Guaranteed Winning Strategy on Horse Betting Odds












3












$begingroup$


Suppose four horses - $A, B, C$, and $D$ - are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $$1$ on $A$, then you receive $$6$ if $A$ wins, or you realize a net gain of $$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?



I found this question from Horse Betting Odds - But Guaranteed Win! and made some revision.



I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.



However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?






probability algebra-precalculus systems-of-equations linear-programming







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Suppose four horses - $A, B, C$, and $D$ - are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $$1$ on $A$, then you receive $$6$ if $A$ wins, or you realize a net gain of $$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?



    I found this question from Horse Betting Odds - But Guaranteed Win! and made some revision.



    I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.



    However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?






    probability algebra-precalculus systems-of-equations linear-programming







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Suppose four horses - $A, B, C$, and $D$ - are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $$1$ on $A$, then you receive $$6$ if $A$ wins, or you realize a net gain of $$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?



      I found this question from Horse Betting Odds - But Guaranteed Win! and made some revision.



      I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.



      However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?






      probability algebra-precalculus systems-of-equations linear-programming







      share|cite|improve this question











      $endgroup$




      Suppose four horses - $A, B, C$, and $D$ - are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $$1$ on $A$, then you receive $$6$ if $A$ wins, or you realize a net gain of $$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?



      I found this question from Horse Betting Odds - But Guaranteed Win! and made some revision.



      I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.



      However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?






      probability algebra-precalculus systems-of-equations linear-programming




      probability algebra-precalculus systems-of-equations linear-programming






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 11:34









      user1551

      73.4k566128




      73.4k566128










      asked Jan 4 at 9:07









      Bratt SwanBratt Swan

      1234




      1234






















          1 Answer
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          active

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          3












          $begingroup$

          Each of the horses has a certain return when they win, A is a $6times$ return, B is $5times$, C $4times$, and D $3times$. In general say you have a set of horses $H_i$ that each have a return $r_i$, and you bet some total amount $B$ with $B_i$ on each horse $H_i$. As long as $B_i ge frac{B}{r_i}$ for all $i$, you will not lose money. That is because one of the $H_i$ will win, and that bet will give you $B_ir_i ge B$, so you end up with at least as much as you started. To maximize your guaranteed winnings, you want to maximize the minimum $B_ir_i$, as $B_ir_i-B$ is your net profit. This will happen when each $B_ir_i$ is equal, which occurs when
          $$B_i = frac{frac{B}{r_i}}{sum frac{1}{r_i}}$$
          and your guaranteed profit is
          $$frac{B}{sum frac{1}{r_i}} - B$$
          This is why in a real horse race $sum frac{1}{r_i}$ will always be greater than $1$ (if for some reason it's not go make a big bet!). In this question it was only $.95$ allowing you to make a nice little profit!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
            $endgroup$
            – Bratt Swan
            Jan 4 at 10:51










          • $begingroup$
            @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
            $endgroup$
            – Erik Parkinson
            Jan 4 at 11:07






          • 1




            $begingroup$
            I.e. this maximises the minimum possible return for a given total bet.
            $endgroup$
            – Henry
            Jan 4 at 23:14











          Your Answer





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          3












          $begingroup$

          Each of the horses has a certain return when they win, A is a $6times$ return, B is $5times$, C $4times$, and D $3times$. In general say you have a set of horses $H_i$ that each have a return $r_i$, and you bet some total amount $B$ with $B_i$ on each horse $H_i$. As long as $B_i ge frac{B}{r_i}$ for all $i$, you will not lose money. That is because one of the $H_i$ will win, and that bet will give you $B_ir_i ge B$, so you end up with at least as much as you started. To maximize your guaranteed winnings, you want to maximize the minimum $B_ir_i$, as $B_ir_i-B$ is your net profit. This will happen when each $B_ir_i$ is equal, which occurs when
          $$B_i = frac{frac{B}{r_i}}{sum frac{1}{r_i}}$$
          and your guaranteed profit is
          $$frac{B}{sum frac{1}{r_i}} - B$$
          This is why in a real horse race $sum frac{1}{r_i}$ will always be greater than $1$ (if for some reason it's not go make a big bet!). In this question it was only $.95$ allowing you to make a nice little profit!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
            $endgroup$
            – Bratt Swan
            Jan 4 at 10:51










          • $begingroup$
            @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
            $endgroup$
            – Erik Parkinson
            Jan 4 at 11:07






          • 1




            $begingroup$
            I.e. this maximises the minimum possible return for a given total bet.
            $endgroup$
            – Henry
            Jan 4 at 23:14
















