Guaranteed Winning Strategy on Horse Betting Odds
$begingroup$
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 $$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?
I found this question from Horse Betting Odds - But Guaranteed Win! and made some revision.
I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.
However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?
probability algebra-precalculus systems-of-equations linear-programming
edited Jan 4 at 11:34
user1551
73.4k566128
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
$endgroup$
add a comment |
$begingroup$
Suppose four horses - $A, B, C$, and $D$ - are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $$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
edited Jan 4 at 11:34
user1551
73.4k566128
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
$endgroup$
add a comment |
$begingroup$
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 $$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you can always make money no matter which horse wins?
I found this question from Horse Betting Odds - But Guaranteed Win! and made some revision.
I think if I bet my money on A, B, C, and D under the following scenario, such as A:B:C:D = 20:15:12:10. I can always make money. These numbers come from the ratios 1/3 : 1/4 : 1/5 : 1/6 multiplied by 60， where 60 is the lowest common multiple for 3,4,5, and 6.
However, I wonder if there is a formal mathematical way to derive this solution. Is this question related to solving the linear inequality system or the probability?
probability algebra-precalculus systems-of-equations linear-programming
edited Jan 4 at 11:34
user1551
73.4k566128
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
$endgroup$
Suppose four horses - $A, B, C$, and $D$ - are entered in a race and the odds on them, respectively, are $5$ to $1$, $4$ to $1$, $3$ to $1$, and $2$ to $1.$ If you bet $$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
edited Jan 4 at 11:34
user1551
73.4k566128
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
edited Jan 4 at 11:34
user1551
73.4k566128
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
edited Jan 4 at 11:34
user1551
73.4k566128
edited Jan 4 at 11:34
user1551
73.4k566128
edited Jan 4 at 11:34
user1551
73.4k566128
73.4k566128
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
asked Jan 4 at 9:07
Bratt SwanBratt Swan
1234
1234
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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!
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
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!
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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!
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
$endgroup$
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!
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
answered Jan 4 at 9:42
Erik ParkinsonErik Parkinson
1,16519
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
add a comment |
$begingroup$
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
add a comment |
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