Death by Quarters

Death by Quarters

By Josh Feldman

It’s been a rough couple of months for me—a running injury has prevented me from hitting the roads this winter. So while the running gang was out there hitting the pavement, I was stuck at home being a lazy bum. My friend The Professor saw that I had a case of the winter doldrums and suggested I head to the local track to witness the big running community event of the season.

It turned out that two local runners, Doc and Hoss, didn’t like each other, and decided to settle their grievances in the most civilized way possible: through a series of quarter-mile races on the track. As the old saying goes, the bigger the conflict, the larger the number of intervals required to settle the dispute!

And what an insane number of intervals these two decided to participate in. In college I might have run 20 quarters in a session, but these two decided to up the stakes by running 50 quarter-mile intervals. Just the thought boggled my mind! And did I mention this would all take place outside during a frigid, windy Midwestern winter day?

The rules of the event were simple: The two combatants would run a lap around the track, and whoever came in first place for that lap would win a point. Following a mere 30-second rest, the two would line up on the starting line again … and again … until they raced each other 50 times. The most points wins! And given the vitriol between the two entrants, it wasn’t enough just to win, as both wanted to embarrass and humiliate the opposition no matter what the score was. Why beat your opponent 27 to 23 when you can demolish your adversary by a bigger margin of victory?

After hearing the details, I quickly headed for the track to watch the festivities. And what a spectacle it was! Doc won the first lap, and after 20 intervals he had a decent-sized lead. But Doc couldn’t pull away, and by interval 45 Hoss looked like he had more left in the tank for the final five laps. Hoss cut the lead to 25-24 going into the final lap, and by this time hundreds in the community heard about the grudge of the century and made their way out in the cold to witness the end of the madness.

Amazingly, Hoss pulled that final quarter out by a whisker, running a world-defying sub-60-second final quarter mile to win the final point. So after 50 laps of extreme competition, both runners were deadlocked at 25 points each … not settling anything at all.

A few minutes after the final quarter mile, fellow onlooker Big T mentioned to me that throughout the entire competition Doc had the lead; it wasn’t until that final lap where Hoss found a way to tie up the score. Knowing I am the supposed math guy, he asked me: What are the odds of something like that occurring? Alas, I couldn’t answer him, as my brain definitely couldn’t function after spending so much time out in the cold and wicked wind. But I bet you my puzzle-solving friends will once again come through for me.

For all of the questions below, feel free to assume that the two racers had equal likelihood of winning each quarter, with there being zero correlation between each race.

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

Solutions may be emailed to cont.puzzles@gmail.com. In order to make the solver list, your solutions must be received by April 1, 2023.

Solutions to Previous Puzzles: Urn Problems

Some people thought these problems were easy—maybe so, but they were a bit tricky.

Problem 1. (Halmos) One hundred balls, 50 red and 50 green, are distributed between two urns. An urn is selected at random; from it, a ball is drawn at random. Does the probability of getting a red ball this way depend on how the balls were distributed between the two urns? Justify your answer.

Answer: Yes.

Solution: If a single red ball is in one urn and the remaining 99 balls are in the other, then P(red)= ½ +(49/99)/2 = 0.7475. Conversely, if a single green ball is in one urn and the remaining 99 balls are in the other, then P(red)= 0.2525.

Problem 2. (Steinhaus) An ichthyologist wanted to estimate the number of fish in a pond that were suitable for catching. She threw a net with regulation-size mesh into the pond; after taking the net out, she found 30 fish in the net. Making sure to keep them alive, she marked each of them with a suitable color and threw them all back into the pond. The next day she threw the same net into the pond and caught 40 fish, of which 2 had been marked. Approximately how many fish were in the pond? Justify.

Answer: 599.5…wait! 600 is better.

Solution: P(2 marked) = . The integral MLEs are 599 and 600. But the likelihood is even larger in an interval around 599.5. However, there are good, practical reasons to round to 600.

Problem 3. (Honsberger) There are 6 red balls and 8 green balls in a bag. Five balls are drawn at random and placed in a red box; the remaining 9 balls are put in a green box. What is the probability that the number of red balls in the green box plus the number of green balls in the red box is not a prime number? Justify.

Answer: 0.213 (did you remember that 1 is not a prime?).

Solution: The red box can have X = 0 thru 5 green balls and thus 5–X red balls. That leaves 6–(5–X) = X+1 red balls in the green box. The total is 2X+1, which takes the values 1,3,5,7,9,11 as X varies from 0 to 5. The total is not prime if it is 1 or 9, i.e., X=0 or 4. P(X=0 or 4) =

Solvers. Mike Blakeney, Joshua Boehme, Scott Boulay, Bob Byrne, Lois Cappellano, William  Carroll, Bob Conger, Andrew Dean, Rui Guo, Steve Itelson, Clive Keatinge, Don Onnen, David Promislow, Gregory Scruton, Noam Segal, Jason Shaw, John Snyder, Al  Spooner, Matt Stephenson, Daniel Wade, and Haiyan Wang.

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