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Question from Sheldon Ross's First Course in Probability, Chapter 2:

If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt two pairs? (This occurs when the cards have denominations $a, a, b, b, c,$ where $a$, $b$, and $c$ are all distinct.)

1st approach:

Out of the 13 types in each deck possible, I select any one card (A,1,2,3...,K,Q,J) by $\binom{13}{1}$ ways, and take 2 of these selected out of 4 possible decks (Clubs, Spades, Diamonds, Heart) using $\binom{4}{2}$ ways. This gets me the first pair.

Similarly, select another card from the remaining 12 types using $\binom{12}{1}$ ways, and make another pair by $\binom{4}{2}$ ways.

Since the 2 types taken above cannot be chosen again, I exclude all the 2 types (8 cards in total), and select the one remaining card in $\binom{44}{1}$ ways.

This gives me the numerator as : $\binom{13}{1} * \binom{4}{2} * \binom{12}{1} * \binom{4}{2} * \binom{44}{1}$ ways which will be divided by $\binom{52}{5}$.

2nd approach:

Out of the 13 types in each deck possible, I select 2 card types (A,1,2,3...,K,Q,J) by $\binom{13}{2}$ ways, and for each type I select 2 cards out of 4 possible decks (Clubs, Spades, Diamonds, Heart) using $\binom{4}{2} * \binom{4}{2}$ ways. This gives me both pairs.

Again, I select the remaining one card in $\binom{44}{1}$ way, by following the previous approach.

The final numerator returns: $\binom{13}{2} * \binom{4}{2} * \binom{4}{2} * \binom{44}{1}$.

Why is both the solution different, and what is the right approach? I want to know the underlying concept underneath it - if I am unconsciously solving the question sequentially or what not.

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    $\begingroup$ Hi. Please add the self-study tag. $\endgroup$ Commented Oct 18, 2023 at 12:15
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    $\begingroup$ @User1865345 Added the tag. $\endgroup$
    – Roy
    Commented Oct 18, 2023 at 16:42

2 Answers 2

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Your first approach has a mistake. You are right that there are $\binom{13}{1}$ ways to pick the first pair card, and then $\binom{12}{1}$ ways to pick the second pair card. However, you are not accounting for the fact that the order of these cards is irrelevant. For instance, if we choose $a=3$ and $b=\text{K}$, then that's equivalent to $a=\text{K}$ and $b=3$ - these two are not distinct combinations. To account for these possible swaps, you need to divide by 2. So instead of $\binom{13}{1}\binom{12}{1}=13\times12$, it becomes $\frac{\binom{13}{1}\binom{12}{1}}{2}=\frac{13\times12}{2}$. And this (as you may have already spotted) turns out to be equal to $\binom{13}{2}$, which makes your first outcome equal to the second.

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  • $\begingroup$ Thank you. Can you please explain a bit further about how the 1st approach is based on ordering? And how is it different from the 2nd approach. I am confused! Thanks $\endgroup$
    – Roy
    Commented Oct 18, 2023 at 16:50
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The difference between the two methods is a factor two and this has to do with the order of picking the two types.

There are 13x12 = 156 ways to pick two different types with counting different orders double (each particular set occurs twice, e.g. 23 and 32):

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
 [1,] ""   "23" "24" "25" "26" "27" "28" "29" "2T" "2J"  "2Q"  "2K"  "2A" 
 [2,] "32" ""   "34" "35" "36" "37" "38" "39" "3T" "3J"  "3Q"  "3K"  "3A" 
 [3,] "42" "43" ""   "45" "46" "47" "48" "49" "4T" "4J"  "4Q"  "4K"  "4A" 
 [4,] "52" "53" "54" ""   "56" "57" "58" "59" "5T" "5J"  "5Q"  "5K"  "5A" 
 [5,] "62" "63" "64" "65" ""   "67" "68" "69" "6T" "6J"  "6Q"  "6K"  "6A" 
 [6,] "72" "73" "74" "75" "76" ""   "78" "79" "7T" "7J"  "7Q"  "7K"  "7A" 
 [7,] "82" "83" "84" "85" "86" "87" ""   "89" "8T" "8J"  "8Q"  "8K"  "8A" 
 [8,] "92" "93" "94" "95" "96" "97" "98" ""   "9T" "9J"  "9Q"  "9K"  "9A" 
 [9,] "T2" "T3" "T4" "T5" "T6" "T7" "T8" "T9" ""   "TJ"  "TQ"  "TK"  "TA" 
[10,] "J2" "J3" "J4" "J5" "J6" "J7" "J8" "J9" "JT" ""    "JQ"  "JK"  "JA" 
[11,] "Q2" "Q3" "Q4" "Q5" "Q6" "Q7" "Q8" "Q9" "QT" "QJ"  ""    "QK"  "QA" 
[12,] "K2" "K3" "K4" "K5" "K6" "K7" "K8" "K9" "KT" "KJ"  "KQ"  ""    "KA" 
[13,] "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "AT" "AJ"  "AQ"  "AK"  ""  

There are 13x12/2 = 78 ways to pick two different types with counting different orders only once:

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
 [1,] ""   ""   ""   ""   ""   ""   ""   ""   ""   ""    ""    ""    ""   
 [2,] "32" ""   ""   ""   ""   ""   ""   ""   ""   ""    ""    ""    ""   
 [3,] "42" "43" ""   ""   ""   ""   ""   ""   ""   ""    ""    ""    ""   
 [4,] "52" "53" "54" ""   ""   ""   ""   ""   ""   ""    ""    ""    ""   
 [5,] "62" "63" "64" "65" ""   ""   ""   ""   ""   ""    ""    ""    ""   
 [6,] "72" "73" "74" "75" "76" ""   ""   ""   ""   ""    ""    ""    ""   
 [7,] "82" "83" "84" "85" "86" "87" ""   ""   ""   ""    ""    ""    ""   
 [8,] "92" "93" "94" "95" "96" "97" "98" ""   ""   ""    ""    ""    ""   
 [9,] "T2" "T3" "T4" "T5" "T6" "T7" "T8" "T9" ""   ""    ""    ""    ""   
[10,] "J2" "J3" "J4" "J5" "J6" "J7" "J8" "J9" "JT" ""    ""    ""    ""   
[11,] "Q2" "Q3" "Q4" "Q5" "Q6" "Q7" "Q8" "Q9" "QT" "QJ"  ""    ""    ""   
[12,] "K2" "K3" "K4" "K5" "K6" "K7" "K8" "K9" "KT" "KJ"  "KQ"  ""    ""   
[13,] "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "AT" "AJ"  "AQ"  "AK"  ""   
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