Probability of selecting a poker hand I am trying to solve a probability problem about five-card poker hand. I have access to the answer which is different from what I had come up with. The question is: What is the probability that a five-card poker hand has exactly two cards of same value, but no other cards duplicated?
My answer to this question was as follows:
$\binom{13}{1} \binom{4}{2} \binom{48}{1}\binom{44}{1} \binom{40}{1}$.
Which means:

*

*First select a card number then select its two suits ie. $\binom{13}{1} \binom{4}{2}$. These will be the two cards of same value.

*Select three other cards which are not duplicate as: $\binom{48}{1}\binom{44}{1} \binom{40}{1}$.

The correct answer doesn't match my answer. This answer is provided in book AOPS and is as: $\binom{13}{1} \binom{4}{2}\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1}$.
So question is, what am I doing wrong? Thanks
 A: the solution of your book is correct. Let' explain the correct brainstorming.
To get exaclty one pair on 5 draws you have:

*

*13 choices to choose the pair {AA,22,33,...}


*for each pair you have $\binom{4}{2}$ choices to choose the suit: hearts, diamonds, clubs or spades


*for the remaining 3 draws you have $\binom{12}{3}$ choices of different cards


*for each of the prevoius choice you have $4^3$ choices for the suit: hearts, diamonds,clubs or spades


*multiply all the prevoius points getting.
$$13\times\binom{4}{2}\times\binom{12}{3}\times4^3$$
A: Suppose you select the hand $7\heartsuit, 7\spadesuit, 5\clubsuit, 9\diamondsuit, J\spadesuit$.  Your method counts this hand $3! = 6$ times, depending on the order in which you select the three singletons.
The order in which the three singletons are selected does not matter, which is why the correct answer selects three ranks from which a single card is drawn and then selects one card from each of those ranks.
Observe that
$$\frac{1}{6}\binom{13}{1}\binom{4}{2}\binom{48}{1}\binom{44}{1}\binom{40}{1} = \binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1}$$
A: Number of possible cases: $ c_p = \binom{52}{5} $.
Number of favorable cases:
Choose the first card suite: $ \binom{13}{1} \binom{4}{2} $.
Note that the first binomial is used to pick a card number, and the second to choose two symbols out of four.
Chose the three distinct card suite: $ \binom{12}{3} \binom{4}{1}^3 $
Note that the first binomial is used to pick three cards, and the second to choose only one symbol for each of the three cards.
Result: $$ p = \frac{\binom{13}{1} \binom{4}{2} \binom{12}{3} \binom{4}{1}^3}{\binom{52}{5}}. $$
In your solution, the last three binomials may provide a suite of three identical cards, because you just choose cards, not symbols.
A: By rule of product, after the first selected card number and its two suits, we need to select $3$ cards with $3$ different values that is $\binom{12}{3}$ and then for each one we can choose among four suits that is $\binom{4}{1}\binom{4}{1}\binom{4}{1}$. By your method the selections $\binom{48}{1}$ and other two subsequent are wrong because you are overcounting them (e.g. $3,5,8$ would be different from $5,3,8$). Therefore, by your way to count, you need to divide by $3!=6$.
A: You and the book count differently how to select the three remaining cards. Your answer is:
$$ \binom{48}{1}\binom{44}{1} \binom{40}{1} = 48 \cdot 44 \cdot 40 = 4^3 \cdot 12 \cdot 11 \cdot 10$$
The book answer is:
$$\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1} = 4^3 \cdot \frac{12\cdot 11\cdot 10}{3!}$$
They differ by a $3!$ factor, which is precisely the number of permutations of three distinct objects. This suggests that you are considering the order of the three remaining cards.
