Picking Fair Swim Teams

Probability Level 5

A group of 9 kids want to hold a swimming competition. To start, each swimmer is given a rank from 1 to 9, with 1 being the fastest swimmer of the group and 9 being the slowest, without repetition.

Let swimmers 1, 2 and 3 be on Team $A$ , $B$ , and $C$ , respectively. These three swimmers each then pick two more kids for their team, making teams of 3.

The competition works as follows:

First, each team will face off against each other individually, meaning Team A will face Team B, Team B will face Team C, and Team C will face Team A. Afterwards, all three teams will compete against each other at the same time. Each of these $4$ meets are scored separately and don't affect one another.
Each meet consists of 3 races. For each race, each team randomly selects a swimmer from their team who has yet to swim a race.
1 point is awarded to the team with the winning swimmer from each race, and no points are given for 2nd and 3rd place.
Assume that a faster ranked swimmer will always beat a slower ranked swimmer.

At first it seems that the teams are picked fairly. In fact, the teams are picked such that Team A is more likely to beat Team B, Team B is more likely to beat Team C, and Team C is more likely to beat Team A! However, while a majority of the time the meet between all three teams will come out in a tie, one team has a higher probability to win than the other two, depending how each team picked their swimmers. If the probability of each team winning exactly 1 race each in the final meet is $\frac{7}{12}$ , then the team that is most likely to win has swimmers with the rankings $R_1$ , $R_2$ , and $R_3$ . What is $R_1R_2R_3$ ?

The answer is 72.

1 solution

Garrett Clarke
Jul 23, 2015

There are $2$ possible arrangements of Teams A, B and C such that A is likely to beat B, B is likely to beat C, and C is likely to beat A:

$\text{Arrangement 1: }A = \{1,5,9\}\text{; }B = \{2,6,7\}\text{; }C = \{3,4,8\}$

$\text{Arrangement 2: }A = \{1,6,8\}\text{; }B = \{2,4,9\}\text{; }C = \{3,5,7\}$

My solution involved using a Java Program to help me list out all the possibilities, and it led me to the following probabilities:

Arrangement 1:

Team 1: 48/216 = 2/9

Team 2: 24/216 = 1/9

Team 3: 24/216 = 1/9

Ties: 120/216 = 5/9

Arrangement 2:

Team A: 24/216 = 1/9

Team B: 48/216 = 2/9

Team C: 18/216 = 1/12

Ties: 126/216 = 7/12

As you can see, Arrangement 2 leads to the desired number of ties. In this arrangement, Team B is most likely to win, therefore our answer must be $2\times4\times9=\boxed{72}$ . If anyone has a more intuitive approach, feel free to post your solution!

A=(2,3,9) ,B=(4,5,6) and C=(1,7,8) also works.

Srikanth Tupurani - 11 months, 1 week ago

While this is valid in terms of breaking transitivity, the problem states that swimmers 1, 2 and 3 be on teams A, B and C, respectively. This solution has swimmers 2 and 3 on the same team, which the problem states isn't possible.

Garrett Clarke - 11 months, 1 week ago

This case may not satisfy the other condition. that 7/12 thing. nice solution. it is interesting. transitivity is voilated. we can select teams A,B,C such that team A has more chance of winning against team B. team B has more chance of winning against team C. team C has more chance of winning against team A. It is little counterintutive. this makes math interesting.

Srikanth Tupurani - 11 months, 1 week ago

I agree, I find non-transitive properties fascinating. Some really cool stuff out there with non-transitive dice, you should check it out!

Garrett Clarke - 11 months, 1 week ago

Picking Fair Swim Teams

The answer is 72.

1 solution

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