Annie’s Table Game

Annie and Becky are playing a game in which the aim is to secure six points first. The Casino randomly rolls a ball onto the table, and the point at which the ball rests is marked. The Casino then rolls another ball at random. If it comes to rest to the left of the initial mark, Annie wins the point; to the right, Becky wins. If Annie is currently leading 5-3, what is the probability that Becky will win the game?

Before we get into statistical methods, we will play the game a million times using R. 

First step: Becky’s chance of winning any game is a random number with uniform probability. If that is the case, the probability of Annie leading 5-3 (or Becky winning 3 out of 8 games) is given by binomial distribution.

beck_pro <- runif(1)
p_5_3 <- dbinom(3, 8, beck_pro)

That gives the probability of having a 5-3 lead for Annie, but that doesn’t mean it will happen for sure. For that to happen, the estimated probability must be greater than a random number between 0 and 1.

  if(runif(1) < p_5_3){
    game_5_3 <- 1
  }else{
    game_5_3 <- 0
  }

Thus, we established instances where Annie was leading 5-3. We will estimate the probability of Becky winning the next three games. 

  if(game_5_3 == 1){
     beck_3_0 <- beck_pro^3
  } else{
     beck_3_0 <- 0
  }

Like we did before, a random number is generated, and if it is less than the probability of Becky winning the next three games, Becky wins; otherwise, Annie wins. 

 if(beck_3_0 == 0){
    win_beck <- "no"
  }else if(runif(1) < beck_3_0){
    win_beck <- "Becky"
  }else{
    win_beck <- "Annie"
  }
  
  })

Calculate this a million times and estimate the proportion of Becky’s win over total wins.

itr <- 1000000
beck_win <- replicate(itr, {
  beck_pro <- runif(1)
  p_5_3 <- dbinom(3, 8, beck_pro)

  if(runif(1) < p_5_3){
    game_5_3 <- 1
  }else{
    game_5_3 <- 0
  }
  
  
  if(game_5_3 == 1){
     beck_3_0 <- beck_pro^3
  } else{
     beck_3_0 <- 0
  }
  
  
  if(beck_3_0 == 0){
    win_beck <- "no"
  }else if(runif(1) < beck_3_0){
    win_beck <- "Becky"
  }else{
    win_beck <- "Annie"
  }
  
  })

sum(beck_win == "Becky")/(sum(beck_win == "Becky") + sum(beck_win == "Annie"))

0.09057296

Introduction to Bayesian Statistics – A Beginner’s Guide: Woody Lewenstein

What is Bayesian statistics?: Nature Biotechnology,  volume 22, 177–1178 (2004)