# Math Help - Card Problems

1. ## Card Problems

1) 5 cards are dealt from a standard 52-card deck. What is the probability that the draws will wield:
a) 3 aces and 2 kings?
b) 3 of a kind and 2 of a kind (a full house)
c) 1 ace, 1 two, 1 three, 1 four, 1 five (a straight)
d) any straight ( 5 ordered cards)

for a) I have that it is $\frac{24}{2598960}$. since it's $\frac{\begin{pmatrix} 4 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}}{\begin{pmatrix} 52 \\ 5 \end{pmatrix}}$

for b) all I know is that it's going to be $\frac{3774}{2598960}$

for c) is it $P^{52}_{5} = \frac{52!}{48!}$ for the denominator and $\binom{4}{1}^5$ for the numerator which equals $\frac{1024}{\frac{52!}{48!}}$

and d) I think that the denominator is the same as c) but the numerator will be 9 times the original, since there are 9 different choices which equals $\frac{(9)(1024)}{\frac{52!}{48!}}$

are these right?

2. Hello, lllll!

You did pretty good . . .

There are: . ${52\choose5} \:=\:2,598,960$ five card Poker hands.
. . This denominator is used throughout the problem.

By the way, use: {52\choose5} or \binom{52}{5}

1) Five cards are dealt from a standard 52-card deck.
What is the probability that the draws will yield:

a) 3 Aces and 2 Kings?
b) 3 of a kind and 2 of a kind (a "Full House")
c) 1 Ace, 1 Two, 1 Three, 1 Four, 1 Five (a "Straight")
d) any Straight ( 5 ordered cards)

for a) I have that it is $\frac{24}{2598960}$, since it is $\frac{ {4\choose2}{4\choose3}}{{52\choose5}}$ . . . . Right!

(b) There are 13 choices for the value of the Triple.
There are 12 choices for the value of the Pair.
Then there are ${4\choose3}$ to get the Triple
. . and ${4\choose2}$ ways to get the Pair.

The number of Full Houses is: . $13\cdot12\cdot 4\cdot 6\;=\;3744$

for c) is it $P^{52}_{5} = \frac{52!}{48!}$ for the denominator . . . . no
and $\binom{4}{1}^5$ for the numerator . . . . yes
The denominator is still: . ${52\choose5}$

d) I think that the denominator is the same as c)
but the numerator will be 9 times the original . . . . not quite

There are ten Straights: .from A-2-3-4-5 to 10-J-Q-K-A

The probability is: . $\frac{(10)(1024)}{{52\choose5}}$