How many games in total would be played in a Chess tournament...

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Zelix224

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Hi, everyone!

I'm having a bit of a problem with a simple math question that I'm too embarrassed to admit that I can't come up with an answer by myself, as I haven't used anything more than basic arithmetic since I graduated years ago.

My question goes like this:

If I'm hosting a chess tournament with a total of 30 players, where each player would have to play against each other 10 times, how many games would be played in total in this tournament?

Naturally, we're not going to have players play against "themselves" as it wouldn't make sense.

So, for example, Player 1 would play against Player 2 10 times, then Player 1 would play against Player 3 10 times, then against Player 4 10 times, and so on until the last player, in which case it would reset it back to Player 2 playing against Player 3 10 times, then against Player 4, then 5, etc.

Then I would simply calculate the total sum of matches played in the tournament, who won the most matches, who lost the most, etc.

I thought the math was simple, just multiply 30 x 10 and there you go, 300 matches, but when I think about it, it just doesn't add up, and so instead of manually trying to count it out, I'm asking for help from you guys in finding the correct way on how to calculate things like this.

Also, apologies if this is posted in the wrong subforum, I was hesitant on whether or not to post this in the "Arithmetic" or "Probability / Statistics" category, so I chose this one, thinking it'd be more appropriate here than there.
 
Zelix224 said:
My question goes like this:
If I'm hosting a chess tournament with a total of 30 players, where each player would have to play against each other 10 times, how many games would be played in total in this tournament?
Click to expand...
With just two players there are only ten games. One pair.
With three players there are three sets of ten games. Three pairs.
With just three players there are only thirty games. three pair.
But With four players there are sixty games because there are six pairs.
Thus you need to know with ten players how many pairs are there. SEE HERE
That is ten choosing 2.
 
Why not deal with the 10 until the very end. For example, IF there was 98 games in total if each player played every other player exactly one time, then shouldn't your answer then be 98*10?

Player 1 plays 29 other players
Player 2 plays 28 other players (since we already counted player 2 and player 1 playing together)
Player 3--27
Player 4---26
...
Player 28--2
Player 29 ---1
Player 30---0
Ignore the 0 at the end.
If you add the 1st and 29th number you get 30.
If you add the 2nd and 28th number you get 30.
Now you continue and figure out the sum.
 
Zelix224 said:
If I'm hosting a chess tournament with a total of 30 players, where each player would have to play against each other 10 times, how many games would be played in total in this tournament?
Click to expand...
Of late there has be a brouhaha over what some said was giving complete solutions. So I held back.
But would appear that Zelix is asking for help a real mathematical question.
When I first read it, I thought it was taken from a a graph theory course.
The solution calls for a complete graph on thirty vertices. So we need [imath]\mathcal{C}_2^{30}=\dbinom{30}{2}=\dfrac{30!}{(30-2)!(2!)}=435[/imath].
That means that as a graph it has [imath]435[/imath] edges in that graph, along which there are ten games.


[imath][/imath][imath][/imath]
 
pka said:
With just two players there are only ten games. One pair.
With three players there are three sets of ten games. Three pairs.
With just three players there are only thirty games. three pair.
But With four players there are sixty games because there are six pairs.
Thus you need to know with ten players how many pairs are there. SEE HERE
That is ten choosing 2.
Click to expand...

Steven G said:
Why not deal with the 10 until the very end. For example, IF there was 98 games in total if each player played every other player exactly one time, then shouldn't your answer then be 98*10?

Player 1 plays 29 other players
Player 2 plays 28 other players (since we already counted player 2 and player 1 playing together)
Player 3--27
Player 4---26
...
Player 28--2
Player 29 ---1
Player 30---0
Ignore the 0 at the end.
If you add the 1st and 29th number you get 30.
If you add the 2nd and 28th number you get 30.
Now you continue and figure out the sum.
Click to expand...
Thanks @pka & @Steven G for the replies!

I asked this question over at Reddit as well and have got an answer like this:
There are 30x29/2=435 pairs of players. If each pair plays ten games, that makes 4350 games.

Another way to arrive at the same number: each of the 30 people plays 290 games, 10 against each of 29 opponents (but don't just multiply 30 x 290 since there are two people in each game.)
Click to expand...

So, now I wonder, how exactly could I create a formula for problems like this in case I would need to solve similar questions in the future?

Number of matches/games a player has to play x (Players - 1)
Click to expand...
* Players-1 because a player wouldn't play against himself.

Also, what would happen if there were an uneven number of players, let's say 35 instead of 30, how would we reach the total number of games played then if we can't make proper pairs?

Here's another example:

Let's say there are a total of 5 players playing against each other 5 times.

So, Player 1 plays against Player 2 5 times (that's 5 games in total), then moves on to Player 3 (10 games in total)... all the way up to Player 5 (where it would be 25 games in total).

