Divide a number in unequal increasing parts according to a dynamic factor












0












$begingroup$


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 money

  • n = number of prizes

  • y = 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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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 money

    • n = number of prizes

    • y = 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







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      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 money

      • n = number of prizes

      • y = 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







      share|cite|improve this question











      $endgroup$




      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 money

      • n = number of prizes

      • y = 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






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      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 31 '18 at 16:39







      Frank

















      asked Dec 31 '18 at 13:17









      FrankFrank

      1033




      1033






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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 is 3.666666666666667. What if I want to avoid o as value for a k?
            $endgroup$
            – 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













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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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 is 3.666666666666667. What if I want to avoid o as value for a k?
            $endgroup$
            – 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


















          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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 is 3.666666666666667. What if I want to avoid o as value for a k?
            $endgroup$
            – 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
















          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 20:29

























          answered Dec 31 '18 at 13:53









          John HughesJohn Hughes

          62.9k24090




          62.9k24090












          • $begingroup$
            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




            $begingroup$
            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








          • 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




          $begingroup$
          @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




          $begingroup$
          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 $n=5,t=8$
          $endgroup$
          – 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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