Boat Race (math question)

Factual Questions
1

I had a dream last night.

I was working at my old job, and we were on a company outing to a seaside resort. Somehow we had managed to aquire two rowboats, and I thought it would be a fun idea have races. I talked around 34-35 people into participating.

We managed to set two pylons out about 100 yards from shore. The goal was to row out from shore, go clockwise around a pylon, and row back to shore, and the first boat to hit the rocks on the beach was the winner of that particular heat.

We also managed to aquire a similar number of team Jerseys, in three colors: Blue, dark green, and lime green/white stripes. This gave us three teams of 11-12 people each.

For some reason, Jennifer Aniston was there, and got on the lime green striped team. We decided to call it Team Flair.

Nevermind that each boat looked rated for only 2 people, or that each one only had 2 oars. Nevermind that running a boat–hard–into rocks is a really bad idea. Nevermind that none of these people had ever been in a boat before, we had no lifejackets, and that we were starting the race at night. In the rain. As a boater, I realize that the only way to improve this situation is to add copious volumes of alcohol. Sometimes in dreams, things don’t always make much sense.

Let’s assume more optimal conditions. Each boat performs at its best with a crew of 5 or 6. That gives us 3 teams, each with 2 sub-teams. So we have:

Blue A
Blue B
Green A
Green B
Team Flair A
Team Flair B

Only two boats may compete in any heat. We have no stopwatch, so the winner each heat is determined by who comes in first.

How do we determine which of the three teams is the winner?

2

Non-math answer:

Couldn’t/wouldn’t you run 3 heats Blue A v. Green B, Green A v. Flair B, and Flair A v. Blue B.

in 75% of the possible outcomes, one team will win both of their heats - they’re the winners. Otherwise, that other 25% of the time, each team will win one heat, and you’ll turn into walruses because it’s a dream.

3

Dream, indeed. My brother-in-law got a ticket for parking a boat in a loading zone in front of a jewelry store. And Courney Cox had a heart made out of chocolate cake. :dubious:

In your scenario, it’s quite possible that each team A will win, leaving us with no clear winner.

4

This is not really a mathematical question, per se. Assuming that you just want to run the race once, the answer depends on how you want to define the winner. Some options:
[ol]
[li]Sum up the times for each team’s sub-teams, and the team with the shortest total time wins.[/li][li]Sum up the ranks (first to finish, second to finish, etc.) of each team’s sub-teams, and the team with the lowest total rank wins.[/li][li]Count only the shortest time of each team (from its two sub-teams), and the shortest time wins.[/li][li]Count only the lowest rank of each team, and the lowest rank wins.[/li][/ol]
Actually, if I were you, I would dream up some caps (in two different colors) and have six different teams! :smiley:

5

Yes, but all of those solutions assume a stopwatch. None is available.

6

This one doesn’t:

That’s the way to go.

7

Not sure I understand.

Blue A vs Blue B, B wins.

Green A vs Green B, A wins.

Flair A vs Flair B, B wins. For example.

2nd round,

Blue B vs Green A, winner plays Flair A?

8

Why do you have teams competing against themselves? Blue A and Blue B are on the same team, so they don’t race each other.

In this example, I’ll apply random winners for demonstration purposes:

Blue A vs Green A (Blue A wins)
Blue A vs Green B (Green B wins)
Blue A vs Flair A (Flair A wins)
Blue A vs Flair B (Blue A wins)

Blue B vs Green A (Green A wins)
Blue B vs Green B (Green B wins)
Blue B vs Flair A (Blue B wins)
Blue B vs Flair B (Flair B wins)

Green A vs Flair A (Green A wins)
Green A vs Flair B (Flair B wins)

Green B vs Flair A (Flair A wins)
Green B vs Flair B (Flair B wins)

So after 12 races, every subteam has raced against every opposing subteam exactly once. (This is known as “round robin” play.) Now we check their records, making it very similar to the regular season in any professional sports league:

75% 3-1 Flair B
50% 2-2 Blue A
50% 2-2 Green A
50% 2-2 Green B
50% 2-2 Flair A
25% 1-3 Blue B

Before declaring Flair B the winning team, we must combine the subteams back into their whole teams by adding the records of both subteams together:

62.5% 5-3 Flair Team
50.0% 4-4 Green Team
37.5% 3-5 Blue Team

And now we have a clear winner, the flair team. (Of course. The other teams were too busy drooling over Jen to focus on winning.)

9

I’d say a simple, double elimination ladder type would work just fine.

One like THIS, just plug in the names on the lines, and create additional brackets if needed.

10

An elimination bracket is guaranteed to terminate in a bounded number of heats, but it’s not inherently fair unless the number of competitors is a power of 2 (as it’s not in this case). If you have a non-power of 2, then at some point, some team gets a bye, and you have to decide somehow which team that should be.

A round-robin tournament is completely fair, but it has no guarantee of terminating in any bounded number of heats. That is to say, it’s possible that after each team has played against each other team, there will be a tie for the best record. One can then have another round among just those players to break the tie, but there’s no guarantee of a single winner there, either. So you have another round yet, and so on.

To illustrate, consider a competition between Rock, Paper, and Scissors. If you have an elimination bracket, then whoever gets the first-round bye is guaranteed to win: For instance, if Rock plays Scissors in the first match, and then Paper plays the winner of that match. This is a clearly unacceptably unfair bias favoring one of the competitors before the match even begins. If, on the other hand, we do a round robin tournament, then each round, each player would win one and lose one, leading to a 3-way tie every single round.

Either of these methods can be used, so long as you recognize their shortfalls and plan ahead for them. For instance, in an elimination bracket, the teams are generally placed on the bracket according to prior performance: The top-ranked teams will be the ones to get the byes. Here, since you have differing team sizes, you might consider giving the small teams the byes as a handicap. Or you could do the initial seeding at random. For a multi-round round robin tournament, you could just set some limit, after which ties are left unbroken, or if a tie persists after some pre-decided limit, it could be decided at random.

But the thing that makes the least sense here, is why was Jennifer Aniston wearing a jersey?

11

Those that read my NFL Division Rankings thread know that I could potentially drone on and on about this subject for far longer than would be interesting. Luckily, Chronos outlined the issues nicely.

You could run enough round robin rounds to eliminate at least one team. If one were left as a clear winner, (like in my example) you’re done. If two were tied for first, you could have each major team run a heat between its two subteams to determine which will represent the team in a tiebreaker, and then have those two representatives compete head to head for the match.

Or, you could play by soccer rules and just accept ties as a valid result. (Damn fools must like kissing their sisters.)