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? 

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. 
