As per the column, we can assume the two racers have equally likely chances of winning each quarter mile lap around the track.

Problem 1

What is the probability that Doc and Hoss would end up tied after running 50 races against each other?

As it doesn’t matter which order that Hoss wins 25 and that Doc wins 25, this is a simple binomial probability of 25 “successes” in 50 trials where a success probability is 1/2.

dbinom(25,50,1/2)
## [1] 0.1122752

This comes out to be about 11.2%.

Problem 2

What is the probability that one of the two runners would be ahead after all 50 laps, and would never be tied or behind on the scoreboard throughout the entire competition?

To solve this, I began creating a version of Pascal’s triangle with the rule that the center value had to be 0. This captures the idea that once an individual takes the lead, we cannot return to a tie. The first few lines looked like this:

1

1 1

1 0 1

1 1 1 1

1 2 0 2 1

1 3 2 2 3 1

1 4 5 0 5 4 1

I built a simple vector that models one half of the triangle at each step. After 50 steps, I summed the vector, multiplied by two, and divided by the total number of possible paths (2^50) to find a surprising answer!!

v <- c(1)

for(i in 2:50){
  if(i%%2==0){
    v <- v + c(0,v)[-(length(v)+1)]
  }
  else{
    v <- c(v,0)+c(0,v)
  }
}

2*sum(v)/2^50
## [1] 0.1122752

The same answer as problem 1! That probably means there is a much more clever way of thinking about this than I could see.

Problem 3

What is the probability that not only would the runners finish the competition tied at 25 points each, but that one of the two competitors would never hold the lead throughout the entirety of the event?

I’ve interpreted this to mean that the other runner does hold the lead throughout the entirity of the event except for the finish, at which point they tie.

For this problem, I did something similar to problem 2. I built a vector that counts paths with the rule that the center value of Pascal’s triangle has to be zero up to and including the 25th race.

After the 25th race a new rule must exist, as one runner cannot continue past 25 wins. For this reason, the vector must now decrease in the number of entries back down to one entry at the 49th race. We then multiply that by 2 and divide by 2^50 for our answer.

v <- c(1)

for(i in 2:25){
  if(i%%2==0){
    v <- v + c(0,v)[-(length(v)+1)]
  }
  else{
    v <- c(v,0)+c(0,v)
  }
}

for(j in 26:49){
  if(j%%2==0){
    v <- v[-1]+v[-length(v)]
  }
  else{
    v <- v + c(v,0)[-1]
  }
}

2 * v / 2^50
## [1] 0.00229133

Here we get 0.00229133 as the solution for Josh Feldman’s cold and frazzled mind after spending so much time out in the cold.