          3












          $begingroup$

          Each of the horses has a certain return when they win, A is a $6times$ return, B is $5times$, C $4times$, and D $3times$. In general say you have a set of horses $H_i$ that each have a return $r_i$, and you bet some total amount $B$ with $B_i$ on each horse $H_i$. As long as $B_i ge frac{B}{r_i}$ for all $i$, you will not lose money. That is because one of the $H_i$ will win, and that bet will give you $B_ir_i ge B$, so you end up with at least as much as you started. To maximize your guaranteed winnings, you want to maximize the minimum $B_ir_i$, as $B_ir_i-B$ is your net profit. This will happen when each $B_ir_i$ is equal, which occurs when
          $$B_i = frac{frac{B}{r_i}}{sum frac{1}{r_i}}$$
          and your guaranteed profit is
          $$frac{B}{sum frac{1}{r_i}} - B$$
          This is why in a real horse race $sum frac{1}{r_i}$ will always be greater than $1$ (if for some reason it's not go make a big bet!). In this question it was only $.95$ allowing you to make a nice little profit!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
            $endgroup$
            – Bratt Swan
            Jan 4 at 10:51










          • $begingroup$
            @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
            $endgroup$
            – Erik Parkinson
            Jan 4 at 11:07






          • 1




            $begingroup$
            I.e. this maximises the minimum possible return for a given total bet.
            $endgroup$
            – Henry
            Jan 4 at 23:14














          3












          3








          3





          $begingroup$

          Each of the horses has a certain return when they win, A is a $6times$ return, B is $5times$, C $4times$, and D $3times$. In general say you have a set of horses $H_i$ that each have a return $r_i$, and you bet some total amount $B$ with $B_i$ on each horse $H_i$. As long as $B_i ge frac{B}{r_i}$ for all $i$, you will not lose money. That is because one of the $H_i$ will win, and that bet will give you $B_ir_i ge B$, so you end up with at least as much as you started. To maximize your guaranteed winnings, you want to maximize the minimum $B_ir_i$, as $B_ir_i-B$ is your net profit. This will happen when each $B_ir_i$ is equal, which occurs when
          $$B_i = frac{frac{B}{r_i}}{sum frac{1}{r_i}}$$
          and your guaranteed profit is
          $$frac{B}{sum frac{1}{r_i}} - B$$
          This is why in a real horse race $sum frac{1}{r_i}$ will always be greater than $1$ (if for some reason it's not go make a big bet!). In this question it was only $.95$ allowing you to make a nice little profit!






          share|cite|improve this answer









          $endgroup$



          Each of the horses has a certain return when they win, A is a $6times$ return, B is $5times$, C $4times$, and D $3times$. In general say you have a set of horses $H_i$ that each have a return $r_i$, and you bet some total amount $B$ with $B_i$ on each horse $H_i$. As long as $B_i ge frac{B}{r_i}$ for all $i$, you will not lose money. That is because one of the $H_i$ will win, and that bet will give you $B_ir_i ge B$, so you end up with at least as much as you started. To maximize your guaranteed winnings, you want to maximize the minimum $B_ir_i$, as $B_ir_i-B$ is your net profit. This will happen when each $B_ir_i$ is equal, which occurs when
          $$B_i = frac{frac{B}{r_i}}{sum frac{1}{r_i}}$$
          and your guaranteed profit is
          $$frac{B}{sum frac{1}{r_i}} - B$$
          This is why in a real horse race $sum frac{1}{r_i}$ will always be greater than $1$ (if for some reason it's not go make a big bet!). In this question it was only $.95$ allowing you to make a nice little profit!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 9:42









          Erik ParkinsonErik Parkinson

          1,16519




          1,16519












          • $begingroup$
            Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
            $endgroup$
            – Bratt Swan
            Jan 4 at 10:51










          • $begingroup$
            @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
            $endgroup$
            – Erik Parkinson
            Jan 4 at 11:07






          • 1




            $begingroup$
            I.e. this maximises the minimum possible return for a given total bet.
            $endgroup$
            – Henry
            Jan 4 at 23:14


















          • $begingroup$
            Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
            $endgroup$
            – Bratt Swan
            Jan 4 at 10:51










          • $begingroup$
            @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
            $endgroup$
            – Erik Parkinson
            Jan 4 at 11:07






          • 1




            $begingroup$
            I.e. this maximises the minimum possible return for a given total bet.
            $endgroup$
            – Henry
            Jan 4 at 23:14
















          $begingroup$
          Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
          $endgroup$
          – Bratt Swan
          Jan 4 at 10:51




          $begingroup$
          Thanks for your explanation. One more question is that why when each $B_ir_i$ is equal, then the profit is maximized?
          $endgroup$
          – Bratt Swan
          Jan 4 at 10:51












          $begingroup$
          @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
          $endgroup$
          – Erik Parkinson
          Jan 4 at 11:07




          $begingroup$
          @BrattSwan If they are not equal, say $B_ir_i >B_jr_j$, then you make more money if $H_i$ wins and less if $H_j$ does. So to maximize your guaranteed winnings no matter what the outcome it, you would want to move some of that $B_i$ money into $B_j$.
          $endgroup$
          – Erik Parkinson
          Jan 4 at 11:07




          1




          1




          $begingroup$
          I.e. this maximises the minimum possible return for a given total bet.
          $endgroup$
          – Henry
          Jan 4 at 23:14




          $begingroup$
          I.e. this maximises the minimum possible return for a given total bet.
          $endgroup$
          – Henry
          Jan 4 at 23:14


















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