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_avoidThere 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%.
