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Gambling to pay off debt?
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Someone told me something interesting today. They said they were going to take their bonus check from work, to the casino because they have "better odds" of paying off more debt then if they would just apply it to the principle balance of their outstanding credit card. Talking some more with the person they indicated they had about $20,000 in debt at about 20% interest.
Lets assume that their bonus check is $1000, they pay 500 a month towards their credit card, and they play a game with a 'better' chance of winning at the casino (craps, baccarat, etc) and have a 0.48 chance to win on the dollar.
We know their expected return is -0.04 per game, but since their owed interest and balance is so high; how would one go about calculating if their statement is true?
probability statistics game-theory gambling
asked May 5 '15 at 13:06
aso118
235
|
show 3 more comments
up vote
0
down vote
favorite
Someone told me something interesting today. They said they were going to take their bonus check from work, to the casino because they have "better odds" of paying off more debt then if they would just apply it to the principle balance of their outstanding credit card. Talking some more with the person they indicated they had about $20,000 in debt at about 20% interest.
Lets assume that their bonus check is $1000, they pay 500 a month towards their credit card, and they play a game with a 'better' chance of winning at the casino (craps, baccarat, etc) and have a 0.48 chance to win on the dollar.
We know their expected return is -0.04 per game, but since their owed interest and balance is so high; how would one go about calculating if their statement is true?
probability statistics game-theory gambling
asked May 5 '15 at 13:06
aso118
235
*than${{{{}}}}$
– Git Gud
May 5 '15 at 13:07
2
Casinoes do their business on the fact that that statement is false. It's how they earn money.
– Arthur
May 5 '15 at 13:10
1
Is there a math-related question here?
– hardmath
May 5 '15 at 13:22
2
What might be true is that they have a positive probability of being able to completely pay off their debt, however having a positive probability and having it be likely are two very different things. As Arthur mentioned, the common phrase to hear is "The house always wins." By that we mean at the end of the day players have lost more money in total than they had won. Sure, there might be some lucky guy who made a fortune, but with how much everyone else lost it more than makes up for it. As such, you should expect to lose money and it is no substitute for proper investment.
– JMoravitz
May 5 '15 at 13:22
Generally, the best money-making strategy is to put everything on a single red/black roulette spin. This has a negative expected value, of course. But if you consider a 47% chance of doubling the bonus (at which point one must walk away) and a 53% chance of the bonus evaporating, then... I guess it's a "good" idea.
– pjs36
May 5 '15 at 13:25
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Someone told me something interesting today. They said they were going to take their bonus check from work, to the casino because they have "better odds" of paying off more debt then if they would just apply it to the principle balance of their outstanding credit card. Talking some more with the person they indicated they had about $20,000 in debt at about 20% interest.
Lets assume that their bonus check is $1000, they pay 500 a month towards their credit card, and they play a game with a 'better' chance of winning at the casino (craps, baccarat, etc) and have a 0.48 chance to win on the dollar.
We know their expected return is -0.04 per game, but since their owed interest and balance is so high; how would one go about calculating if their statement is true?
probability statistics game-theory gambling
asked May 5 '15 at 13:06
aso118
235
Someone told me something interesting today. They said they were going to take their bonus check from work, to the casino because they have "better odds" of paying off more debt then if they would just apply it to the principle balance of their outstanding credit card. Talking some more with the person they indicated they had about $20,000 in debt at about 20% interest.
Lets assume that their bonus check is $1000, they pay 500 a month towards their credit card, and they play a game with a 'better' chance of winning at the casino (craps, baccarat, etc) and have a 0.48 chance to win on the dollar.
We know their expected return is -0.04 per game, but since their owed interest and balance is so high; how would one go about calculating if their statement is true?
probability statistics game-theory gambling
probability statistics game-theory gambling
asked May 5 '15 at 13:06
aso118
235
asked May 5 '15 at 13:06
aso118
235
edited May 5 '15 at 13:40
asked May 5 '15 at 13:06
aso118
235
asked May 5 '15 at 13:06
aso118
235
asked May 5 '15 at 13:06
aso118
235
235
*than${{{{}}}}$
– Git Gud
May 5 '15 at 13:07
2
Casinoes do their business on the fact that that statement is false. It's how they earn money.
