A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is frozen if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently eight spaces behind player B, what is the probability that player B will be frozen after player A rolls?  Pick.
A) 1/12 B) 5/36 C) 1/6 D) 1/3 E) 17/36
no. of possible outcomes = 6*6= 
<gadget id="calculator">6*6</gadget>
<output>36</output>
36   no. of outcomes that result a total of 8 (as A is 8 spaces behind 5/36)= 5 ( (2,6), (3,5), (4,4), (5,3), (6,2) )   So, the probability = 5/36 (Option 5/36)
<result>B</result>
A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is frozen if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently eight spaces behind player B, what is the probability that player B will be frozen after player A rolls?