FA Cup
In four years 2001 to 2004 Arsenal have been drawn against Chelsea
in the FA cup and have beaten Chelsea every time. What was the
probability of this? Lots of fractions in the calculations!
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In the FA Cup 64 teams play a knockout tournament. Over the four years 2001, 2002, 2003 and 2004 Arsenal have been drawn against Chelsea each year and have beaten Chelsea every time. What is the probability of that happening?
Let's say 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.
You'll need to work out, for each round, the probability that Arsenal play Chelsea and also the probability that they do not play against each other but both survive to play in the following round. A tree diagram is useful in thinking through this problem.
Well done Roy M. 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
This requires an understanding of the simple rules of probability and a lot of calculating with fractions.