Chess tournament winning streaks
$begingroup$
On lichess.org, they use a points system for keeping track of who is winning in a tournament. A win is worth two points, a draw is worth one point, and a loss worth zero points. Once a player has amassed two wins in a row, the following games will be worth double of what they are normally worth until there is a draw or a loss to break the winning streak. Furthermore, a player may "berserk" and cut their time in a game in half in order to add one extra tournament point with a win.
Here's an example: loss=0, win=2, win(berserked)=3, win=4, win(berserked)=5, win=4, drawbreaking the streak=2, draw(berserked)=1, loss=0, draw=1, represented as 0234542101, for a grand total of 22 tournament points. You can also find the official lichess explanation here.
To see if you understand, find a simple way to score 15 points in 4 games as a warm-up:
Win all four games, and berserk any three of them - for example, all but the first one, for a pattern of 2355.
Part One: (easy)
A player played 18 games, won two-thirds of them, and "berserked" half of them.
What is the minimum and maximum number of tournament points he could have received?
Part Two (Hard)
How many ways are there to score seven points in seven games?
Note that permutations of a "way" are not additional "ways": 0202021 is the exact same solution as 0022012 or 2020201, but not the same solution as 3020200.
mathematics combinatorics chess
asked 4 hours ago
Brandon_JBrandon_J
1,864230
$endgroup$
add a comment |
$begingroup$
On lichess.org, they use a points system for keeping track of who is winning in a tournament. A win is worth two points, a draw is worth one point, and a loss worth zero points. Once a player has amassed two wins in a row, the following games will be worth double of what they are normally worth until there is a draw or a loss to break the winning streak. Furthermore, a player may "berserk" and cut their time in a game in half in order to add one extra tournament point with a win.
Here's an example: loss=0, win=2, win(berserked)=3, win=4, win(berserked)=5, win=4, drawbreaking the streak=2, draw(berserked)=1, loss=0, draw=1, represented as 0234542101, for a grand total of 22 tournament points. You can also find the official lichess explanation here.
To see if you understand, find a simple way to score 15 points in 4 games as a warm-up:
Win all four games, and berserk any three of them - for example, all but the first one, for a pattern of 2355.
Part One: (easy)
A player played 18 games, won two-thirds of them, and "berserked" half of them.
What is the minimum and maximum number of tournament points he could have received?
Part Two (Hard)
How many ways are there to score seven points in seven games?
Note that permutations of a "way" are not additional "ways": 0202021 is the exact same solution as 0022012 or 2020201, but not the same solution as 3020200.
mathematics combinatorics chess
asked 4 hours ago
Brandon_JBrandon_J
1,864230
$endgroup$
1
$begingroup$
I'm not sure if the combinatorics tag is correct. If you have any insight into this, please let me know and drop an edit if necessary.
$endgroup$
– Brandon_J
4 hours ago
add a comment |
$begingroup$
On lichess.org, they use a points system for keeping track of who is winning in a tournament. A win is worth two points, a draw is worth one point, and a loss worth zero points. Once a player has amassed two wins in a row, the following games will be worth double of what they are normally worth until there is a draw or a loss to break the winning streak. Furthermore, a player may "berserk" and cut their time in a game in half in order to add one extra tournament point with a win.
Here's an example: loss=0, win=2, win(berserked)=3, win=4, win(berserked)=5, win=4, drawbreaking the streak=2, draw(berserked)=1, loss=0, draw=1, represented as 0234542101, for a grand total of 22 tournament points. You can also find the official lichess explanation here.
To see if you understand, find a simple way to score 15 points in 4 games as a warm-up:
Win all four games, and berserk any three of them - for example, all but the first one, for a pattern of 2355.
Part One: (easy)
A player played 18 games, won two-thirds of them, and "berserked" half of them.
What is the minimum and maximum number of tournament points he could have received?
Part Two (Hard)
How many ways are there to score seven points in seven games?
Note that permutations of a "way" are not additional "ways": 0202021 is the exact same solution as 0022012 or 2020201, but not the same solution as 3020200.
mathematics combinatorics chess
asked 4 hours ago
Brandon_JBrandon_J
1,864230
$endgroup$
On lichess.org, they use a points system for keeping track of who is winning in a tournament. A win is worth two points, a draw is worth one point, and a loss worth zero points. Once a player has amassed two wins in a row, the following games will be worth double of what they are normally worth until there is a draw or a loss to break the winning streak. Furthermore, a player may "berserk" and cut their time in a game in half in order to add one extra tournament point with a win.
