Question

Two persons each make a single throw with a pair of dice. The probability that the throws are equal is

• A. $\frac{55}{1296}$
• B. $\frac{73}{648}$
• C. $\frac{71}{648}$
• D. none of these

Answer 1

The correct option is B $\frac{73}{648}$

There are $36$ outcomes for one roll of a pair of dice, so the total number of outcomes for the two persons making one throw each is $36\times 36=1296$.
Now let $a_i$, with $2\le i\le 12$, be the number of ways to get a sum of $i$ showing on the pair of dice when they are rolled.
Then $a_2=a_{12}=1,a_3=a_{11}=2,a_4=a_{10}=3,a_5=a_9=4,a_6=a_8=5,a_7=6$
Each player can throw an $i$ in $a_i$ ways, so both of them will throw an $i$ is ${a_i}^2$ ways.
Summing over all values of $i$, we see that the number of ways the throws of the two persons will be equal is
${a_2}^2+{a_3}^2+...+{a_{12}}^2=2({a_2}^2+{a_3}^2+{a_4}^2+{a_5}^2+{a_6}^2)+{a_7}^2=2(1^2+2^2+3^2+4^2+5^2)+6^2.$
Using the result
$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$
the number of favorable ways is
$2\frac{5(5+1)(2\left(5\right)+1)}{6}+36=146$
So that the required probability is $\frac{146}{1296}=\frac{73}{648}$