After we've finished with Player 1, we would move on to Player 2 starting a match against Player 3 (since he already played against Player 1, there's no need to add him to the equation)... and so on we go in a where each next sequential player would play against the next one, but not the previous ones until we arrive at the last match of Player 4 vs. Player 5.
 
I have to admit that I am not sure what the question is even asking.

Thirty players means [imath]\dbinom{30}{2} = 435[/imath] distinct pairs. If each pair is to play 10 games, that is 4350 games.

If that is indeed the question, the answering formula is amazingly simple. With n > 1 players and m games for each pair

[math]\dbinom{n}{2} * m = \dfrac{m * n!}{2! * (n - 2)! } \text { where}\\ n! = 1 \text { if } n = 0 \text { and}\\ n! = n * (n - 1)! \text { if } n \text { is an integer} > 0.[/math]
There is no issue with respect to an odd number of players.
 
Last edited by a moderator:
JeffM said:
I have to admit that I am not sure what the question is even asking.

Thirty players means [imath]\dbinom{30}{2} = 435[/imath] distinct pairs. If each pair is to play 10 games, that is 4350 games.

If that is indeed the question, the answering formula is amazingly simple. With n > 1 players and m games for each pair

[math]\dbinom{n}{2} * m = \dfrac{m * n!}{2! * (n - 2)}, \text { where}\\ n! = 1 \text { if } n = 0 \text { and}\\ n! = n * (n - 1)! \text { if } n \text { is an integer} > 0.[/math]
There is no issue with respect to an odd number of players.
Click to expand...
Hi, Jeff!

Sorry for the confusion in my question, but yes, that is indeed what I was asking, and thank you for providing the formula for it :)

Now, I don't want to sound dumb, but would there be a way to even further simplify this formula, not by using pairs, but by the number of players, and the matches being played in total? Or am I misunderstanding here, and that's actually what you wrote?
 
Zelix224 said:
Would there be a way to even further simplify this formula, not by using pairs, but by the number of players, and the matches being played in total? Or am I misunderstanding here, and that's actually what you wrote?
Click to expand...
The general formula is: If [imath]N~\&~n[/imath] are positive integers such that [imath]N\ge n[/imath]
then there are [imath]\dbinom{N}{n}=\dfrac{N!}{(N-n)!\cdot(n!)}[/imath] unique selections of [imath]n[/imath] each.
Thus [imath]\dbinom{30}{2}=\dfrac{30!}{(28)!\cdot(2!)}=435[/imath].
For students of computer science this is known as the handshake question.
There are [imath]435[/imath] connections needed for [imath]30[/imath] stations to talk directly to each other.

[imath][/imath][imath][/imath][imath][/imath]
 
Yes, that is actually what I wrote. The letter n stands for the number of players (not pairs), and m stands for the number of games that you want each possible pair of players to play against each other.

Obviously, a game of chess is played by a pair of players so we must compute the total number of distinct pairs that can be formed from n players. If we call the number of distinct pairs p, then the number of games to be played is basic arithmetic:

[math]p \times m.[/math]
With me to here?

So, the part of your question that involves more than basic arithmetic is finding the number of distinct pairs from n distinct people. This kind of counting problem comes up so often that it has its own mathematical symbol. Remembering that we have said p is the number of pairs and n is the number of players, we get

[math]p = \dbinom{n}{2} = \dfrac{n!}{2! \times (n - 2)!} = n \times (n -1) \div 2.[/math]
We can eliminate p and combine the formulas to a single formula:

[math]\text {total number of games} = m \times n \times (n - 1) \div 2, \ \text { where }\\ m = \text {number of games each distinct pair is to play, and}\\ n = \text {total number of players.}[/math]
I said in my previous post that having an odd number of players does not affect the number of games. That is true. The only effect of having an odd number of players is that one player must sit out each round.

Suppose we have 5 players: Alan, Betsy, Charles, Debby, and Ed. Our formulas say we have
[imath]5 \times (5 - 1) \div 2 = 5 \times 4 \div 2 = 10[/imath] pairs, namely
Alan-Betsy
Alan-Charles
Alan-Debby
Alan-Ed
Betsy-Charles
Betsy-Debby
Betsy-Ed
Charles-Debby
Charles-Ed
Debby-Ed

But 5 is an odd number so one player must sit out each round. We could leave out Alan for the first round.

Betsy-Charles and Debby-Ed in round 1. Takes care of 2 pairs.

Now leave Betsy out.

Alan-Debby and Charles-Ed in round 2. Takes care of 2 more pairs.

Now take Charles out.

Alan-Ed and Betsy-Debby. Takes care of 2 more pairs.

Now take Debby out.

Alan-Charles and Betsy-Ed. Takes care of 2 more pairs.

Now take Ed out (notice that Ed has played each player in a previous round).

Alan and Betsy have not yet played each other so pair them. And Charles and Debbie have not yet played so pair them.