– Arthur
May 5 '15 at 13:10
1
Is there a math-related question here?
– hardmath
May 5 '15 at 13:22
2
What might be true is that they have a positive probability of being able to completely pay off their debt, however having a positive probability and having it be likely are two very different things. As Arthur mentioned, the common phrase to hear is "The house always wins." By that we mean at the end of the day players have lost more money in total than they had won. Sure, there might be some lucky guy who made a fortune, but with how much everyone else lost it more than makes up for it. As such, you should expect to lose money and it is no substitute for proper investment.
– JMoravitz
May 5 '15 at 13:22
Generally, the best money-making strategy is to put everything on a single red/black roulette spin. This has a negative expected value, of course. But if you consider a 47% chance of doubling the bonus (at which point one must walk away) and a 53% chance of the bonus evaporating, then... I guess it's a "good" idea.
– pjs36
May 5 '15 at 13:25
|
show 3 more comments
*than${{{{}}}}$
– Git Gud
May 5 '15 at 13:07
2
Casinoes do their business on the fact that that statement is false. It's how they earn money.
– Arthur
May 5 '15 at 13:10
1
Is there a math-related question here?
– hardmath
May 5 '15 at 13:22
2
What might be true is that they have a positive probability of being able to completely pay off their debt, however having a positive probability and having it be likely are two very different things. As Arthur mentioned, the common phrase to hear is "The house always wins." By that we mean at the end of the day players have lost more money in total than they had won. Sure, there might be some lucky guy who made a fortune, but with how much everyone else lost it more than makes up for it. As such, you should expect to lose money and it is no substitute for proper investment.
– JMoravitz
May 5 '15 at 13:22
Generally, the best money-making strategy is to put everything on a single red/black roulette spin. This has a negative expected value, of course. But if you consider a 47% chance of doubling the bonus (at which point one must walk away) and a 53% chance of the bonus evaporating, then... I guess it's a "good" idea.
– pjs36
May 5 '15 at 13:25
*than${{{{}}}}$
– Git Gud
May 5 '15 at 13:07
*than${{{{}}}}$
– Git Gud
May 5 '15 at 13:07
2
2
Casinoes do their business on the fact that that statement is false. It's how they earn money.
– Arthur
May 5 '15 at 13:10
Casinoes do their business on the fact that that statement is false. It's how they earn money.
– Arthur
May 5 '15 at 13:10
1
1
Is there a math-related question here?
– hardmath
May 5 '15 at 13:22
Is there a math-related question here?
– hardmath
May 5 '15 at 13:22
2
2
What might be true is that they have a positive probability of being able to completely pay off their debt, however having a positive probability and having it be likely are two very different things. As Arthur mentioned, the common phrase to hear is "The house always wins." By that we mean at the end of the day players have lost more money in total than they had won. Sure, there might be some lucky guy who made a fortune, but with how much everyone else lost it more than makes up for it. As such, you should expect to lose money and it is no substitute for proper investment.
– JMoravitz
May 5 '15 at 13:22
What might be true is that they have a positive probability of being able to completely pay off their debt, however having a positive probability and having it be likely are two very different things. As Arthur mentioned, the common phrase to hear is "The house always wins." By that we mean at the end of the day players have lost more money in total than they had won. Sure, there might be some lucky guy who made a fortune, but with how much everyone else lost it more than makes up for it. As such, you should expect to lose money and it is no substitute for proper investment.
– JMoravitz
May 5 '15 at 13:22
Generally, the best money-making strategy is to put everything on a single red/black roulette spin. This has a negative expected value, of course. But if you consider a 47% chance of doubling the bonus (at which point one must walk away) and a 53% chance of the bonus evaporating, then... I guess it's a "good" idea.
– pjs36
May 5 '15 at 13:25
Generally, the best money-making strategy is to put everything on a single red/black roulette spin. This has a negative expected value, of course. But if you consider a 47% chance of doubling the bonus (at which point one must walk away) and a 53% chance of the bonus evaporating, then... I guess it's a "good" idea.
– pjs36
May 5 '15 at 13:25
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
Clearly, this is not a wise decision in general, assuming constant utility per dollar. But let's explore it in brief, anyway.
