5 cards from a deck of 52 , how many ways of four of a kind can be dealt ?
I have (13c1) to determine the rank of card
so (13c1)(4c4) is all i can think of until now
the answer is 2(13c2)(4c4)(4c1)
can someone explain how does this work ?
[Math] probability of 4 of a kind from a deck of 52
probability
Related Solutions
I have a feeling your teacher/professor intended for you to learn the following (we first introduce some notation):
For two sets $A, B$ in a common universe $U$, define their union as
$$ A \cup B = \{x \in U : x \in A \text { or } x \in B\}. $$
Define their intersection as:
$$ A \cap B = \{x \in U : x \in A \text{ and } x \in B\}. $$
It is important not to get too wrapped in the English here. Being in $A \cup B$ simply means being in one of the two sets (or possibly both). Being in $A \cap B$ simply means being in set $A$ and being in set $B$ at the same time.
Finally, a finite set $A$ with $k$ elements ($k$ things in the set) has cardinality $k$, and this is written as $|A| = k$, or $\#A = k$ or even sometimes, $n(A) = k$.
Therefore:
$$ \{1,3,5\} \cup \{1,2,3\} = \{1,2,3,5\}, $$
while
$$ \{1 ,3 , 5 \} \cap \{ 1 , 2, 3 \} = \{ 1, 3\}. $$
Also, $$ | \{ 1,3,5\}| = 3, $$
while
$$ | \{ 1, 3\} | = 2. $$
Now, what your teacher probably wanted you to learn was the following "rule":
$$ |A \cup B| = |A| + |B| - |A \cap B|. $$
This is easy to see it is true, since to count the number of elements that are in either $A$ or $B$ (or possibly both $A$ and $B$), you count the number of elements in $A$, add to it the number of elements in $B$, and then subtract the stuff you double counted, which is precisely the elements in $A \cap B$.
Therefore, if you want to find the number of sides of a die that are (say) even or prime, you count the number of sides which are even (there are $3$ such sides - namely $2$ , $4$, $6$) add to it the number of sides which are prime (again, there are $3$ such sides - namely $2$, $3$, $5$), and then subtract the sides which we double counted (we counted the side with the number $2$ twice).
Therefore, there are $3 + 3 - 1 = 5$ sides of a die which are even or prime.
Now, you can take this strategy and count the number of cards in a deck which are either non face cards or clubs.
I am not familiar with Pinochle decks. So for the count I will assume that for example the two $\heartsuit$ Queens are identical.
Then there are $3$ types of $5$-card hands: (i) all cards distinct; (ii) there is $1$ pair of identicals; (ii) there are $2$ pairs of identicals. For the count, we calculate the number of each type, and add up.
(i) This is easiest. There are $24$ different cards, and there are $\binom{24}{5}$ ways to choose $5$.
(ii) There are $24$ different cards. We choose $1$ of the kinds to have $2$ of, and then $3$ different cards for the rest. That gives total $\binom{24}{1}\binom{23}{3}$.
(iii) This can be tricky. We pick $2$ of the kinds to have $2$ each of, and then an odd card. That gives $\binom{24}{2}\binom{22}{1}$.
Now for poker hand probabilities, we have to forget about the count we just made. For the different hands that we counted above are not equally likely. The Type (i) jands are all equally likely, as are the type (ii) hands, as are the type (iii) hands, but we do not have equal likelihood between different types.
There are two ways to proceed. We can take account of the different probabilities for each type. But my inclination is to imagine that the "repeats" are coloured red and blue, making the $48$ cards distinct. Then there are $\binom{48}{5}$ equally likely hands.
For each type of poker hand, count how many ways there are to produce it, taking colouring into account. I will leave the (unpleasant) details to you. We need to define carefully what we mean by each poker hand. For example, five of a kind is now possible.
Let's do a relatively straightforward one, $1$ pair. The type of card we have $2$ of can be chosen in $6$ ways, $9$ to Ace. Suppose for example it is $2$ Kings. They can be either $2$ of the same suit ($4$ choices) or of different suits. If so, the different suits can be chosen in $\binom{4}{2}$ ways, and then the actual cards in $2^2$ ways (all combinations of red and blue), for a total of $4+(6)(4)=28$ ways. Now we choose $3$ types from the remaining $5$, and for each choose a colour, red or blue. This gives a total of $(6)(28)(80)$. Check it. Now divide by $\binom{48}{5}$ for the probability.
Best Answer
There are indeed $\binom{13}{1}$ ways of choosing the kind we have $4$ of. And once that is done, that part of the hand is determined. But then there is the useless fifth card, which can be chosen in $\binom{48}{1}$ ways.
So the number of $4$ of a kind hands is $(13)(48)$. For the probability, divide by $\binom{52}{5}$.
Remark: The person who did the counting you quote did it in I think less clear way. The $4$ of a kind hand will have two denominations, which can be chosen in $\binom{13}{2}$ ways.
For each choice, we have $2$ choices as to which denomination we will have $1$ of. And then the actual card can be chosen in $\binom{4}{1}$ ways, for a total of $2\binom{13}{2}\binom{4}{1}$.