Question

Poker dice is played by simultaneously rolling $5$dice. Show that

(a) P{no two alike}

(b) P{one pair}

(c) P{two pair}

(d) P{three alike}

(e) P{full house}

(f) P{four alike}

(g) P{five alike}

Short answer

Hence proved.

Step-by-step solution

Step 1

Since we are rolling 5 dice, the total number of events is$6^5$.

a) P{no two alike}

No two alike i.e. we want $5$different numbers from $5$dice.

There are $\left(\begin{array}{c}6\\ 5\end{array}\right)$ways to choose $5$different numbers from $1$to $6$.

5$!$for rolling of $5$different dice.

Thus, number of no two alike is $5!\times \left(\begin{array}{c}6\\ 5\end{array}\right)$$=720$

Therefore, probability is $\frac{720}{6^5}$$=.0962$

b) P{one pair}

There are $\left(\begin{array}{c}6\\ 1\end{array}\right)\times \left(\begin{array}{c}5\\ 2\end{array}\right)$ways to choose a pair and $\left(\begin{array}{c}5\\ 3\end{array}\right)\times 3!$ways to choose which is not a pair.

Thus, number of one pair is$\left(\begin{array}{c}6\\ 1\end{array}\right)\times \left(\begin{array}{c}5\\ 2\end{array}\right)\times \left(\begin{array}{c}5\\ 3\end{array}\right)\times 3!=3600$.

Therefore, probability is$\frac{3600}{6^5}$$=.4630$

c) P{two pair}

The number of two pairs is $\left(\begin{array}{c}6\\ 2\end{array}\right)\times \left(\begin{array}{c}5\\ 2\end{array}\right)\times \left(\begin{array}{c}3\\ 2\end{array}\right)\times \left(\begin{array}{c}4\\ 1\end{array}\right)=1800$.

Thus, probability is$\frac{1800}{6^5}$$=.2315$

d) P{three alike}

First choose common number and dice having common number then choose other two numbers from five choisces.

The number of three alike is $\left(\begin{array}{c}6\\ 1\end{array}\right)\times \left(\begin{array}{c}5\\ 3\end{array}\right)\times \left(\begin{array}{c}5\\ 2\end{array}\right)\times 2!=1200$.

Thus, probability is $\frac{1200}{6^5}=.1543$

e) P{full house}

Full house is three alike with pair

The number of full house is $\left(\begin{array}{c}6\\ 1\end{array}\right)\times \left(\begin{array}{c}5\\ 3\end{array}\right)\times \left(\begin{array}{c}5\\ 1\end{array}\right)=300$.

Thus, probability is$\frac{300}{6^5}=.0386$

f) P{four alike}

The number of four alike is $\left(\begin{array}{c}6\\ 1\end{array}\right)\times \left(\begin{array}{c}5\\ 4\end{array}\right)\times \left(\begin{array}{c}5\\ 1\end{array}\right)=150$.

Thus, probability is $\frac{150}{6^5}=.0193$

g) P{five alike}

The number of five alike is $\left(\begin{array}{c}6\\ 1\end{array}\right)$$=$$6$.

Thus, probability is$\frac{6}{6^5}=.0008$