We'll make some more assumptions. They put the entire grand on a single even-money bet, and win another grand with probability $0.48$, and lose their bet with probability $0.52$. That's it; then they walk home. (Obviously, this is not the way real people bet, but there's a limit to how much we can model. We just want to see if there's a non-linear effect one can take advantage of.)
On the debt side, they have debt at $20$ percent interest, and we'll assume zero inflation. (Now we're really fictional!) With those assumptions, then a debt of $D$ will get paid off in $n$ months according to
$$
frac{500(1-r^n)}{1-r} = D
$$
$$
500(1-r^n) = D(1-r)
$$
$$
500r^n = 500-D(1-r)
$$
$$
r^n = 1-frac{D(1-r)}{500}
$$
with $r = 60/61$. For $D = 20000$, this yields $n doteq 64.513$; for $D = 19000$, $n doteq 61.699$; for $D = 18000$, $n doteq 59.009$. Paying down a thousand dollars of debt immediately saves your co-worker about $1357$ dollars; paying another thousand of winnings saves them only another $1345$ dollars. So there's a non-linearity, but it goes the wrong way. (Not surprising, really, since the further in debt you are, the more additional debt hurts you.)
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Clearly, this is not a wise decision in general, assuming constant utility per dollar. But let's explore it in brief, anyway.
We'll make some more assumptions. They put the entire grand on a single even-money bet, and win another grand with probability $0.48$, and lose their bet with probability $0.52$. That's it; then they walk home. (Obviously, this is not the way real people bet, but there's a limit to how much we can model. We just want to see if there's a non-linear effect one can take advantage of.)
On the debt side, they have debt at $20$ percent interest, and we'll assume zero inflation. (Now we're really fictional!) With those assumptions, then a debt of $D$ will get paid off in $n$ months according to
$$
frac{500(1-r^n)}{1-r} = D
$$
$$
500(1-r^n) = D(1-r)
$$
$$
500r^n = 500-D(1-r)
$$
$$
r^n = 1-frac{D(1-r)}{500}
$$
with $r = 60/61$. For $D = 20000$, this yields $n doteq 64.513$; for $D = 19000$, $n doteq 61.699$; for $D = 18000$, $n doteq 59.009$. Paying down a thousand dollars of debt immediately saves your co-worker about $1357$ dollars; paying another thousand of winnings saves them only another $1345$ dollars. So there's a non-linearity, but it goes the wrong way. (Not surprising, really, since the further in debt you are, the more additional debt hurts you.)
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
add a comment |
up vote
2
down vote
Clearly, this is not a wise decision in general, assuming constant utility per dollar. But let's explore it in brief, anyway.
We'll make some more assumptions. They put the entire grand on a single even-money bet, and win another grand with probability $0.48$, and lose their bet with probability $0.52$. That's it; then they walk home. (Obviously, this is not the way real people bet, but there's a limit to how much we can model. We just want to see if there's a non-linear effect one can take advantage of.)
On the debt side, they have debt at $20$ percent interest, and we'll assume zero inflation. (Now we're really fictional!) With those assumptions, then a debt of $D$ will get paid off in $n$ months according to
$$
frac{500(1-r^n)}{1-r} = D
$$
$$
500(1-r^n) = D(1-r)
$$
$$
500r^n = 500-D(1-r)
$$
$$
r^n = 1-frac{D(1-r)}{500}
$$
with $r = 60/61$. For $D = 20000$, this yields $n doteq 64.513$; for $D = 19000$, $n doteq 61.699$; for $D = 18000$, $n doteq 59.009$. Paying down a thousand dollars of debt immediately saves your co-worker about $1357$ dollars; paying another thousand of winnings saves them only another $1345$ dollars. So there's a non-linearity, but it goes the wrong way. (Not surprising, really, since the further in debt you are, the more additional debt hurts you.)
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
add a comment |
up vote
2
down vote
up vote
2
down vote
Clearly, this is not a wise decision in general, assuming constant utility per dollar. But let's explore it in brief, anyway.
We'll make some more assumptions. They put the entire grand on a single even-money bet, and win another grand with probability $0.48$, and lose their bet with probability $0.52$. That's it; then they walk home. (Obviously, this is not the way real people bet, but there's a limit to how much we can model. We just want to see if there's a non-linear effect one can take advantage of.)
