February 2023

We have seen how three English teams could avoid playing each other in the Champions League quarter-finals. What about if four teams reach; this happened on three occasions – 2008, 2009 and 2019 – in the last four decades. It’s impossible to avoid one another. Or is it? Let’s calculate.

Let’s use the same code we used earlier by adding another team, E4.

team <- c("E1", "E2", "E3", "E4", "R5", "R6", "R7", "R8")

itr <- 1000000

draw <- replicate(itr, {

      dr <- sample(team, 8, replace = FALSE)  

     dr1 <- paste(dr[1], dr[2])
     dr2 <- paste(dr[3], dr[4])
     dr3 <- paste(dr[5], dr[6])
     dr4 <- paste(dr[7], dr[8])

     dr_all <- c(str_count(dr1, "E"), str_count(dr2, "E"), str_count(dr3, "E"), str_count(dr4, "E")) 


     if(any(dr_all == 2)){
         counter <- 1
     }else{
        counter <- 0
     }


})

P_avoid <- 1 - mean(draw)
P_avoid

There is a 22.8% chance to avoid – not so bad!

We use the same approach for the analytical solution. The number of ways to form four pairs from eight teams is 8C2 x 6C2 x 4C2 /4! = 105. To avoid the team from the same league, the four English clubs must pair with the other four. The first team has 4, the second team has 3, the third has 2, and the last has one choice. So total = 4 x 3 x 2 = 24. The probability to avoid = 24/105 = 0.228 or 22.8%.

Three teams from the English premier league have reached the last season’s (2021-22) champions league last eight. But strangely, they avoided each other in the knockouts. The draw for the quarter-final is determined by picking names, at random, in a bowl containing all eight. The first plays with the second, third with the fourth etc. So, how likely are the three EPL teams to avoid each other in the lot?

The simplest way is to run the lot a million times and estimate the average time it happened. Let’s do that in R.

team <- c("E1", "E2", "E3", "R4", "R5", "R6", "R7", "R8")

itr <- 1000000

draw <- replicate(itr, {

      dr <- sample(team, 8, replace = FALSE)  

     dr1 <- paste(dr[1], dr[2])
     dr2 <- paste(dr[3], dr[4])
     dr3 <- paste(dr[5], dr[6])
     dr4 <- paste(dr[7], dr[8])

     dr_all <- c(str_count(dr1, "E"), str_count(dr2, "E"), str_count(dr3, "E"), str_count(dr4, "E")) 


     if(any(dr_all == 2)){
         counter <- 1
     }else{
        counter <- 0
     }


})

P_avoid <- 1 - mean(draw)
P_avoid

The answer turned out to be around 57%.

The next step is to solve this problem analytically. The number of ways to form the first pair is estimated using combinations formula ₈C₂. For the second group, it is ₆C₂, the third is ₄C₂, and the last is ₂C₂ or 1. This gives a total number = ₈C₂.x ₆C₂ x ₄C₂, of which 4! are the same, as the order of arrangement doesn’t matter. So, after correcting, it becomes ₈C₂.x ₆C₂ x ₄C₂ / 4! = 105

choose(8,2)* choose(6 ,2)*choose(4,2)/ factorial(4)

Now, calculate the number of ways the three English can pair with the other five. For the first team, it’s 5 ways; for the second team, it’s 4; for the third, it’s 3. So total = 5 x 4 x 3 = 60. The probability to avoid = 60/105 = 0.57 or 57%.