See the story and numerical experiments in the previous post. Annie is currently leading 5-3; what is the probability that Becky will reach six and win the game?

Frequentist solution 

Based on the results so far, 3 wins in 8 matches, Beck’s probability of winning is (3/8). That means the probability of Becky winning the next three games (to reach 6 before Annie wins another, which will make her 6) is (3/8)³ = 0.053.

Bayesian solution 

First, we write down the Bayes equation. The probability of B winning the game, given A is leading 5-3, 

Let’s start with the denominator. If p represents the probability of Becky winning a game, the probability of a 3-5 or 3 in 8 result is obtained by applying the binomial equation. The denominator is the sum of all possible scenarios (p values from 0 to 1). In other words, The denominator is,

We use an R shortcut to evaluate the integral as the area under the curve using the ‘AUC’ function from the ‘DescTools’ library.

library(DescTools)
x <- seq(0,1,0.01)
AUC(x, choose(8,3)*x^3*(1-x)^5, from = min(x, na.rm = TRUE), to = max(x, na.rm = TRUE)) 

0.1111

And here is the area. 

plot(x, choose(8,3)*x^3*(1-x)^5, type="l", col="blue", ylab = "")
polygon(x, choose(8,3)*x^3*(1-x)^5, col = "lightblue")

The numerator is the multiplication of the likelihood function with the prior probability of Becky winning three games. We will use a binomial equation as the prior function and p x p x p = p³ as P(B). 

Again, the area under the curve all possible p values (0 -1),

AUC(x, choose(8,3)*x^6*(1-x)^5, from = min(x, na.rm = TRUE), to = max(x, na.rm = TRUE)) 

0.0101

The required Bayesian posterior is 0.0101/0.1111 = 0.0909

To sum up

The Bayesian estimate is almost double compared to the frequentist. The Bayesian estimate is much closer to the results we saw earlier. Does this make Bayesian the true winner? We’ll see next. 

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)