Well done Roy Madar of Allerton High School for the following
excellent solution.
To calculate the probability of Arsenal and Chelsea playing each
other in the FA cup in four consecutive years and Arsenal winning
each time we have to make reasonable assumptions about the
probabilities of these teams winning their games. So we assume that
whenever Arsenal and Chelsea play the probability of Arsenal
winning is $0.6$ and otherwise, throughout the tournament, both
these teams have a probability of winning of $0.7$ in the first
round, $0.6$ in the second round and $0.5$ in the subsequent
rounds.
To go about this problem I decided to find the chance of Arsenal
playing and beating Chelsea round by round.
Round One: The chance of Arsenal being drawn against Chelsea and
beating them is: $(1/63)\times(3/5)= 1/105$.
Round Two: The chance of Arsenal not being drawn against Chelsea in
round one, and both teams winning their round one matches,
multiplied by the chance of them playing each other in round two
and Arsenal winning is:
$(62/63)\times(7/10)\times(7/10)\times(1/31)\times(3/5)=
7/750$
Continuing in this fashion, the probability of Arsenal beating
Chelsea in any given round is equal to the product of the chances
of the two teams not being drawn against each other in any former
round, times the chances of the two teams beating any team they
were drawn against in all former rounds, times the chances of
Arsenal being drawn against Chelsea in the given round, times the
chance of Arsenal winning.
The odds for Arsenal beating Chelsea in rounds 1 to 6 (6 being the
final) is as follows:
Round 
Probability A plays C and wins 
$\quad$ Probability A, C don't
meet, both win $\quad$ 
1 
$\frac{1}{63}.\frac{3}{5}=\frac{1}{105}$

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

2 
$\frac{1}{31}.\frac{62}{63}.{\frac{7}{10}}^2.\frac{3}{5}=\frac{7}{750}$

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

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

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

4 
$\frac{1}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2{\frac{3}{5}}^2.{\frac{1}{2}}^2.\frac{3}{5}=\frac{21}{6250}$

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

5 
$\frac{1}{3}.\frac{6}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^4.\frac{3}{5}=\frac{21}{12500}$

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

6 
$\quad
\frac{2}{3}.\frac{6}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^6.\frac{3}{5}=\frac{21}{25000}
\quad$


Adding all these probabilities together (i.e. finding the chances
of one of them happening) you get: $${1101\over 35000}$$ which is
roughly $3.15$ per cent. So for the odds given, there is a $3.15$
per cent chance of Arsenal playing and beating Chelsea in the FA
cup.
However the question asks for the probability of it happening four
years in a row. So you must multiply this result by itself four
times $$\left({1101\over 35000}\right)^4=9.792 \times 10^{7}$$
(giving the result to 4 s.f.) which gives roughly 1 in a million
chance of it happening or a probability of $0.000098$ per
cent.
So for a bet of £1 you could be laughing all the way to
the bank with a sum of just over £1m