Mixing
Divide a number in unequal increasing parts according to a dynamic factor
$begingroup$
There is a game where you need to gather money from people participating to it and when you have the total you need to divide it in order to put a "reasonable amount" for each of the n prizes they will be able to win.
E.g. in general we usually make the calcution approximately, so in case you have a total of 8$ (t = 8) and 5 prizes (n = 5) we could divide the total into:
- prize 1:
0.8$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
1.9$
- prize 5:
2.6$
As you can see it would be important to have prize n to be "more important" than prize 1, in general.
I would like to find a rule to obtain those prize values dynamically according to:
t = total amount of moneyn = number of prizesy = a variable factor to provide in order to abtain different combinations of prize values, in a way that a lower 'y' value means being more "fair" in giving values to prizes, on the contrary a higher 'y' value means increasing the difference between prize 1 and prize 'n'
E.g. in case I want to be more "fair" a possibility would be:
- prize 1:
1.2$
- prize 2:
1.4$
- prize 3:
1.6$
- prize 4:
1.8$
- prize 5:
2.0$
or
- prize 1:
1.1$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
2.0$
- prize 5:
2.2$
but I really have no idea about what formula or algorithm I could use to make that calculation by simply having as input (y, n and t) and how to be able to provide a y factor to use in order to vary the resulting prize values.
Any idea about it?
Thanks
arithmetic
asked Dec 31 '18 at 13:17
FrankFrank
1033
$endgroup$
add a comment |
$begingroup$
There is a game where you need to gather money from people participating to it and when you have the total you need to divide it in order to put a "reasonable amount" for each of the n prizes they will be able to win.
E.g. in general we usually make the calcution approximately, so in case you have a total of 8$ (t = 8) and 5 prizes (n = 5) we could divide the total into:
- prize 1:
0.8$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
1.9$
- prize 5:
2.6$
As you can see it would be important to have prize n to be "more important" than prize 1, in general.
I would like to find a rule to obtain those prize values dynamically according to:
t = total amount of moneyn = number of prizesy = a variable factor to provide in order to abtain different combinations of prize values, in a way that a lower 'y' value means being more "fair" in giving values to prizes, on the contrary a higher 'y' value means increasing the difference between prize 1 and prize 'n'
E.g. in case I want to be more "fair" a possibility would be:
- prize 1:
1.2$
- prize 2:
1.4$
- prize 3:
1.6$
- prize 4:
1.8$
- prize 5:
2.0$
or
- prize 1:
1.1$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
2.0$
- prize 5:
2.2$
but I really have no idea about what formula or algorithm I could use to make that calculation by simply having as input (y, n and t) and how to be able to provide a y factor to use in order to vary the resulting prize values.
Any idea about it?
Thanks
arithmetic
asked Dec 31 '18 at 13:17
FrankFrank
1033
$endgroup$
add a comment |
$begingroup$
There is a game where you need to gather money from people participating to it and when you have the total you need to divide it in order to put a "reasonable amount" for each of the n prizes they will be able to win.
E.g. in general we usually make the calcution approximately, so in case you have a total of 8$ (t = 8) and 5 prizes (n = 5) we could divide the total into:
- prize 1:
0.8$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
1.9$
- prize 5:
2.6$
As you can see it would be important to have prize n to be "more important" than prize 1, in general.
I would like to find a rule to obtain those prize values dynamically according to:
t = total amount of moneyn = number of prizesy = a variable factor to provide in order to abtain different combinations of prize values, in a way that a lower 'y' value means being more "fair" in giving values to prizes, on the contrary a higher 'y' value means increasing the difference between prize 1 and prize 'n'
E.g. in case I want to be more "fair" a possibility would be:
- prize 1:
1.2$
- prize 2:
1.4$
- prize 3:
1.6$
- prize 4:
1.8$
- prize 5:
2.0$
or
- prize 1:
1.1$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
2.0$
- prize 5:
2.2$
but I really have no idea about what formula or algorithm I could use to make that calculation by simply having as input (y, n and t) and how to be able to provide a y factor to use in order to vary the resulting prize values.
Any idea about it?
Thanks
arithmetic
asked Dec 31 '18 at 13:17
FrankFrank
1033
$endgroup$
There is a game where you need to gather money from people participating to it and when you have the total you need to divide it in order to put a "reasonable amount" for each of the n prizes they will be able to win.
E.g. in general we usually make the calcution approximately, so in case you have a total of 8$ (t = 8) and 5 prizes (n = 5) we could divide the total into:
- prize 1:
0.8$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
1.9$
- prize 5:
2.6$
As you can see it would be important to have prize n to be "more important" than prize 1, in general.
I would like to find a rule to obtain those prize values dynamically according to:
t = total amount of moneyn = number of prizesy = a variable factor to provide in order to abtain different combinations of prize values, in a way that a lower 'y' value means being more "fair" in giving values to prizes, on the contrary a higher 'y' value means increasing the difference between prize 1 and prize 'n'
E.g. in case I want to be more "fair" a possibility would be:
- prize 1:
1.2$
- prize 2:
1.4$
- prize 3:
1.6$
- prize 4:
1.8$
- prize 5:
2.0$
or
- prize 1:
1.1$
- prize 2:
1.2$
- prize 3:
1.5$
- prize 4:
2.0$
- prize 5:
2.2$
but I really have no idea about what formula or algorithm I could use to make that calculation by simply having as input (y, n and t) and how to be able to provide a y factor to use in order to vary the resulting prize values.
