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Combinations and permutations - more math than chess
Combinations and permutations - more math than chess
Two chess players can be paired only 1 way: (1-2)
Four players can be paired 3 ways: (1-2 and 3-4) (1-3 and 2-4) (1-4 and 2-3)
Six players can be paired 15 ways: I'll spare my typing finger, and not list them; trust me.
Eight players? How many ways can eight players be paired? Does anyone know the formula for the possible combinations for n players, where n is even? How about where n is uneven?
Four players can be paired 3 ways: (1-2 and 3-4) (1-3 and 2-4) (1-4 and 2-3)
Six players can be paired 15 ways: I'll spare my typing finger, and not list them; trust me.
Eight players? How many ways can eight players be paired? Does anyone know the formula for the possible combinations for n players, where n is even? How about where n is uneven?
Hi John. Since colour is an important part of the pairing process (2-1 and 1-2), why aren't you interested in in permutations? For combinations, I think the number you're looking for is 28 (but I forget the formula). :)
Last edited by Peter McKillop; Tuesday, 26th January, 2016, 08:30 PM.
"We hang the petty thieves and appoint the great ones to public office." - Aesop
"Only the dead have seen the end of war." - Plato
"If once a man indulges himself in murder, very soon he comes to think little of robbing; and from robbing he comes next to drinking and Sabbath-breaking, and from that to incivility and procrastination." - Thomas De Quincey
Four players can be paired 3 ways: (1-2 and 3-4) (1-3 and 2-4) (1-4 and 2-3)
Six players can be paired 15 ways: I'll spare my typing finger, and not list them; trust me.
Eight players? How many ways can eight players be paired? Does anyone know the formula for the possible combinations for n players, where n is even? How about where n is uneven?
So you would have the same sorts of pairings with 13, 14, etc. all the way to 18 (seven different groups); 7 x 15 = 105.
Since I had forgotten the formula for the number of combinations of eight things combined two at a time (it's been half a century since grade 11 math), here's how I came up with 28:
Player 1 can be paired with each of 7 other players.
Player 2 can be paired with each of 6 other players (the pairing of 1 and 2 was counted above).
Player 3 .... 5 other players.
.... and so on.
i.e. 7 + 6 +5 +4 +3 + 2 + 1 = 28
"We hang the petty thieves and appoint the great ones to public office." - Aesop
"Only the dead have seen the end of war." - Plato
"If once a man indulges himself in murder, very soon he comes to think little of robbing; and from robbing he comes next to drinking and Sabbath-breaking, and from that to incivility and procrastination." - Thomas De Quincey
Re: Combinations and permutations - more math than chess
John, n would always be even because you have in reserve, for uneven situations, the player called 'bye'.
"We hang the petty thieves and appoint the great ones to public office." - Aesop
"Only the dead have seen the end of war." - Plato
"If once a man indulges himself in murder, very soon he comes to think little of robbing; and from robbing he comes next to drinking and Sabbath-breaking, and from that to incivility and procrastination." - Thomas De Quincey
Since I had forgotten the formula for the number of combinations of eight things combined two at a time (it's been half a century since grade 11 math), here's how I came up with 28:
Player 1 can be paired with each of 7 other players.
Player 2 can be paired with each of 6 other players (the pairing of 1 and 2 was counted above).
Player 3 .... 5 other players.
.... and so on.
i.e. 7 + 6 +5 +4 +3 + 2 + 1 = 28
No mathematician am I, but, I always thought the formula would be (# of players) x (# of players less 1) divided by 2.
So, 8 players would be 8 x 7 \ 2 = 28
A 15 player event would be 15 x 14 \ 2 = 105.
This is assuming you meant a round robin of x number of players playing each participant once to obtain the number of games that would be played.
I might add that I don't see any difference between say 7 or 8 players. Wouldn't it be the same number of possible pairings (i.e. 105)?
Thanks Tom, I came up with ths same number, but I couldn't prove it.
Before I retired from chess organising, I developed a pairing system for kids which does not require that all round start at the same time. In fact, we don't even have rounds. We get 1400+ kids at our big event each year. We don't assign colours, the kids pick colours (chosing a pawn) themselves. I'm writing a pairing program to computerise this system, by way of keeping my brain active. It's an adventure.
No mathematician am I, but, I always thought the formula would be (# of players) x (# of players less 1) divided by 2.
So, 8 players would be 8 x 7 \ 2 = 28
A 15 player event would be 15 x 14 \ 2 = 105.
This is assuming you meant a round robin of x number of players playing each participant once to obtain the number of games that would be played.
Yes, you're right about the formula for n things combined 2 at a time.
"We hang the petty thieves and appoint the great ones to public office." - Aesop
"Only the dead have seen the end of war." - Plato
"If once a man indulges himself in murder, very soon he comes to think little of robbing; and from robbing he comes next to drinking and Sabbath-breaking, and from that to incivility and procrastination." - Thomas De Quincey
Re: Combinations and permutations - more math than chess
n is any even number of players (just add a bye for odd numbers to make them all even). Then P is the number of possible pairing arrangements for any individual round (assuming your players then randomize for white & black).
P(n) = 1 * 3 * 5 * ... * (n-1)
This gives us:
P(2) = 1
P(4) = 1*3 = 3
P(6) = 1*3*5 = 15
P(8) = 1*3*5*7 = 105
P(10) = 1*3*5*7*9 = 945
etc.
Tom O'Donnell's post does a good job of explaining the reasoning.
This is different than the number of ways to select a group of two people out of a group of n, which as mentioned already is n(n-1)/2, and gives you the total number of games that would be played in a single round robin tournament.
n is any even number of players (just add a bye for odd numbers to make them all even). Then P is the number of possible pairing arrangements for any individual round (assuming your players then randomize for white & black).
P(n) = 1 * 3 * 5 * ... * (n-1)
This gives us:
P(2) = 1
P(4) = 1*3 = 3
P(6) = 1*3*5 = 15
P(8) = 1*3*5*7 = 105
P(10) = 1*3*5*7*9 = 945
etc.
Tom O'Donnell's post does a good job of explaining the reasoning.
This is different than the number of ways to select a group of two people out of a group of n, which as mentioned already is n(n-1)/2, and gives you the total number of games that would be played in a single round robin tournament.
Thanks, Andrew. I misunderstood what John was looking for.
"We hang the petty thieves and appoint the great ones to public office." - Aesop
"Only the dead have seen the end of war." - Plato
"If once a man indulges himself in murder, very soon he comes to think little of robbing; and from robbing he comes next to drinking and Sabbath-breaking, and from that to incivility and procrastination." - Thomas De Quincey
Before I retired from chess organising, I developed a pairing system for kids which does not require that all round start at the same time. In fact, we don't even have rounds. We get 1400+ kids at our big event each year. We don't assign colours, the kids pick colours (chosing a pawn) themselves. I'm writing a pairing program to computerise this system, by way of keeping my brain active. It's an adventure.
Hi John, I would be interested in such a program. Thanks and regards, Aris.
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