How Uefa Champions League Draw 16 Probability Calculated

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cldraw1 A number of news outlets have reported a peculiar quirk that arose during Friday's Champions League describe. Apparently, the sport's European governing body, UEFA, ran a trial run the day before the primary outcome, and the schedule chosen during this result was identical to that of the actual draw on Friday.

Given this foreign coincidence, a number of people take been expressing the various odds of this occurrence. For example, the author of this newspaper article claimed that 'bookies' calculated the odds at v,000 to i. In other words, the probability of this event was 0.0002.

The aforementioned article also says that the probability of this event (2 random draws being identical) occurring is non as depression as one might think. However, this article does not give the probability or odds of this result occurring. The oblivious reason for this is that such a calculation is difficult. Since teams from the aforementioned domestic league and teams from the same country cannot play each other, such a calculation involves using conditional probabilities over a variety of scenarios.

Despite my training in Mathematics and interest in quantitative pursuits, I have always struggled to calculate the probability of multiple provisional events. Given that in that location are many different ways in which two identical draws tin can be made, such a calculation is, unfortunately, beyond my admittedly limited power.

Thankfully, there's a cheats style to getting a rough answer: use Monte Carlo simulation. The code beneath shows how to write up a function in R that performs synthetic draws for the Champions League given the aforementioned conditions. With this part, I performed two draws 200,000 times, and calculated that the probability of the identical draws is: 0.00011, so the odds are around: 1 in 9,090. This probability is subject to some sampling fault, notwithstanding getting a more than accurate measure out via simulation would require more than computing power like that enabled by Rcpp (which I really need to learn). Even so, the answer is clearly lower than that proposed either by the 'bookies' or the newspaper article's writer.

# cl draw rm(list=ls()) setwd("C:/Users/Alan/Desktop") dat <- read.csv("cldraw.csv")  #============================= > dat team iso pos grouping 1       Galatasaray TUR  RU     H 2           Schalke GER  WI     B iii            Celtic SCO  RU     G iv          Juventus ITA  WI     East 5           Arsenal ENG  RU     B 6            Bayern GER  WI     F vii  Shakhtar Donetsk UKR  RU     Eastward 8          Dortmund GER  WI     D 9             Milan ITA  RU     C 10        Barcelona ESP  WI     M xi      Real Madrid ESP  RU     D 12      Man. United ENG  WI     H thirteen         Valencia ESP  RU     F 14              PSG FRA  WI     A 15            Porto POR  RU     A 16           Malaga ESP  WI     C #=============================  draw <- function(x){   fixtures <- matrix(NA,nrow=viii,ncol=2)   p <- 0   while(p==0){     for(j in ane:8){       k <- 0       n <- 0       while(k==0){         n <- n + 1         if(northward>l){break}         aa <- 10[x[,"pos"]=="RU",]         t1 <- aa[sample(1:dim(aa)[1],one),]         bb <- 10[10[,"pos"]=="WI",]         t2 <- bb[sample(ane:dim(bb)[one],1),]         k <- ifelse(t1[,"iso"]!=t2[,"iso"] & t1[,"grouping"]!=t2[,"grouping"],ane,0)       }       fixtures[j,1] <- as.character(t1[,"team"])       fixtures[j,2] <- equally.character(t2[,"team"])       x <- x[!(10[,"team"] %in% c(as.grapheme(t1[,"team"]),                                  as.graphic symbol(t2[,"team"]))),]     }     if(n>l){p <- 0}         p <- ifelse(sum(equally.numeric(is.na(fixtures)))==0,i,0)   }   return(fixtures) }  drawtwo <- office(10){    f1 <- every bit.vector(unlist(ten))   joinup <- data.frame(squad=f1[one:16], iso=f1[17:32],                        pos=f1[33:48], group=f1[49:64])   check1 <- data.frame(describe(joinup))   check2 <- information.frame(describe(joinup))   rightdraw <- ifelse(sum(na.omit(check1[society(check1),2])==     na.omit(check2[guild(check2),2]))==8, 1, 0)   return(rightdraw) }  drawtwo(dat)  dat2 <- rbind(equally.vector(unlist(dat)),               equally.vector(unlist(dat))) dat3 <- dat2[rep(i,g),]  vals <- 0 for(i in ane:200){   yy <- apply(dat3, 1, drawtwo)   vals <- sum(yy) + vals } #============================= # Probability  > vals/200000 [1] 0.00011 # Odds > one/(vals/200000)-1 [ane] 9089.909 #=============================