Question

If it is assumed that all$\left(\begin{array}{c}52\\ 5\end{array}\right)$ poker hands are equally likely, what is the probability of being dealt

$\left(a\right)$a flush? (A hand is said to be a flush if all $5$cards are of the same suit.)

$\left(b\right)$one pair? (This occurs when the cards have denominations $a,a,b,c,d,$where $a,b,c,$and$d$are all distinct.)

$\left(c\right)$two pairs? (This occurs when the cards have denominations $a,a,b,b,c,$where $a,b,$and $c$are all distinct.)

$\left(d\right)$three of a kind? (This occurs when the cards have denominations $a,a,a,b,c,$where $a,b,$and $c$are all distinct.)

$\left(e\right)$four of a kind? (This occurs when the cards have denominations$a,a,a,a,b$)

Short answer

$$\begin{gathered} a)0.0019 \\ b)0.4225 \\ c)0.04754 \\ d)0.0211 \\ e)0.0002 \end{gathered}$$

Step-by-step solution

Given Information.

If it is assumed that all $\left(\begin{array}{c}52\\ 5\end{array}\right)$poker hands are equally likely,

Part (a) Explanation.

$P\left(flush\right)=\left[\right(4C1)*(13C5\left)\right]/52C5=0.0019$.

Part (b) Explanation.

$P(onepair)=\left[\right(13C1)*(4C2)*)\left(12C3\right)*\left(4C1\right)*\left(4C1\right)*\left(4C1\right)]/(52C5)=0.4225$.

Part (c) Explanation.

$P(twopairs)=\left[\right(13C1)*(4C2)*(4C2)*(11C1)*(4C1\left)\right]/(52C5)=0.04754$.

Part (d) Explanation.

$P(threeofakind)=\left[\right(13\mathrm{Cl})*(4\mathrm{C}3)*(12\mathrm{C}2)*(4\mathrm{Cl})*(4\mathrm{Cl}\left)\right]/\left(52\mathrm{C}5\right)=0.0211$.

Part (e) Explanation.

$P(fourofakind)=\left[\right(13\mathrm{Cl})*(4\mathrm{C}4)*(12\mathrm{C}1)*(4\mathrm{Cl}\left)\right]/\left(52\mathrm{C}5\right)=0.0002$.