We first assume that the probability of Arsenal meeting Chelsea in the first round is 1/63 as there is an equal chance of Arsenal meeting any one of the other teams.
ccc Round Probability A and C meet and A wins xxx xxxxxxxx Probability A and C don't meet but survive
1

\frac{1}{63} . \frac{3}{5}

0.00952

\frac{62}{63} . {\frac{7}{10}}^{2}

\frac{1}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . \frac{3}{5}

0.00933

\frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2}

\frac{1}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . \frac{3}{5}

0.00672

\frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{2}

\frac{1}{7} . \frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{2} . \frac{3}{5}

0.00336

\frac{6}{7} . \frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{4}

\frac{1}{3} . \frac{6}{7} . \frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{4} . \frac{3}{5}

0.00168

\frac{2}{3} . \frac{6}{7} . \frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{6}

\frac{2}{3} . \frac{6}{7} . \frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{6} . \frac{3}{5}

0.00084
Total probability 0.0314571

The probability of this happening in one year is roughly a little over 3 in a hundred and we raise this to the power 4 to get the probability of it happening in 4 consecutive years. This gives

9.8 \times 10^{- 7}

or roughly 1 in a million. Alternative approach:

Multiplying the probabilities on the tree diagram below we get for each round $f_{k} (y) = a + by$ where $a$ and $b$ are the probabilities on the lateral and vertical stems respectively. The probability of A meeting C and winning is thus

$f_{1} (f_{2} (f_{3} (f_{4} (f_{5} (3 / 5))))) .$

This is computed to be

$\frac{1}{63} . \frac{3}{5} + \frac{62}{63} . {\frac{7}{10}}^{2} (\frac{1}{31} . \frac{3}{5} + \frac{30}{31} . {\frac{3}{5}}^{2} (\frac{1}{15} . \frac{3}{5} + \frac{14}{15} . {\frac{1}{2}}^{2} (\frac{1}{7} . \frac{3}{5} + \frac{6}{7} . {\frac{1}{2}}^{2} (\frac{1}{3} . \frac{3}{5} + \frac{2}{3} . {\frac{1}{2}}^{2} (\frac{3}{5}))))) = 0.0314571$