Q. 2.15 - A First Course in Probability

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$a, a, a, a, b$)

Step-by-Step Solution

Answer

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

Step 1 Given Information.

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

Step 2 Part (a) Explanation.

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

Step 3 Part (b) Explanation.

$P(one pair)=\left[\right(13C1) *(4C2)*)\left(12C3\right)*\left(4C1\right)*\left(4C1\right)*\left(4C1\right)] /( 52 C 5)=0.4225$.

Step 4 Part (c) Explanation.

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

Step 5 Part (d) Explanation.

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

Step 6 Part (e) Explanation.

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