## Second Number Larger Than The First

Suppose a six-sided die has sides numbered $1$ through $6$. If a person throws the die two times, what is the probability that the second number will be larger than the first?
Source: NCTM Mathematics Teacher, September 2006

Solution
The desirable outcomes are
$1:\left \{2,3,4,5,6\right \}$ where $1$ is the first number and $2,3,4,5,6$ are the second number
$2:\left \{3,4,5,6\right \}$
$3:\left \{4,5,6\right \}$
$4:\left \{5,6\right \}$
$5:\left \{6\right \}$
$6:\left \{\textrm{not possible}\right\}$
Total number of desirable outcomes
$5+4+3+2+1=15$
Number of possible outcomes
$6\times 6=36$
Probability that the second number is larger than the first = $15/36$

Answer: $15/36$

Alternative solution
First, the die is thrown and the first number $\left \{1,2,3,4,5,6\right \}$ comes up each with a probability equal $1/6$. Then, the die is thrown again and we want the second number larger than the first.
$1:\left \{2,3,4,5,6\right \}$ probability = $5/6$
$2:\left \{3,4,5,6\right \}$ probability = $4/6$
$3:\left \{4,5,6\right \}$ probability = $3/6$
$4:\left \{5,6\right \}$ probability =$2/6$
$5:\left \{6\right \}$ probability = $1/6$
$6:\left \{\textrm{not possible}\right \}$ probability = $0/6$
Probability of rolling the die two times and the second number is larger than the first
$\dfrac{1}{6}\times \dfrac{5}{6}+\dfrac{1}{6}\times \dfrac{4}{6}+\dfrac{1}{6}\times \dfrac{3}{6}+\dfrac{1}{6}\times \dfrac{2}{6}+\dfrac{1}{6}\times \dfrac{1}{6}+\dfrac{1}{6}\times \dfrac{0}{6}=\dfrac{1}{6}\left (\dfrac{5+4+3+2+1+0}{6}\right )$
$=\dfrac{1}{6}\times \dfrac{15}{6}$
$=15/36$