Here's an example: loss=0, win=2, win(berserked)=3, win=4, win(berserked)=5, win=4, drawbreaking the streak=2, draw(berserked)=1, loss=0, draw=1, represented as 0234542101, for a grand total of 22 tournament points. You can also find the official lichess explanation here.
To see if you understand, find a simple way to score 15 points in 4 games as a warm-up:
Win all four games, and berserk any three of them - for example, all but the first one, for a pattern of 2355.
Part One: (easy)
A player played 18 games, won two-thirds of them, and "berserked" half of them.
What is the minimum and maximum number of tournament points he could have received?
Part Two (Hard)
How many ways are there to score seven points in seven games?
Note that permutations of a "way" are not additional "ways": 0202021 is the exact same solution as 0022012 or 2020201, but not the same solution as 3020200.
mathematics combinatorics chess
mathematics combinatorics chess
asked 4 hours ago
Brandon_JBrandon_J
1,864230
asked 4 hours ago
Brandon_JBrandon_J
1,864230
edited 3 hours ago
Brandon_J
asked 4 hours ago
Brandon_JBrandon_J
1,864230
asked 4 hours ago
Brandon_JBrandon_J
1,864230
asked 4 hours ago
Brandon_JBrandon_J
1,864230
1,864230
1
$begingroup$
I'm not sure if the combinatorics tag is correct. If you have any insight into this, please let me know and drop an edit if necessary.
$endgroup$
– Brandon_J
4 hours ago
add a comment |
1
$begingroup$
I'm not sure if the combinatorics tag is correct. If you have any insight into this, please let me know and drop an edit if necessary.
$endgroup$
– Brandon_J
4 hours ago
1
1
$begingroup$
I'm not sure if the combinatorics tag is correct. If you have any insight into this, please let me know and drop an edit if necessary.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
I'm not sure if the combinatorics tag is correct. If you have any insight into this, please let me know and drop an edit if necessary.
$endgroup$
– Brandon_J
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Part One.
The minimum is
27
To get the minimum
Make sure there are no winning streaks, and to berserk on as many losses as possible. Only 3 winning games will be berserked.
220220220230230230
The maximum is
60
To get the maximum:
Make sure to only have draws and no losses, berserk only on wins, also have the biggest winning streak you can, and - this is important - have at least one draw after the winning streak to increase that draw score by 1.
111112245555555552
Part Two.
The number of ways to score 7 points in 7 games is
10
Since permutations are not counted, then
we only need to find out the number of different scenarios to get 7 points. There are 15 ways to disassemble 7 into a sum of smaller integers. However, we cannot have a 4 or a 5, as that is only possible on a third consecutive win, and if there are three consecutive wins the total will be 8. So we are only left with 8 ways:
1111111
211111
22111
31111
2221
3211
322
331
The thing to notice here is
3 can only be achieved by berserking a win, and 1 can only be achieved by a draw. However, 2 is either a win or a draw after two consecutive wins. So the sequence 2221 can happen in two ways: either one draw and there wins - not in a row; or two wins in a row followed by a draw and another draw. Same thing with the sequence 322: three wins, or two wins in a row followed by a draw. So that brings the total number of ways to 8+2=10.
answered 3 hours ago
AmorydaiAmorydai
5218
$endgroup$
2
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
add a comment |
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$begingroup$
Part One.
The minimum is
27
To get the minimum
Make sure there are no winning streaks, and to berserk on as many losses as possible. Only 3 winning games will be berserked.
220220220230230230
The maximum is
60
To get the maximum:
Make sure to only have draws and no losses, berserk only on wins, also have the biggest winning streak you can, and - this is important - have at least one draw after the winning streak to increase that draw score by 1.
111112245555555552
Part Two.
The number of ways to score 7 points in 7 games is
10
Since permutations are not counted, then
we only need to find out the number of different scenarios to get 7 points. There are 15 ways to disassemble 7 into a sum of smaller integers. However, we cannot have a 4 or a 5, as that is only possible on a third consecutive win, and if there are three consecutive wins the total will be 8. So we are only left with 8 ways:
1111111
211111
22111
31111
2221
3211
322
331
The thing to notice here is
3 can only be achieved by berserking a win, and 1 can only be achieved by a draw. However, 2 is either a win or a draw after two consecutive wins. So the sequence 2221 can happen in two ways: either one draw and there wins - not in a row; or two wins in a row followed by a draw and another draw. Same thing with the sequence 322: three wins, or two wins in a row followed by a draw. So that brings the total number of ways to 8+2=10.
answered 3 hours ago
AmorydaiAmorydai
5218
$endgroup$
2
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
add a comment |
$begingroup$
Part One.
