Coin Game with infinite paradox Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Flip a Fair CoinAnother probabilistic coin puzzleLeft coin, Right coin, Last coin- A fair versionCoin Flipping Game with the DevilCoin division puzzleEccentric Millionaire Probability ParadoxWhat is the value of this other coin flipping game?The asymmetric coin gameFind Coin's unfairness with only 4 tossesThe Monster and the Chessboard
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Coin Game with infinite paradox
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Flip a Fair CoinAnother probabilistic coin puzzleLeft coin, Right coin, Last coin- A fair versionCoin Flipping Game with the DevilCoin division puzzleEccentric Millionaire Probability ParadoxWhat is the value of this other coin flipping game?The asymmetric coin gameFind Coin's unfairness with only 4 tossesThe Monster and the Chessboard
$begingroup$
Here’s the puzzle/problem:
Let’s presume we are best friends. I own the house next to you. I make a gaming proposition along the following lines:
You throw a coin over and over again. So long as it comes up heads, you keep throwing. When it comes up tails you stop and I pay you $$$ — depending on how many heads you threw.
If you threw a tails to start, you get $0.
If you threw a heads then a tails, you get $1.
If you threw 2 heads then a tails, you get $2.
If you threw 3 heads, you get $4.
If you threw 4 heads, you get $8.
If you threw 5 heads, you get $16.
And on and on. The payoff doubles everytime you add an extra heads.
How much will you pay me to play this game (once)?
Note: This is not a loophole question, you will know when you have the right answer.
logical-deduction probability coins
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Here’s the puzzle/problem:
Let’s presume we are best friends. I own the house next to you. I make a gaming proposition along the following lines:
You throw a coin over and over again. So long as it comes up heads, you keep throwing. When it comes up tails you stop and I pay you $$$ — depending on how many heads you threw.
If you threw a tails to start, you get $0.
If you threw a heads then a tails, you get $1.
If you threw 2 heads then a tails, you get $2.
If you threw 3 heads, you get $4.
If you threw 4 heads, you get $8.
If you threw 5 heads, you get $16.
And on and on. The payoff doubles everytime you add an extra heads.
How much will you pay me to play this game (once)?
Note: This is not a loophole question, you will know when you have the right answer.
logical-deduction probability coins
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox
$endgroup$
– Tojrah
2 hours ago
add a comment |
$begingroup$
Here’s the puzzle/problem:
Let’s presume we are best friends. I own the house next to you. I make a gaming proposition along the following lines:
You throw a coin over and over again. So long as it comes up heads, you keep throwing. When it comes up tails you stop and I pay you $$$ — depending on how many heads you threw.
If you threw a tails to start, you get $0.
If you threw a heads then a tails, you get $1.
If you threw 2 heads then a tails, you get $2.
If you threw 3 heads, you get $4.
If you threw 4 heads, you get $8.
If you threw 5 heads, you get $16.
And on and on. The payoff doubles everytime you add an extra heads.
How much will you pay me to play this game (once)?
Note: This is not a loophole question, you will know when you have the right answer.
logical-deduction probability coins
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Here’s the puzzle/problem:
Let’s presume we are best friends. I own the house next to you. I make a gaming proposition along the following lines:
You throw a coin over and over again. So long as it comes up heads, you keep throwing. When it comes up tails you stop and I pay you $$$ — depending on how many heads you threw.
If you threw a tails to start, you get $0.
If you threw a heads then a tails, you get $1.
If you threw 2 heads then a tails, you get $2.
If you threw 3 heads, you get $4.
If you threw 4 heads, you get $8.
If you threw 5 heads, you get $16.
And on and on. The payoff doubles everytime you add an extra heads.
How much will you pay me to play this game (once)?
