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Assume that X and Y are the only two stations on an ethernet. Each has a steady
queue of frames to send. Both X and Y attempt to transmit a frame, collide and Y
wins the first backoff race. At the end of this successful transmission by Y, both

X and Y attempt to transmit and collide. The probability that X wins the second
backoff race is

I think it should be 0.03125 or 1/32. But the answer given is 0.125 or 1/8.

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After 1st collision both X and Y have slot (0,1)

X Y
0 0
0 1
1 0
1 1

Probability of Y winning is 1/4 when X=1 and Y=0.

Now X still want to send its first  frame while Y is ready with second and collide 

now as per  algorithm  X can send in slot (0,1,2,3) while Y can send in (0,1)

X Y
0 0
0 1
1 0
1 1
2 0
2 1
3 0
3 1

Now probability of X winning is 1/8 when X=0 and Y=1 

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