Round Robin Tennis Game

[as2528]

The question is as follows:

A round-robin tournament is being held with n tennis players; this means that every player will play against every other player exactly once.
(a) How many possible outcomes are there for the tournament (the outcome lists out who won and who lost for each game)?
(b) How many games are played in total?

I attempted to solve part a.
I thought that for a game involving 4 players: {A,B,C,D}, there would be 12 games. I made a few sets as follows listing the winner of the match and then the loser of the match:

{Winner, Loser}
{A,B}
{B,A}
{A,C}
{C,A}
{A,D}
{D,A}

{B,C}
{C,B}
{B,D}
{D,B}

{C,D}
{D,C}

From what I saw in the sets, the number of games each player plays reduces. For example, A does not play B in the sets with B. Also, there are two possible outcomes for each game. So, I thought the total possible outcomes is this:

3*(2C1)+2*(2C1)+1*(2C1)

The (2C1) is meant to be the two possible outcomes for each game. 3,2, and 1 are there for the games each player plays. So A has 6 possible outcomes, B has 4, C has 2 possible outcomes.

This gave me the summation: [math]\sum_{n=0}^{n-1} n*2C1[/math]
This sum is my final answer, but the actual answer is [math]2^{(n*(n-1))/2}[/math]
Where is my logic going wrong? I do not understand why my equation is wrong, and while I try to make sense of it I think I'm missing something important.

 

[Dr.Peterson]

I thought that for a game involving 4 players: {A,B,C,D}, there would be 12 games.

No, there would be 6 games: AB, AC, AD, BC, BD, CD. Order doesn't matter in a game. (It does matter if you choose to list the players in each game with the winner first, but I think doing so may have misled you.)

I would think about this in two steps: First, how many games are played; second, how many different ways are there to assign a winner to each of those games? The latter could be compared to listing each game as an ordered pair, and choosing a subset of them in which the first player won.

This gave me the summation: [math]\sum_{n=0}^{n-1} n*2C1[/math]
This sum is my final answer, but the actual answer is [math]2^{(n*(n-1))/2}[/math]
Where is my logic going wrong? I do not understand why my equation is wrong, and while I try to make sense of it I think I'm missing something important.

Now, your answer has some of the right pieces in it; the summation is valid (and shows up, in a different form, in the correct answer), but the multiplication by 2C1 should not be a multiplication.

 

[as2528]

No, there would be 6 games: AB, AC, AD, BC, BD, CD. Order doesn't matter in a game. (It does matter if you choose to list the players in each game with the winner first, but I think doing so may have misled you.)

I would think about this in two steps: First, how many games are played; second, how many different ways are there to assign a winner to each of those games? The latter could be compared to listing each game as an ordered pair, and choosing a subset of them in which the first player won.

Now, your answer has some of the right pieces in it; the summation is valid (and shows up, in a different form, in the correct answer), but the multiplication by 2C1 should not be a multiplication.

Thank you! I'll try it again with that in mind.

 

[Dr.Peterson]

Now, your answer has some of the right pieces in it; the summation is valid (and shows up, in a different form, in the correct answer), but the multiplication by 2C1 should not be a multiplication.

I should add that in saying the summation was valid, I only meant the general idea; what you actually wrote is nonsense. (What does n mean?)

This gave me the summation: [math]\sum_{n=0}^{n-1} n*2C1[/math]

 

[NotJeffM]

I do not in any way disagree with Dr. Peterson. I just would attack the problem in a slightly different way.

Your idea of starting with a simple example is fine, but it can help to consider the simplest possible example first and then progressively consider several more complex example.

If there were two players A and B, there would be one match, AB.

If there were three players A, B, and C, there would be three matches, AB, AC, and BC.

If there were four players, A, B, C, and D, there would be six matches, AB, AC, AD, BC, BD, and CD.

How might we generalize that result? If we have n distinct players, how many distinct ways can we choose two of them? Do you know a formula for that kind of general question?

Now think about the different possible outcomes.

If n = 2, there are two, AB or BA

If n = 3, there are eight, AB, AC, BC or AB, AC, CB or AB, CA, BC or AB, CA, CB, or
BA, AC, BC or BA, AC, CB or BA, CA, BC or BA, CA, CB

A pattern seems to be emerging. Can you logically justify that pattern?