Any idea about it?
Thanks
arithmetic
arithmetic
asked Dec 31 '18 at 13:17
FrankFrank
1033
asked Dec 31 '18 at 13:17
FrankFrank
1033
edited Dec 31 '18 at 16:39
Frank
asked Dec 31 '18 at 13:17
FrankFrank
1033
asked Dec 31 '18 at 13:17
FrankFrank
1033
asked Dec 31 '18 at 13:17
FrankFrank
1033
1033
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's a possibility:
Set the $k$th prize to be
$$
C r^{k-1}
$$
for some value of $C$,
where $r = 1 + y$. For $y = 0$, you get all prizes the same; for $y = 1$, each prize is double the previous one. For $y = 0.1$, you get something like the values you had.
Of course, to complete this, I have to give a value for $C$. But we know that the sum of all the prizes has to be
$$
C(1 + r + ldots + r^{n-1}) = C frac{1-r^n}{1-r}
$$
and we want this sum to be the total $t$. That means that
$$
C = t frac{1-r}{1-r^n}.
$$
The formula for the $k$th prize, where $k$ goes from $1$ to $n$ is thus
begin{align}
p_k &= t, r^{k-1} frac{1-r}{1-r^n}\
&= t, (1+y)^{k-1} frac{y}{(1+y)^n - 1}
end{align}
If I were writing a program, I'd compute $r = 1+y$ first, and then use the first form rather than the second, but that's a matter of taste.
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
$endgroup$
$begingroup$
thank you! I tried to compute it (in javascript):var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money);but the result is:[0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667]and total money is3.666666666666667. What if I want to avoidoas value for ak?
$endgroup$
– Frank
Dec 31 '18 at 14:27
1
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
$endgroup$
– Frank
Dec 31 '18 at 14:41
$begingroup$
In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
1
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
$endgroup$
– Ross Millikan
Dec 31 '18 at 17:03
|
show 4 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a possibility:
Set the $k$th prize to be
$$
C r^{k-1}
$$
for some value of $C$,
where $r = 1 + y$. For $y = 0$, you get all prizes the same; for $y = 1$, each prize is double the previous one. For $y = 0.1$, you get something like the values you had.
Of course, to complete this, I have to give a value for $C$. But we know that the sum of all the prizes has to be
$$
C(1 + r + ldots + r^{n-1}) = C frac{1-r^n}{1-r}
$$
and we want this sum to be the total $t$. That means that
$$
C = t frac{1-r}{1-r^n}.
$$
The formula for the $k$th prize, where $k$ goes from $1$ to $n$ is thus
begin{align}
p_k &= t, r^{k-1} frac{1-r}{1-r^n}\
&= t, (1+y)^{k-1} frac{y}{(1+y)^n - 1}
end{align}
If I were writing a program, I'd compute $r = 1+y$ first, and then use the first form rather than the second, but that's a matter of taste.
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
$endgroup$
$begingroup$
thank you! I tried to compute it (in javascript):var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money);but the result is:[0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667]and total money is3.666666666666667. What if I want to avoidoas value for ak?
$endgroup$
– Frank
Dec 31 '18 at 14:27
1
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
$endgroup$
– Frank
Dec 31 '18 at 14:41
$begingroup$
In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
1
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
$endgroup$
– Ross Millikan
Dec 31 '18 at 17:03
|
show 4 more comments
$begingroup$
Here's a possibility:
Set the $k$th prize to be
$$
C r^{k-1}
$$
for some value of $C$,
where $r = 1 + y$. For $y = 0$, you get all prizes the same; for $y = 1$, each prize is double the previous one. For $y = 0.1$, you get something like the values you had.
Of course, to complete this, I have to give a value for $C$. But we know that the sum of all the prizes has to be
$$
C(1 + r + ldots + r^{n-1}) = C frac{1-r^n}{1-r}
$$
and we want this sum to be the total $t$. That means that
$$
C = t frac{1-r}{1-r^n}.
$$
The formula for the $k$th prize, where $k$ goes from $1$ to $n$ is thus
begin{align}
p_k &= t, r^{k-1} frac{1-r}{1-r^n}\
&= t, (1+y)^{k-1} frac{y}{(1+y)^n - 1}
end{align}
If I were writing a program, I'd compute $r = 1+y$ first, and then use the first form rather than the second, but that's a matter of taste.
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
$endgroup$
$begingroup$
thank you! I tried to compute it (in javascript):var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money);but the result is:[0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667]and total money is3.666666666666667. What if I want to avoidoas value for ak?