Note: This is not a loophole question, you will know when you have the right answer.
logical-deduction probability coins
logical-deduction probability coins
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
SpqrTiangSpqrTiang
161
161
New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
SpqrTiang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
2
$begingroup$
This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox
$endgroup$
– Tojrah
2 hours ago
add a comment |
2
$begingroup$
This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox
$endgroup$
– Tojrah
2 hours ago
2
2
$begingroup$
This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox
$endgroup$
– Tojrah
2 hours ago
$begingroup$
This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox
$endgroup$
– Tojrah
2 hours ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
This gambling problem is the famous St. Petersburg paradox. It is a paradox because
the wager has infinite expected value regardless of the amount paid to play. A rational agent should pay
any amount of money to play this game, and over many rounds of the game, they can expect to make a profit.
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the world GDP (~$55 trillion), the expected value of the lottery is
around $45.
This analysis also assumes that each dollar won has equal utility, which is a measure of what a dollar means to you personally. Utility of money tends to flatten out for large amounts - a $10 million, for example, is a life-changing amount of money, but the next $10 million won't have as big an impact on your life as the first $10 million. So, $20 million is not really "worth" twice as much as $10 million. The "infinite wager" solution to the St. Petersburg paradox assumes constant utility of money, even though that's not a realistic model of how people behave.
$endgroup$
add a comment |
$begingroup$
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win $$X$ in the game is roughly proportional to $frac aa+X$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is $$w$ and your pay-or-don't parameter is $$a$, the "correct" price $$p$ is such that
$$frac12logfracw-pw+sum_k=0^inftyfrac12^k+2fracaa+2^klogfracw-p+2^kw=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
- Suppose your $a$ parameter is ($)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about $4.87.
- Suppose my net wealth remains at $1M but now $a$ is $$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... $5.46.
- Suppose you remain a billionaire but I am much poorer, having only $1000 to my name. Then the amount I'm willing to pay goes down to $2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.)
- Suppose we're both poor, so I'm at $1000 and your $a$ parameter is the same. Then I am willing to pay about $2.39.
- Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about $12.80.
So I'm fairly comfortable answering the original question as follows: I am willing to pay somewhere between about $2 and about $13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.
$endgroup$
add a comment |
$begingroup$
I would pay you
All my money, and then be sad when I lost it all on the first flip
because
Let the amount I pay you be represented as C. After paying C, you have a 1/2 chance of winning 0, a 1/4 chance of winning 1, a 1/8 chance of winning 2, a 1/16 chance of winning 4 etc. Your expected payout would be $$-C + sum_i=0^infty frac12^i+22^i = -C + sum_i=0^infty frac14 = -C + infty$$ Theoretically if you played forever and ever, you would profit no matter how much you paid per game. Realistically, you would probably not profit no matter how much you paid.
$endgroup$
add a comment |
$begingroup$
One dollar. It's a nice round amount.
$endgroup$
add a comment |
$begingroup$
I will pay you
One dollar
to play this game, because
The amount you pay me has no bearing on what I pay you.
If there was a single throw there would be 50-50 chance of getting my dollar back.
But the game can go on, so there is better than 50-50 of getting my dollar back.
Your bankroll is the limit — your house against my $1.
$endgroup$
1
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
add a comment |
$begingroup$
To play once I would pay:
50 cents, since it is a doubling game and one heads is a dollar.
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This gambling problem is the famous St. Petersburg paradox. It is a paradox because
the wager has infinite expected value regardless of the amount paid to play. A rational agent should pay
any amount of money to play this game, and over many rounds of the game, they can expect to make a profit.
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the world GDP (~$55 trillion), the expected value of the lottery is
around $45.
This analysis also assumes that each dollar won has equal utility, which is a measure of what a dollar means to you personally. Utility of money tends to flatten out for large amounts - a $10 million, for example, is a life-changing amount of money, but the next $10 million won't have as big an impact on your life as the first $10 million. So, $20 million is not really "worth" twice as much as $10 million. The "infinite wager" solution to the St. Petersburg paradox assumes constant utility of money, even though that's not a realistic model of how people behave.
