Riddler Fall Classic

Riddler Express

“If a baseball team is truly .500, meaning it has a 50 percent chance of winning each game, what’s the probability that it has won two of its last four games and four of its last eight games?”

We can answer this question in several different ways. I will cover a few. We are trying to find a probability of a compound event (that word “and” lets you know). Breaking this up, let’s let $A$ be the event that the team won two of its last four games and $B$ be the event that the team won four of its last eight games. Then, we’re after $$P(A\cap B).$$

This can be solved by using conditional probability. We condition on one of the events. This is the two ways I will solve the problem. First, by conditioning on $A$, and second, by conditioning on $B$.

Here is what I mean by conditioning on $A$. I find the probability of $A$ occurring without considering $B$ at all, and then find the probability that $B$ will occur given that $A$ has already occurred. What this looks like notationally is $$P(A \cap B) = P(A)\times P(B|A).$$

To find $P(A)$, the probality that the team won two of its last four games, is to consider all of the possibilities. Here are the equally likely outcomes of the last four games: $$\{LLLL,LLLW,LLWL,LWLL,WLLL,LLWW,LWLW,WLLW, \\ LWWL,WLWL,WWLL,LWWW,WLWW,WWLW,WWWL,WWWW\}$$

We can see that 6 of these 16 come up with two wins and two losses. That means that $P(A) = 6/16 = 3/8 = 0.375$. For those that recognize this as a binomial distribution with $n=4$ and $p=0.5$, we could calculate this by

dbinom(2,4,0.5)

## [1] 0.375

Now, we pretend this has already happened. What is the probability $B$ occurs? Again, we’re given that $A$ already had occurred, so we just need for it to happen again for $B$ to occur! This happens with the same probability of 0.375. Now, using the formula we have

$$P(A \cap B) = P(A)\times P(B|A) = (0.375)\times(0.375) = 0.140625.$$

If we tackled this from the other direction, we would first find $P(B)$. This would be a bit more involved, as we would have to list out 256 outcomes rather than the 16. Of those 256 outcomes, we would find that 70 of them had 4 wins and 4 losses.

This means $P(B) = 70/256 = 0.2734375$. Now, under the condition that $B$ has occurred (that is, one of those 70 outcomes in which we have 4 losses and 4 wins occurred), what is the probability that $A$ occurs.

This could take quite a bit of work, sifting through all 70 outcomes and counting up those that have 2 wins and 2 losses in the last four games (which ends up being 36). Using counting techniques or a hypergeometric distribution could get you there faster.

$$P(A \cap B) = P(B)\times P(A|B) = \frac{70}{256} \times \frac{36}{70} = \frac{36}{256} = 0.140625$$

As this is the same answer as the previous value, this provided a good check.

Riddler Classic

"Riddler League Baseball, also known as the RLB, consists of three teams: the Mississippi Moonwalkers, the Delaware Doubloons and the Tennessee Taters.

"Each time a batter for the Moonwalkers comes to the plate, they have a 40 percent chance of getting a walk and a 60 percent chance of striking out. Each batter for the Doubloons, meanwhile, hits a double 20 percent percent of the time, driving in any teammates who are on base, and strikes out the remaining 80 percent of the time. Finally, each batter for the Taters has a 10 percent chance of hitting a home run and a 90 percent chance of striking out.

“During the RLB season, each team plays an equal number of games against each opponent. Games are nine innings long and can go into extra innings just like in other baseball leagues. Which of the three teams is most likely to have the best record at the end of the season?”

I will answer this problem using simulation. To simulate a season of the Riddler Baseball League, I will need to first break down the simulation into simpler steps. First, I’ll simulate an at-bat for someone on team Moonwalker.

At Bat

We will input a situation, which will be a 3-tuple of 0’s or 1’s, representing whether a runner is on 1st, 2nd, or 3rd base. Next, we’ll input the number of runs. Last, we’ll input the number of outs. Then, with the specified probabilities given in the problem, the output will be the resulting situation, runs, and outs.

MW_atBat <- function(situation, runs, outs){
  p <- sample(0:1, 1, prob = c(0.6,0.4))
  if(p==0){
    return(list(situation = situation, runs = runs, outs = outs+1))
  }
  if(p==1){
    if(situation[3]==1){runs = runs+1}
    situation[2:3] <- situation[1:2]
    situation[1] <- 1
    return(list(situation = situation, runs = runs, outs = outs))
  }
}

set.seed(9876)
MW_atBat(c(1,0,0), 2, 1)

## $situation
## [1] 1 1 0
## 
## $runs
## [1] 2
## 
## $outs
## [1] 1

MW_atBat(c(0,0,0), 0, 2)

## $situation
## [1] 0 0 0
## 
## $runs
## [1] 0
## 
## $outs
## [1] 3

For the at-bat functions for the Doubloons, the situation will be a guy on second or not, so we just need a variable that takes on 0 or 1. For the Taters, there is never a situation on the bases since they’ll either hit a homerun or not.

Dbl_atBat <- function(On2nd, runs, outs){
  p <- sample(0:1, 1, prob = c(0.8,0.2))
  if(p==0){
    return(list(On2nd = On2nd, runs = runs, outs = outs+1))
  }
  if(p==1){
    if(On2nd == 1){runs <- runs+1}
    return(list(On2nd = 1, runs = runs, outs = outs))
  }
}

Tat_atBat <- function(runs, outs){
  p <- sample(0:1, 1, prob = c(0.9,0.1))
  if(p==0){
    return(list(runs = runs, outs = outs+1))
  }
  if(p==1){
    return(list(runs = runs + 1, outs = outs))
  }
}


Dbl_atBat(0, 1, 2)

## $On2nd
## [1] 0
## 
## $runs
## [1] 1
## 
## $outs
## [1] 3

Dbl_atBat(1, 2, 0)

## $On2nd
## [1] 1
## 
## $runs
## [1] 2
## 
## $outs
## [1] 1

Tat_atBat(0,0)

## $runs
## [1] 0
## 
## $outs
## [1] 1

Tat_atBat(1,2)

## $runs
## [1] 1
## 
## $outs
## [1] 3