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