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It is given on page #21 of U.G. book : Combinatorics - A guided Tour, by David Mazur, here an example detailing chances of getting three-of-a-kind in 5 card hands.

The problem statement is : From a standard 52-card deck, how many different five-card hands are possible? How likely is it that you will be dealt three-of-a-kind?

Solution approach for the second part is:

Encode each hand as a 5-list : A B C D E; where :

  1. A is the denomination of the three-of-a-kind
    $\implies$ 13 ways to choose

  2. B is the 3-set of suits for the three-of-a-kind
    $\implies$ $\binom{4}{3}$

  3. C is a 2-set of denominations for the other two cards
    $\implies$ $\binom{12}{2}$

  4. D is the suit of the smaller denomination in C
    $\implies$ $4$ ways to choose D

  5. E is the suit of the larger denomination in C
    $\implies$ $4$ ways to choose E

The chances are given by the product rule.


My doubt is in question #24 following that. It asks to find wrong with the below alternate approach:

There are $52$ ways to select the first card that is a part of the three-of-a-kind. Then there are $\binom{3}{2}$ ways to pick the other two cards to make up the three-of-a-kind. Finally, there are $\binom{48}{2}$ ways to pick the other two cards (any cards except those of the denomination of the three-of-a-kind). By the product principle there are $52\cdot \binom{3}{2} \cdot \binom{48}{2}$.

My answer needs vetting, and is stated below in two parts. The first part concerns the first two steps:

(i) To have $52$ ways for the first card seems fine, & so is the next step to have $\binom{3}{2}$ ways to pick the other two cards.
But the issue is that the product of first steps (by the first approach) leads to: $13\cdot \binom{4}{3} = 13\cdot 4 = 52$, while the second approach's first two steps lead to : $52\cdot \binom{3}{2} = 52\cdot 3=156$.

In order to explain the multiplication factor of $3$, consider a smaller example of smaller number of denominations (let, $3$) & the same number of (i.e., $4$) suits; as it is the # of suits that matter in terms of picking three-of-a-kind. $$ \left[ \begin{array}{c|c|c} &A&B&C&D\\ 1&1A&1B&1C&1D\\ 2&2A&2B&2C&2D\\ 3&3A&3B&3C&3D\\ \end{array} \right] $$

The duplication of choices, by factor of $\binom{3}{2}=3$, is shown below for the above small case, by considering only the first denomination (i.e. $1$), with 1A1B1C, 1A1B1D, 1A1C1D, 1B1C1D occuring $3$ times each.
I have shown only $4$ cases for the first row (i.e, denomination = $1$), but it can be easily shown that there are a total of $12(=4*3)\cdot 3= 36$ cases, of which only $12$ are original, explaining the overcount's multiplier factor of $\binom{3}{2}=3$.

$$ \left[ \begin{array}{c|c|c|c} 1A&1A1B1C&1A1B1D&1A1C1D\\ 1B&1B1C1A&1B1A1D&1B1C1D\\ 1C&1C1A1B&1C1A1D&1C1B1D\\ 1D&1D1A1B&1D1A1C&1D1B1C\\ \end{array} \right] $$


(ii) Next comes the difference between the first approach's last 3 steps, & the second approach taking $\binom{48}{2}$.
The first approach leads to: $(\binom{12}{2}=66) \cdot (\binom{4}{1}=4) (\binom{4}{1}=4) = 66*16 = 1056$,
while the second approach leads to: $\binom{48}{2}=24*47= 1128$.

Second part analysis: The definition of three-of-a-kind considers the denomination of the rest two cards to be different, irrespective of their suits.
If take $\binom{48}{2}$, then also consider the cases where the third card is of the same denomination as second. This leads to an extra number of cases, that need be subtracted.
I am stating here subtraction, as the number of chances for second case are found by application of sum rule.

To show how it works, by considering the original example; have extra $1128-1056 = 72$ cases.
Only $12$ denominations are available for the 4th & 5th cards, but the two cards need have different denominations. So, either can be of higher denomination than the other.

Assuming, that denomination $1$ is taken up by the three-of-a-kind; then $12$ denominations are left. In each denomination there are still $3$ more suits (for the over-calculation to occur) leading to $12*3=36$ cases, which by ordering should lead to $36*2=72$ cases.

a) Fourth card denomination = $2$, fifth card denomination choices are = $3, 4, \cdots, 13$, a total of $12$ choices.
b) Similarly, the chances based on other denominations of fourth card are : $11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1$.
This sums up to : $12 \cdot 6 = 72$ cases.

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  • $\begingroup$ Too long; didn't read. But I don't like step 2 in the book solution. Where does the 6 in $6\choose3$ come from? $\endgroup$ – Gerry Myerson Nov 12 '18 at 4:59
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    $\begingroup$ @GerryMyerson Sorry for error. Edited the OP. $\endgroup$ – jiten Nov 12 '18 at 5:05
  • $\begingroup$ 52 is wrong as there are only 13 ranks. Combin(48;2) is wrong as it could include pairs. $\endgroup$ – paparazzo Nov 12 '18 at 15:04
  • $\begingroup$ @paparazzo I hope that the first error introduces a multiplicative factor of $3$, while the second error introduces an additive factor of $72$; as shown above. $\endgroup$ – jiten Nov 13 '18 at 0:20
  • $\begingroup$ @jiten You hope for error? $\endgroup$ – paparazzo Nov 13 '18 at 2:19
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For the first part, we can cut down the multiple counting by first converting the problem to a problem that consider order to matter and then remove it.

$$52 \cdot \binom32 \cdot \frac{2!}{3!}=52.$$

For the second part, to reduce those that are from the same denomination:

$$12 \cdot \binom43=72$$

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  • $\begingroup$ Thanks, but request elaboration of the ratio $\frac{2!}{3!}$, as seemingly it is just reverse of $3C2$. $\endgroup$ – jiten Nov 14 '18 at 4:58
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    $\begingroup$ How would you answer the question, what are the number of possibility to pick $3$ cards out of $52$ such that $3$ of them share the same denomination where order matters? After you answer that, divide it by $3!$. $\endgroup$ – Siong Thye Goh Nov 14 '18 at 5:11
  • $\begingroup$ I feel that your last comment has an error in the last word, i.e. should end with $3$. $\endgroup$ – jiten Nov 14 '18 at 6:15
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    $\begingroup$ for having $3$ cards where order matters is $52 \times 3 \times 2$ right? to remove the order factor, divide by $3!$. $\endgroup$ – Siong Thye Goh Nov 14 '18 at 6:50
  • $\begingroup$ If you want to remove just the order factor of $3!$, then why multiply by $2!$ also. $\endgroup$ – jiten Nov 14 '18 at 8:07

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