The question is:

A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is

A) 0.5

B) 0.625

C) 0.75

D) 1.0

The solution given: 0.625

p(A wins the second backoff race)

=(A chooses 0)*(B chooses 1or2or3)+

(A chooses 1)*(B chooses 2or3)

=(1/2)(3/4)+(1/2)(2/4)

=5/8

=0.625

but my question is, when A chooses 1, why cant B choose 0 along with 2 and 3? Why are there only 2 choices for B ?

Is it given in the algorithm that B has to choose a value greater than B?