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 six spaces behind player B, what is the probability that player B will be frozen after player A rolls?
Answers:
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 6 (as A is 6 spaces behind 5/36)= 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )   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 six spaces behind player B, what is the probability that player B will be frozen after player A rolls?