Probability: Combination and Permutation in Chess

Question

Question: In a chess match, there are 16 contestants. Every player has to be each other player (like a round-robin). The player with the most wins/points wins the tournament.

a) How many games must be played until there is a victor?

b) If every player has to team up with each other player to play doubles chess. How many games must now be played until one of the teams is a victor?

My Attempt:

a) Each of the 16 player would have to verse 15 other people, but Player 1 vs Player 16 is same as Player 16 vs Player 1. Hence, $(16*15)/2$

b) No idea

Official Answer:

a) ${}^{16}C_2$

b) ${}^{16}C_2/2$

My Problem:

a) ${}^{16}C_2$ is the same as my answer, however I thought combination would mean, how many different ways you can choose 2 people out of 16 people. However the question asks how many games have to played, so how does how many games have to be played mean the same thing as how many ways you can choose 2 people out of 16?

b) No idea

Answer 1

a) If I choose any two people there has to be one game between them, so the number of games is the same as the number of ways to choose two people. (The reason you use combination instead of permutation here is that it doesn't matter who is black and who is white.)

b) I can't see how the answer given is plausible for any interpretation of the question.

If the intended meaning was that every player teams up with one other player and a round-robin between the teams is played, there would be $8$ contestants (the teams) and so by the same argument as (a) you need ${}^8C_2={}^{16/2}C_2$ games.

On the other hand, if they really mean all possible teams are competing, there are ${}^{16}C_2$ possible teams, and each has to play each other, so it looks like you need ${}^{{}^{16}C_2}C_2$. However, this includes games where there is one player on both teams, which doesn't make much sense, so perhaps a better interpretation is that every team has to play every other team that neither of its players are on. This would give ${}^{16}C_2\times{}^{14}C_2/2$ games, because for each team there are ${}^{14}C_2$ teams they can play taken from the other $14$ players, and then you can swap the two teams to get the same game.

Answer 2

For the first part for a better clarity begin with some smaller cases.

Let's consider there are 3 people playing the chess. The what will be the number of matches until a Victor is decided. Through simple counting we get this as 3( which is simply $\binom {3}{2}$ )

Let's consider 4 people to get a stronger hold over the assumption. Again by simple counting you see that the number of matches would be 6 which is simply $\binom {4}{2}$

While counting these cases you might have noticed that what you are doing is nothing but finding the number of ways you can select two people from $16$ contestants.

For the second part though I have a doubt over the official answer. If they meant to find the matches between all possible teams among themselves. Then we must form two teams. For the first team we select 2 people out of 16. (ways = $\binom {16}{2}$ ). Now for the second team we select 2 people from remaining 14 people( ways $=\binom {14}{2}$ )

But during this process we do overcounting and allow two same teams compete twice. Hence we divide the answer by 2.

Hence according to me the answer to the second part should be $$\frac {\binom {16}{2}\binom {14}{2}}{2}$$