With 30 players, I might use a computer routine to do the pairings.

Hope this helps.
 
You got as far as wanting to add 1 + 2 + 3 + ... + 29
Here is a formula. 1 + 2 + 3 + ... + n = n(n+1)/2.

So 1 + 2 + 3 + ... + 29 = 29(29+1)/2= 29(30)/2 = 29(15) = 435-----435*10 = 4350
 
pka said:
The general formula is: If [imath]N~\&~n[/imath] are positive integers such that [imath]N\ge n[/imath]
then there are [imath]\dbinom{N}{n}=\dfrac{N!}{(N-n)!\cdot(n!)}[/imath] unique selections of [imath]n[/imath] each.
Thus [imath]\dbinom{30}{2}=\dfrac{30!}{(28)!\cdot(2!)}=435[/imath].
For students of computer science this is known as the handshake question.
There are [imath]435[/imath] connections needed for [imath]30[/imath] stations to talk directly to each other.

[imath][/imath][imath][/imath][imath][/imath]
Click to expand...
JeffM said:
Yes, that is actually what I wrote. The letter n stands for the number of players (not pairs), and m stands for the number of games that you want each possible pair of players to play against each other.

Obviously, a game of chess is played by a pair of players so we must compute the total number of distinct pairs that can be formed from n players. If we call the number of distinct pairs p, then the number of games to be played is basic arithmetic:

[math]p \times m.[/math]
With me to here?

So, the part of your question that involves more than basic arithmetic is finding the number of distinct pairs from n distinct people. This kind of counting problem comes up so often that it has its own mathematical symbol. Remembering that we have said p is the number of pairs and n is the number of players, we get

[math]p = \dbinom{n}{2} = \dfrac{n!}{2! \times (n - 2)!} = n \times (n -1) \div 2.[/math]
We can eliminate p and combine the formulas to a single formula:

[math]\text {total number of games} = m \times n \times (n - 1) \div 2, \ \text { where }\\ m = \text {number of games each distinct pair is to play, and}\\ n = \text {total number of players.}[/math]
I said in my previous post that having an odd number of players does not affect the number of games. That is true. The only effect of having an odd number of players is that one player must sit out each round.

Suppose we have 5 players: Alan, Betsy, Charles, Debby, and Ed. Our formulas say we have
[imath]5 \times (5 - 1) \div 2 = 5 \times 4 \div 2 = 10[/imath] pairs, namely
Alan-Betsy
Alan-Charles
Alan-Debby
Alan-Ed
Betsy-Charles
Betsy-Debby
Betsy-Ed
Charles-Debby
Charles-Ed
Debby-Ed

But 5 is an odd number so one player must sit out each round. We could leave out Alan for the first round.

Betsy-Charles and Debby-Ed in round 1. Takes care of 2 pairs.

Now leave Betsy out.

Alan-Debby and Charles-Ed in round 2. Takes care of 2 more pairs.

Now take Charles out.

Alan-Ed and Betsy-Debby. Takes care of 2 more pairs.

Now take Debby out.

Alan-Charles and Betsy-Ed. Takes care of 2 more pairs.

Now take Ed out (notice that Ed has played each player in a previous round).

Alan and Betsy have not yet played each other so pair them. And Charles and Debbie have not yet played so pair them.

With 30 players, I might use a computer routine to do the pairings.

Hope this helps.
Click to expand...
Steven G said:
You got as far as wanting to add 1 + 2 + 3 + ... + 29
Here is a formula. 1 + 2 + 3 + ... + n = n(n+1)/2.

So 1 + 2 + 3 + ... + 29 = 29(29+1)/2= 29(30)/2 = 29(15) = 435-----435*10 = 4350
Click to expand...

Yep, I understand it better now. Thank you to all three of you for the explanations you've given me, they were really helpful :)
 
JeffM said:
With 30 players, I might use a computer routine to do the pairings.
Click to expand...
@Zelix224 there's a graphical way to do the pairings. This is especially good if you own a compass, protractor, and split pin :D...

nodes7.gifnodes8.gifnodes9.gifnodes10.gif

Just move the wheel one position clockwise to obtain the next round's matches.

NOTE: there must be an odd number of players on the outer circle (and obviously all of them must be evenly spaced so be careful marking the angles). If you have an even number of players then just put one in the centre.
 
Here's a visualisation of why the [imath]n \times (n -1) \div 2[/imath] part of the formula works. Imagine an n=5 tournament. Can you see that the results for ALL games could be written only on the grey squares below (the players have been assigned numbers 1 to 5)...

results.png

The number of grey squares is half of the total number of squares within the outlined rectangle. Don't count them - just observe the symmetry. The area of a rectangle is given by the height times width. Therefore, for n=5, the number of grey squares will be...
[math]5 \times 4 \div 2[/math]Compare this to the formula
[math]n \times (n-1) \div 2[/math]This would hold for any size of n greater than 1
 
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