$endgroup$
add a comment |
$begingroup$
This gambling problem is the famous St. Petersburg paradox. It is a paradox because
the wager has infinite expected value regardless of the amount paid to play. A rational agent should pay
any amount of money to play this game, and over many rounds of the game, they can expect to make a profit.
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the world GDP (~$55 trillion), the expected value of the lottery is
around $45.
This analysis also assumes that each dollar won has equal utility, which is a measure of what a dollar means to you personally. Utility of money tends to flatten out for large amounts - a $10 million, for example, is a life-changing amount of money, but the next $10 million won't have as big an impact on your life as the first $10 million. So, $20 million is not really "worth" twice as much as $10 million. The "infinite wager" solution to the St. Petersburg paradox assumes constant utility of money, even though that's not a realistic model of how people behave.
$endgroup$
add a comment |
$begingroup$
This gambling problem is the famous St. Petersburg paradox. It is a paradox because
the wager has infinite expected value regardless of the amount paid to play. A rational agent should pay
any amount of money to play this game, and over many rounds of the game, they can expect to make a profit.
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the world GDP (~$55 trillion), the expected value of the lottery is
around $45.
This analysis also assumes that each dollar won has equal utility, which is a measure of what a dollar means to you personally. Utility of money tends to flatten out for large amounts - a $10 million, for example, is a life-changing amount of money, but the next $10 million won't have as big an impact on your life as the first $10 million. So, $20 million is not really "worth" twice as much as $10 million. The "infinite wager" solution to the St. Petersburg paradox assumes constant utility of money, even though that's not a realistic model of how people behave.
$endgroup$
This gambling problem is the famous St. Petersburg paradox. It is a paradox because
the wager has infinite expected value regardless of the amount paid to play. A rational agent should pay
any amount of money to play this game, and over many rounds of the game, they can expect to make a profit.
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the world GDP (~$55 trillion), the expected value of the lottery is
around $45.
This analysis also assumes that each dollar won has equal utility, which is a measure of what a dollar means to you personally. Utility of money tends to flatten out for large amounts - a $10 million, for example, is a life-changing amount of money, but the next $10 million won't have as big an impact on your life as the first $10 million. So, $20 million is not really "worth" twice as much as $10 million. The "infinite wager" solution to the St. Petersburg paradox assumes constant utility of money, even though that's not a realistic model of how people behave.
edited 2 hours ago
answered 2 hours ago
Nuclear WangNuclear Wang
1,351616
1,351616
add a comment |
add a comment |
$begingroup$
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win $$X$ in the game is roughly proportional to $frac aa+X$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is $$w$ and your pay-or-don't parameter is $$a$, the "correct" price $$p$ is such that
$$frac12logfracw-pw+sum_k=0^inftyfrac12^k+2fracaa+2^klogfracw-p+2^kw=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
- Suppose your $a$ parameter is ($)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about $4.87.
- Suppose my net wealth remains at $1M but now $a$ is $$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... $5.46.
- Suppose you remain a billionaire but I am much poorer, having only $1000 to my name. Then the amount I'm willing to pay goes down to $2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.)
- Suppose we're both poor, so I'm at $1000 and your $a$ parameter is the same. Then I am willing to pay about $2.39.
- Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about $12.80.
So I'm fairly comfortable answering the original question as follows: I am willing to pay somewhere between about $2 and about $13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.
$endgroup$
add a comment |
$begingroup$
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win $$X$ in the game is roughly proportional to $frac aa+X$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is $$w$ and your pay-or-don't parameter is $$a$, the "correct" price $$p$ is such that
$$frac12logfracw-pw+sum_k=0^inftyfrac12^k+2fracaa+2^klogfracw-p+2^kw=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
- Suppose your $a$ parameter is ($)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about $4.87.
- Suppose my net wealth remains at $1M but now $a$ is $$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... $5.46.
- Suppose you remain a billionaire but I am much poorer, having only $1000 to my name. Then the amount I'm willing to pay goes down to $2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.)