The minimum is
27
To get the minimum
Make sure there are no winning streaks, and to berserk on as many losses as possible. Only 3 winning games will be berserked.
220220220230230230
The maximum is
60
To get the maximum:
Make sure to only have draws and no losses, berserk only on wins, also have the biggest winning streak you can, and - this is important - have at least one draw after the winning streak to increase that draw score by 1.
111112245555555552
Part Two.
The number of ways to score 7 points in 7 games is
10
Since permutations are not counted, then
we only need to find out the number of different scenarios to get 7 points. There are 15 ways to disassemble 7 into a sum of smaller integers. However, we cannot have a 4 or a 5, as that is only possible on a third consecutive win, and if there are three consecutive wins the total will be 8. So we are only left with 8 ways:
1111111
211111
22111
31111
2221
3211
322
331
The thing to notice here is
3 can only be achieved by berserking a win, and 1 can only be achieved by a draw. However, 2 is either a win or a draw after two consecutive wins. So the sequence 2221 can happen in two ways: either one draw and there wins - not in a row; or two wins in a row followed by a draw and another draw. Same thing with the sequence 322: three wins, or two wins in a row followed by a draw. So that brings the total number of ways to 8+2=10.
answered 3 hours ago
AmorydaiAmorydai
5218
$endgroup$
2
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
add a comment |
$begingroup$
Part One.
The minimum is
27
To get the minimum
Make sure there are no winning streaks, and to berserk on as many losses as possible. Only 3 winning games will be berserked.
220220220230230230
The maximum is
60
To get the maximum:
Make sure to only have draws and no losses, berserk only on wins, also have the biggest winning streak you can, and - this is important - have at least one draw after the winning streak to increase that draw score by 1.
111112245555555552
Part Two.
The number of ways to score 7 points in 7 games is
10
Since permutations are not counted, then
we only need to find out the number of different scenarios to get 7 points. There are 15 ways to disassemble 7 into a sum of smaller integers. However, we cannot have a 4 or a 5, as that is only possible on a third consecutive win, and if there are three consecutive wins the total will be 8. So we are only left with 8 ways:
1111111
211111
22111
31111
2221
3211
322
331
The thing to notice here is
3 can only be achieved by berserking a win, and 1 can only be achieved by a draw. However, 2 is either a win or a draw after two consecutive wins. So the sequence 2221 can happen in two ways: either one draw and there wins - not in a row; or two wins in a row followed by a draw and another draw. Same thing with the sequence 322: three wins, or two wins in a row followed by a draw. So that brings the total number of ways to 8+2=10.
answered 3 hours ago
AmorydaiAmorydai
5218
$endgroup$
Part One.
The minimum is
27
To get the minimum
Make sure there are no winning streaks, and to berserk on as many losses as possible. Only 3 winning games will be berserked.
220220220230230230
The maximum is
60
To get the maximum:
Make sure to only have draws and no losses, berserk only on wins, also have the biggest winning streak you can, and - this is important - have at least one draw after the winning streak to increase that draw score by 1.
111112245555555552
Part Two.
The number of ways to score 7 points in 7 games is
10
Since permutations are not counted, then
we only need to find out the number of different scenarios to get 7 points. There are 15 ways to disassemble 7 into a sum of smaller integers. However, we cannot have a 4 or a 5, as that is only possible on a third consecutive win, and if there are three consecutive wins the total will be 8. So we are only left with 8 ways:
1111111
211111
22111
31111
2221
3211
322
331
The thing to notice here is
3 can only be achieved by berserking a win, and 1 can only be achieved by a draw. However, 2 is either a win or a draw after two consecutive wins. So the sequence 2221 can happen in two ways: either one draw and there wins - not in a row; or two wins in a row followed by a draw and another draw. Same thing with the sequence 322: three wins, or two wins in a row followed by a draw. So that brings the total number of ways to 8+2=10.
answered 3 hours ago
AmorydaiAmorydai
5218
answered 3 hours ago
AmorydaiAmorydai
5218
answered 3 hours ago
AmorydaiAmorydai
5218
answered 3 hours ago
AmorydaiAmorydai
5218
5218
2
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
add a comment |
2
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
2
2
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Very nice! I'll wait and see if anyone else has any input, but you likely have a check-mark coming your way.
$endgroup$
– Brandon_J
3 hours ago
add a comment |
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$begingroup$
I'm not sure if the combinatorics tag is correct. If you have any insight into this, please let me know and drop an edit if necessary.
$endgroup$
– Brandon_J
4 hours ago