On the debt side, they have debt at $20$ percent interest, and we'll assume zero inflation. (Now we're really fictional!) With those assumptions, then a debt of $D$ will get paid off in $n$ months according to
$$
frac{500(1-r^n)}{1-r} = D
$$
$$
500(1-r^n) = D(1-r)
$$
$$
500r^n = 500-D(1-r)
$$
$$
r^n = 1-frac{D(1-r)}{500}
$$
with $r = 60/61$. For $D = 20000$, this yields $n doteq 64.513$; for $D = 19000$, $n doteq 61.699$; for $D = 18000$, $n doteq 59.009$. Paying down a thousand dollars of debt immediately saves your co-worker about $1357$ dollars; paying another thousand of winnings saves them only another $1345$ dollars. So there's a non-linearity, but it goes the wrong way. (Not surprising, really, since the further in debt you are, the more additional debt hurts you.)
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
Clearly, this is not a wise decision in general, assuming constant utility per dollar. But let's explore it in brief, anyway.
We'll make some more assumptions. They put the entire grand on a single even-money bet, and win another grand with probability $0.48$, and lose their bet with probability $0.52$. That's it; then they walk home. (Obviously, this is not the way real people bet, but there's a limit to how much we can model. We just want to see if there's a non-linear effect one can take advantage of.)
On the debt side, they have debt at $20$ percent interest, and we'll assume zero inflation. (Now we're really fictional!) With those assumptions, then a debt of $D$ will get paid off in $n$ months according to
$$
frac{500(1-r^n)}{1-r} = D
$$
$$
500(1-r^n) = D(1-r)
$$
$$
500r^n = 500-D(1-r)
$$
$$
r^n = 1-frac{D(1-r)}{500}
$$
with $r = 60/61$. For $D = 20000$, this yields $n doteq 64.513$; for $D = 19000$, $n doteq 61.699$; for $D = 18000$, $n doteq 59.009$. Paying down a thousand dollars of debt immediately saves your co-worker about $1357$ dollars; paying another thousand of winnings saves them only another $1345$ dollars. So there's a non-linearity, but it goes the wrong way. (Not surprising, really, since the further in debt you are, the more additional debt hurts you.)
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
answered May 5 '15 at 17:04
Brian Tung
25.6k32553
25.6k32553
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
add a comment |
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
So with a probability of 0.52 to end up with 0 and a 0.48 chance to end up 4702 (1000 + 1000 + 1357 + 1345) would give us an expected value of 0.52 * 0 + 0.48 * 3702 of 2256.96? Versus if he applied it directly which would give us an expected value of 2357 (1000 + 1357). So we can expect a theoretical loss of of approximately 0.0424 (1 - 2256.96/2357)?
– aso118
May 5 '15 at 19:40
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
No; when we assume the winnings are applied entirely to debt, the $1357$ and $1345$ are the positive impact of the $2000$. So if he gambles, he has a $48$ percent chance of ending up $1357+1345 = 2702$ better off, whereas if he doesn't gamble, he has a $100$ percent chance of ending up $1357$ better off. The theoretical loss would be $1357-2702cdot 0.48 doteq 60$. (I hasten to emphasize that this is "theoretical" only in a very loose sense.)
– Brian Tung
May 5 '15 at 20:11
add a comment |
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*than${{{{}}}}$
– Git Gud
May 5 '15 at 13:07
2
Casinoes do their business on the fact that that statement is false. It's how they earn money.
– Arthur
May 5 '15 at 13:10
1
Is there a math-related question here?
– hardmath
May 5 '15 at 13:22
2
What might be true is that they have a positive probability of being able to completely pay off their debt, however having a positive probability and having it be likely are two very different things. As Arthur mentioned, the common phrase to hear is "The house always wins." By that we mean at the end of the day players have lost more money in total than they had won. Sure, there might be some lucky guy who made a fortune, but with how much everyone else lost it more than makes up for it. As such, you should expect to lose money and it is no substitute for proper investment.
– JMoravitz
May 5 '15 at 13:22
Generally, the best money-making strategy is to put everything on a single red/black roulette spin. This has a negative expected value, of course. But if you consider a 47% chance of doubling the bonus (at which point one must walk away) and a 53% chance of the bonus evaporating, then... I guess it's a "good" idea.
– pjs36
May 5 '15 at 13:25