$endgroup$
– Frank
Dec 31 '18 at 14:27
1
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
$endgroup$
– Frank
Dec 31 '18 at 14:41
$begingroup$
In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
1
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
$endgroup$
– Ross Millikan
Dec 31 '18 at 17:03
|
show 4 more comments
$begingroup$
Here's a possibility:
Set the $k$th prize to be
$$
C r^{k-1}
$$
for some value of $C$,
where $r = 1 + y$. For $y = 0$, you get all prizes the same; for $y = 1$, each prize is double the previous one. For $y = 0.1$, you get something like the values you had.
Of course, to complete this, I have to give a value for $C$. But we know that the sum of all the prizes has to be
$$
C(1 + r + ldots + r^{n-1}) = C frac{1-r^n}{1-r}
$$
and we want this sum to be the total $t$. That means that
$$
C = t frac{1-r}{1-r^n}.
$$
The formula for the $k$th prize, where $k$ goes from $1$ to $n$ is thus
begin{align}
p_k &= t, r^{k-1} frac{1-r}{1-r^n}\
&= t, (1+y)^{k-1} frac{y}{(1+y)^n - 1}
end{align}
If I were writing a program, I'd compute $r = 1+y$ first, and then use the first form rather than the second, but that's a matter of taste.
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
$endgroup$
Here's a possibility:
Set the $k$th prize to be
$$
C r^{k-1}
$$
for some value of $C$,
where $r = 1 + y$. For $y = 0$, you get all prizes the same; for $y = 1$, each prize is double the previous one. For $y = 0.1$, you get something like the values you had.
Of course, to complete this, I have to give a value for $C$. But we know that the sum of all the prizes has to be
$$
C(1 + r + ldots + r^{n-1}) = C frac{1-r^n}{1-r}
$$
and we want this sum to be the total $t$. That means that
$$
C = t frac{1-r}{1-r^n}.
$$
The formula for the $k$th prize, where $k$ goes from $1$ to $n$ is thus
begin{align}
p_k &= t, r^{k-1} frac{1-r}{1-r^n}\
&= t, (1+y)^{k-1} frac{y}{(1+y)^n - 1}
end{align}
If I were writing a program, I'd compute $r = 1+y$ first, and then use the first form rather than the second, but that's a matter of taste.
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
edited Jan 1 at 20:29
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
answered Dec 31 '18 at 13:53
John HughesJohn Hughes
62.7k24090
62.7k24090
$begingroup$
thank you! I tried to compute it (in javascript):var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money);but the result is:[0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667]and total money is3.666666666666667. What if I want to avoidoas value for ak?
$endgroup$
– Frank
Dec 31 '18 at 14:27
1
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
$endgroup$
– Frank
Dec 31 '18 at 14:41
$begingroup$
In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
1
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
$endgroup$
– Ross Millikan
Dec 31 '18 at 17:03
|
show 4 more comments
$begingroup$
thank you! I tried to compute it (in javascript):var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money);but the result is:[0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667]and total money is3.666666666666667. What if I want to avoidoas value for ak?
$endgroup$
– Frank
Dec 31 '18 at 14:27
1
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
$endgroup$
– Frank
Dec 31 '18 at 14:41
$begingroup$
In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
1
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
$endgroup$
– Ross Millikan
Dec 31 '18 at 17:03
$begingroup$
thank you! I tried to compute it (in javascript):
var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money); but the result is: [0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667] and total money is 3.666666666666667. What if I want to avoid o as value for a k?$endgroup$
– Frank
Dec 31 '18 at 14:27
$begingroup$
thank you! I tried to compute it (in javascript):
var prizes = 5; var factor = 0.2; var prizeValues = ; var money = 0; for (var k = 1; k <= prizes; k++) { var calc = (prizes * ((1 + factor) ^ (k - 1))) * (factor / (((1 + factor) ^ prizes) - 1)); prizeValues.push(calc); money += calc; } console.warn(prizeValues); console.warn(money); but the result is: [0.3333333333333333, 0, 1, 0.6666666666666666, 1.6666666666666667] and total money is 3.666666666666667. What if I want to avoid o as value for a k?$endgroup$
– Frank
Dec 31 '18 at 14:27
1
1
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
You seem to be assuming that the total of the prizes is the same as the number of prizes because you use $n$ for both. I don't think that is intended. It would be good to use a different variable for one of them.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:35
$begingroup$
@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
$endgroup$
– Frank
Dec 31 '18 at 14:41
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@RossMillikan sorry I did not understand your point. What is my mistake? Thank you
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– Frank
Dec 31 '18 at 14:41
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In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
$begingroup$
In the first display equation you have $n$ terms in the sum on the left, which represents $n$ prizes. Then you say you want the total to be $n$. In the original post there were $5$ prizes totaling $8$, so one of the two uses of $n$ should be changed to a different variable. OP uses $t$ for the total of the prizes. This just scales the equation by $frac tn$, not a big deal.
$endgroup$
– Ross Millikan
Dec 31 '18 at 14:57
1
1
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
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– Ross Millikan
Dec 31 '18 at 17:03
$begingroup$
Yes, it becomes $C=tfrac {1-r}{1-r^n}$ with $n=5,t=8$
$endgroup$
– Ross Millikan
Dec 31 '18 at 17:03
|
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