#### You may also like ### Rain or Shine

Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry. ### Knock-out

Before a knockout tournament with 2^n players I pick two players. What is the probability that they have to play against each other at some point in the tournament? ### Squash

If the score is 8-8 do I have more chance of winning if the winner is the first to reach 9 points or the first to reach 10 points?

# FA Cup

##### Age 16 to 18Challenge Level

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