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 Probability A, C don't meet, both win

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 $\frac{1}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . \frac{3}{5} = \frac{21}{3125}$ $\frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{2}$

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

Adding all these probabilities together (i.e. finding the chances of one of them happening) you get:

$\frac{1101}{35000}$

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

${(\frac{1101}{35000})}^{4} = 9.792 \times 10^{- 7}$

(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

Round	Probability A plays C and wins	Probability A, C don't meet, both win
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	$\frac{1}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . \frac{3}{5} = \frac{21}{3125}$	$\frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{2}$
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	$\frac{2}{3} . \frac{6}{7} . \frac{14}{15} . \frac{30}{31} . \frac{62}{63} . {\frac{7}{10}}^{2} . {\frac{3}{5}}^{2} . {\frac{1}{2}}^{6} . \frac{3}{5} = \frac{21}{25000}$