- Suppose we're both poor, so I'm at $1000 and your $a$ parameter is the same. Then I am willing to pay about $2.39.
- Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about $12.80.
So I'm fairly comfortable answering the original question as follows: I am willing to pay somewhere between about $2 and about $13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.
$endgroup$
add a comment |
$begingroup$
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win $$X$ in the game is roughly proportional to $frac aa+X$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is $$w$ and your pay-or-don't parameter is $$a$, the "correct" price $$p$ is such that
$$frac12logfracw-pw+sum_k=0^inftyfrac12^k+2fracaa+2^klogfracw-p+2^kw=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
- Suppose your $a$ parameter is ($)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about $4.87.
- Suppose my net wealth remains at $1M but now $a$ is $$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... $5.46.
- Suppose you remain a billionaire but I am much poorer, having only $1000 to my name. Then the amount I'm willing to pay goes down to $2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.)
- Suppose we're both poor, so I'm at $1000 and your $a$ parameter is the same. Then I am willing to pay about $2.39.
- Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about $12.80.
So I'm fairly comfortable answering the original question as follows: I am willing to pay somewhere between about $2 and about $13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.
$endgroup$
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win $$X$ in the game is roughly proportional to $frac aa+X$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is $$w$ and your pay-or-don't parameter is $$a$, the "correct" price $$p$ is such that
$$frac12logfracw-pw+sum_k=0^inftyfrac12^k+2fracaa+2^klogfracw-p+2^kw=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
- Suppose your $a$ parameter is ($)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about $4.87.
- Suppose my net wealth remains at $1M but now $a$ is $$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... $5.46.
- Suppose you remain a billionaire but I am much poorer, having only $1000 to my name. Then the amount I'm willing to pay goes down to $2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.)
- Suppose we're both poor, so I'm at $1000 and your $a$ parameter is the same. Then I am willing to pay about $2.39.
- Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about $12.80.
So I'm fairly comfortable answering the original question as follows: I am willing to pay somewhere between about $2 and about $13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.
answered 1 hour ago
Gareth McCaughan♦Gareth McCaughan
67.9k3172264
67.9k3172264
add a comment |
add a comment |
$begingroup$
I would pay you
All my money, and then be sad when I lost it all on the first flip
because
Let the amount I pay you be represented as C. After paying C, you have a 1/2 chance of winning 0, a 1/4 chance of winning 1, a 1/8 chance of winning 2, a 1/16 chance of winning 4 etc. Your expected payout would be $$-C + sum_i=0^infty frac12^i+22^i = -C + sum_i=0^infty frac14 = -C + infty$$ Theoretically if you played forever and ever, you would profit no matter how much you paid per game. Realistically, you would probably not profit no matter how much you paid.
$endgroup$
add a comment |
$begingroup$
I would pay you
All my money, and then be sad when I lost it all on the first flip
because
Let the amount I pay you be represented as C. After paying C, you have a 1/2 chance of winning 0, a 1/4 chance of winning 1, a 1/8 chance of winning 2, a 1/16 chance of winning 4 etc. Your expected payout would be $$-C + sum_i=0^infty frac12^i+22^i = -C + sum_i=0^infty frac14 = -C + infty$$ Theoretically if you played forever and ever, you would profit no matter how much you paid per game. Realistically, you would probably not profit no matter how much you paid.
$endgroup$
add a comment |
$begingroup$
I would pay you
All my money, and then be sad when I lost it all on the first flip
because
Let the amount I pay you be represented as C. After paying C, you have a 1/2 chance of winning 0, a 1/4 chance of winning 1, a 1/8 chance of winning 2, a 1/16 chance of winning 4 etc. Your expected payout would be $$-C + sum_i=0^infty frac12^i+22^i = -C + sum_i=0^infty frac14 = -C + infty$$ Theoretically if you played forever and ever, you would profit no matter how much you paid per game. Realistically, you would probably not profit no matter how much you paid.
