @alex.jordan Part (a) is correct according to the solutions. $\endgroup$
    – user1527227
    Commented May 9, 2013 at 3:08
  • $\begingroup$ Ok thanks @vadim123. So then I thought about $\binom{25}{2}\binom{24}{13}\cdot 2^{13}$ for the numerator but that gives a probability of 0.415. I treated the old maid as a pair but only raised 2 to the 13'th because there are 13 pairs remaining... $\endgroup$
    – user1527227
    Commented May 9, 2013 at 3:10
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    $\begingroup$ By including $2^{13}$ you are always making $13$ binary choices. You need to use the sum principle depending on whether or not the old maid is chosen. $\endgroup$
    – vadim123
    Commented May 9, 2013 at 3:16

2 Answers 2

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Let’s split the calculation into two cases, hands with the Old Maid and hands without.

A hand with the Old Maid that has exactly two pairs contains the Old Maid, two pairs, and $12$ singletons. Such a hand therefore has cards from $12+2=14$ of the $25$ denominations. There are $\binom{25}2$ ways to choose the denominations of the two pairs, and there are then $\binom{23}{12}$ ways to choose the denominations of the $12$ singletons. For each singleton there are $2$ ways to choose which member of the pair we get. Thus, there are $2^{12}\binom{25}2\binom{23}{12}$ hands of this type.

Now see if you can modify that calculation to get the number of hands of the desired type that do not include the Old Maid.

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  • $\begingroup$ Thanks @Brian M. Scott!! I added the full answer below, but I credit you :). $\endgroup$
    – user1527227
    Commented May 9, 2013 at 5:04
  • $\begingroup$ @user1527227: You’re welcome! I just upvoted your answer. $\endgroup$
    – Brian M. Scott
    Commented May 9, 2013 at 5:24
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I was able to answer this with Brian M. Scott's post (thanks Brian)! This problem must first be decomposed into 2 cases:

(1) Hand with the old maid card: Pair1, Pair2, Old Maid, 12 denominations.
(2) Hand without the old maid card: Pair1, Pair2, 13 denominations.

Case 1:

Number of denominations= 12 singles + 2 pairs = 14 denominations. The two pairs can be chosen $\binom{25}{2}$ ways. The remaining 12 single denominations can be chosen $\binom{23}{12}$ ways by the partition principle. However, for each single denomination, you have 2 choices so you multiply by $2^{12}$.

$\binom{25}{2} \binom{23}{12} 2^{12}$

Case 2:

Number of denominations= 13 singles + 2 pairs = 16 denominations.

By the same logic, the number of outcomes is= $\binom{25}{2}\binom{23}{13} 2^{13}$

Solution:

$Pr=\cfrac{\binom{25}{2} \binom{23}{12} 2^{12} + \binom{25}{2}\binom{23}{13} 2^{13}}{\binom{51}{17}}=0.3028278$

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