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
1
63
 .3/5

0.00952
62
63
 . 7
10
 2

 


2
1
31
 . 62
63
 . 7
10
 2

 
 .3/5

0.00933
30
31
 . 62
63
 . 7
10
 2

 
 .3/52


3
1
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.3/5

0.00672
14
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.1/22


4
1/7. 14
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.1/22.3/5

0.00336
6/7. 14
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.1/24


5
1/3.6/7. 14
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.1/24.3/5

0.00168
2/3.6/7. 14
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.1/26


6
2/3.6/7. 14
15
 . 30
31
 . 62
63
 . 7
10
 2

 
 .3/52.1/26.3/5

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 ×10-7 or roughly 1 in a million. Alternative approach:

Multiplying the probabilities on the tree diagram below we get for each round fk(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
f1(f2(f3(f4(f5(3/5))))).
This is computed to be
1
63
 . 3
5
 + 62
63
 . 7
10
 2

 
 æ
ç
è 		1
31
 . 3
5
 + 30
31
 . 3
5
 2

 
 æ
ç
è 		1
15
 . 3
5
 + 14
15
 . 1
2
 2

 
 æ
ç
è 		1
7
 . 3
5
 + 6
7
 . 1
2
 2

 
 æ
ç
è 		1
3
 . 3
5
 + 2
3
 . 1
2
 2

 
 æ
ç
è 		3
5
 ö
÷
ø 		ö
÷
ø 		ö
÷
ø 		ö
÷
ø 		ö
÷
ø 		= 0.0314571