Probability that the second throw of a fair die exceeds the first A player throws an ordinary die and records the score $A$.  The player then throws the die again and again records the score, $B$. if $B>A$ then we set a score for this player.  What is the probability that a player didn't get any score?  The short answer given in the  Last Chapter of my book says $\frac{7}{12}$. 
Anyone can help me to solve this question? 
 A: You can make a table. The names of the rows are the outcomes of the first dice (d1). The names of the columns are the outcomes of the second dice. In total we have 36 possible outcomes. And the cells marked with $\color{red}x$ are the outcomes where the second dice has a greater outcome than the first dice.
$ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|}  \hline \text{ d1/d2 }  & 1 &2 &3 &4 &5 &6  \\ \hline \hline \hline 1 & &\color{red}x &\color{red}x &\color{red}x &\color{red}x &\color{red}x \\  \hline  2&  & &\color{red}x &\color{red}x &\color{red}x&\color{red}x  \\ \hline 3& & & &\color{red}x &\color{red}x&\color{red}x \\ \hline 4 & & &&&\color{red}x&\color{red}x  \\ \hline  5 & &&&&&\color{red}x  \\ \hline  6&&&&&&  \\ \hline \end{array}$
Count the red x´s. And the probabilty that the player does not win is 1 minus the probability that the player does win. 
A: The straight forward, but somewhat tedious solution is to simply list the outcomes:


*

*First throw 1 then throw 1, no score

*First throw 1 then throw 2, score

*First throw 1 then throw 3, score


etc. 
Then it's just a matter of counting scores vs no scores (every outcome is equally probable). I get $15$ outcomes that scores and $21$ that doesn't (there's $36$ in total). This gives the probability $21/36 = 7/12$.
You can also make it more compact by just considering the probabily for no-score given the first throw. If thrown a $1$ the probability for no-score is $1/6$, if thrown a $2$ the probability is $2/6$ and so on. For each first throw the probability is $1/6$ so the probability becomes:
$${1\over6}{1\over6}+
{1\over6}{2\over6}+  
{1\over6}{3\over6}+  
{1\over6}{4\over6}+  
{1\over6}{5\over6}+  
{1\over6}{6\over6} = 21/36 = 7/12
$$
A: The probability of a tied result on a fair die: $\mathsf P(A=B) ~=~ \tfrac 1 6$
By symmetry: $\mathsf P(A>B)~=~\mathsf P(B>A)$.
By total probability: $~\color{blue}{\mathsf P(B>A)}~$ $\color{blue}{=~ \tfrac 12(1-\mathsf P(A=B)) ~\\=~ \dfrac 5{12}}$

Note: if you wanted a tie or greater:  $~\mathsf P(B\geq A) ~$$=~ \mathsf P(A=B)+\mathsf P(B>A) ~\\=~ \dfrac 7 {12}$
A: Since you want multifarious methods, out of $36$ possible outcomes, $A=B$ in $6$.
By symmetry, $15$ of the remaining $30$ must have $B>A$, thus $P(\;set\;score) = \frac{15}{36}$
and P(not set score) $= 1 - \frac{15}{36}$
