Question

A and B each throws a die. Then it is 7:5 that A's throw is not greater than B's. If this is true enter 1, else enter 0.

Answer 1

Let (p,q) denote a typical throw in which first number is thrown by A and second by B.
Then A can throw a higher number than B in the following 15 ways:
$(2,1),(3,2),(3,1),(4,3),(4,2),(4,1),(5,4),$
$(5,3),(5,2),(5,1),(6,5),(6,4),(6,3),(6,2),(6,1),$
Hence the number of ways in which A's throw is greater than B's $=36-15=21$
Hence odds in favor of A not throwing a number greater than B are as $21:15$, i.e. $7:5$.