$endgroup$
I would pay you
All my money, and then be sad when I lost it all on the first flip
because
Let the amount I pay you be represented as C. After paying C, you have a 1/2 chance of winning 0, a 1/4 chance of winning 1, a 1/8 chance of winning 2, a 1/16 chance of winning 4 etc. Your expected payout would be $$-C + sum_i=0^infty frac12^i+22^i = -C + sum_i=0^infty frac14 = -C + infty$$ Theoretically if you played forever and ever, you would profit no matter how much you paid per game. Realistically, you would probably not profit no matter how much you paid.
answered 2 hours ago
PunPun1000PunPun1000
37615
37615
add a comment |
add a comment |
$begingroup$
One dollar. It's a nice round amount.
$endgroup$
add a comment |
$begingroup$
One dollar. It's a nice round amount.
$endgroup$
add a comment |
$begingroup$
One dollar. It's a nice round amount.
$endgroup$
One dollar. It's a nice round amount.
answered 33 mins ago
AcccumulationAcccumulation
554111
554111
add a comment |
add a comment |
$begingroup$
I will pay you
One dollar
to play this game, because
The amount you pay me has no bearing on what I pay you.
If there was a single throw there would be 50-50 chance of getting my dollar back.
But the game can go on, so there is better than 50-50 of getting my dollar back.
Your bankroll is the limit — your house against my $1.
$endgroup$
1
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
add a comment |
$begingroup$
I will pay you
One dollar
to play this game, because
The amount you pay me has no bearing on what I pay you.
If there was a single throw there would be 50-50 chance of getting my dollar back.
But the game can go on, so there is better than 50-50 of getting my dollar back.
Your bankroll is the limit — your house against my $1.
$endgroup$
1
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
add a comment |
$begingroup$
I will pay you
One dollar
to play this game, because
The amount you pay me has no bearing on what I pay you.
If there was a single throw there would be 50-50 chance of getting my dollar back.
But the game can go on, so there is better than 50-50 of getting my dollar back.
Your bankroll is the limit — your house against my $1.
$endgroup$
I will pay you
One dollar
to play this game, because
The amount you pay me has no bearing on what I pay you.
If there was a single throw there would be 50-50 chance of getting my dollar back.
But the game can go on, so there is better than 50-50 of getting my dollar back.
Your bankroll is the limit — your house against my $1.
answered 2 hours ago
Weather VaneWeather Vane
2,485113
2,485113
1
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
add a comment |
1
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
1
1
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer?
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
@NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists.
$endgroup$
– Weather Vane
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same.
$endgroup$
– Arnaud Mortier
1 hour ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
$begingroup$
@ArnaudMortier the game is not being offered to me for $1.01 or 2 and it is not Nuclear Wang's puzzle. The question was: How much will I pay? And that was my answer.
$endgroup$
– Weather Vane
56 mins ago
add a comment |
$begingroup$
To play once I would pay:
50 cents, since it is a doubling game and one heads is a dollar.
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
add a comment |
$begingroup$
To play once I would pay:
50 cents, since it is a doubling game and one heads is a dollar.
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
add a comment |
$begingroup$
To play once I would pay:
50 cents, since it is a doubling game and one heads is a dollar.
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
To play once I would pay:
50 cents, since it is a doubling game and one heads is a dollar.
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
likwidfire2klikwidfire2k
13
13
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
likwidfire2k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
add a comment |
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
$begingroup$
Brutal with the downvotes when updating answers...
$endgroup$
– likwidfire2k
1 hour ago
add a comment |
SpqrTiang is a new contributor. Be nice, and check out our Code of Conduct.
SpqrTiang is a new contributor. Be nice, and check out our Code of Conduct.
SpqrTiang is a new contributor. Be nice, and check out our Code of Conduct.
SpqrTiang is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox
$endgroup$
– Tojrah
2 hours ago