Mixing
Players A and B repeatedly flip a (possibly unfair) coin until one loses everything. Each starting with a different amount of money
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
There are two players $A$ and $B$. At the beginning $A$ has $a_0 ∈ [0,1]$ amount of money and $B$ has $b_0 = 1 - a_0$ amount of money. They have one possibly unfair coin which they flip repeatedly. $A$ always guesses $Heads$ and $B$ always guesses $Tails$. Before each flip, they both bet a minimum of what each of them has (i.e. $bet_i = min(a_i, b_i)$). After the coin is tossed, the winner takes the $bet_i$ and they play again until one of them has no more money to play with.
I've been thinking about it for some time now and was unable to find a closed formula nor find it mentioned anywhere. So any more info would be appreciated.
The best I could do was a simulation. Interestingly it seems that when the coin is fair, the probability of $A$ winning is proportional to $a_0$ but when the coin is unfair, the graph looks more peculiar:
Here, the horizontal axis represents $a_0$ and the vertical one represents $P(A wins given that the coin is fair)$
In the next graph, the horizontal axis also represents $a_0$ and the vertical one represents $P(A wins given that the coin ends up Heads 1 in 3 times)$.
The code for the simulation can be found here.
probability economics
asked Aug 1 at 21:31
Peter Jankuliak
1454
 |Â
show 1 more comment
up vote
6
down vote
favorite
There are two players $A$ and $B$. At the beginning $A$ has $a_0 ∈ [0,1]$ amount of money and $B$ has $b_0 = 1 - a_0$ amount of money. They have one possibly unfair coin which they flip repeatedly. $A$ always guesses $Heads$ and $B$ always guesses $Tails$. Before each flip, they both bet a minimum of what each of them has (i.e. $bet_i = min(a_i, b_i)$). After the coin is tossed, the winner takes the $bet_i$ and they play again until one of them has no more money to play with.
I've been thinking about it for some time now and was unable to find a closed formula nor find it mentioned anywhere. So any more info would be appreciated.
The best I could do was a simulation. Interestingly it seems that when the coin is fair, the probability of $A$ winning is proportional to $a_0$ but when the coin is unfair, the graph looks more peculiar:
Here, the horizontal axis represents $a_0$ and the vertical one represents $P(A wins given that the coin is fair)$
In the next graph, the horizontal axis also represents $a_0$ and the vertical one represents $P(A wins given that the coin ends up Heads 1 in 3 times)$.
The code for the simulation can be found here.
probability economics
asked Aug 1 at 21:31
Peter Jankuliak
1454
As far as I can tell the only question you have is a request for more information. I think you'll need to be more specific. The idea behind your questions is intriguing, though.
– Theoretical Economist
Aug 1 at 21:37
Mainly whether there exist a closed formula, If only for some special cases (e.g. it's easy if $a_0 = b_0$). Or whether the problem has a name by which I could google more. Also, the area under the curve seems to be equal to $P(coin ends up Heads)$ which is interesting to me.
– Peter Jankuliak
Aug 1 at 21:51
The way you describe it, the player with the smaller amount is “all in†and this game should terminate the first time that player loses their bet.
– Laars Helenius
Aug 1 at 22:19
1
@PeterJankuliak the fact that "if the coin is fair, your odds of winning are proportional to your starting income" is a consequence of what is known as martingale theory (in particular, a consequence of optional stopping theorem). As for the second one, this looks quite interesting, though one might be able to do what is known as a doob decomposition and recover a similar result; the jumps look quite strange though)
– E-A
Aug 1 at 23:02
1
You will find a copy of the same graph in an answer to another question and some similar graphs in a third
– Henry
Aug 1 at 23:13
 |Â
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
There are two players $A$ and $B$. At the beginning $A$ has $a_0 ∈ [0,1]$ amount of money and $B$ has $b_0 = 1 - a_0$ amount of money. They have one possibly unfair coin which they flip repeatedly. $A$ always guesses $Heads$ and $B$ always guesses $Tails$. Before each flip, they both bet a minimum of what each of them has (i.e. $bet_i = min(a_i, b_i)$). After the coin is tossed, the winner takes the $bet_i$ and they play again until one of them has no more money to play with.
I've been thinking about it for some time now and was unable to find a closed formula nor find it mentioned anywhere. So any more info would be appreciated.
The best I could do was a simulation. Interestingly it seems that when the coin is fair, the probability of $A$ winning is proportional to $a_0$ but when the coin is unfair, the graph looks more peculiar:
Here, the horizontal axis represents $a_0$ and the vertical one represents $P(A wins given that the coin is fair)$
In the next graph, the horizontal axis also represents $a_0$ and the vertical one represents $P(A wins given that the coin ends up Heads 1 in 3 times)$.
The code for the simulation can be found here.
probability economics
asked Aug 1 at 21:31
Peter Jankuliak
1454
There are two players $A$ and $B$. At the beginning $A$ has $a_0 ∈ [0,1]$ amount of money and $B$ has $b_0 = 1 - a_0$ amount of money. They have one possibly unfair coin which they flip repeatedly. $A$ always guesses $Heads$ and $B$ always guesses $Tails$. Before each flip, they both bet a minimum of what each of them has (i.e. $bet_i = min(a_i, b_i)$). After the coin is tossed, the winner takes the $bet_i$ and they play again until one of them has no more money to play with.
I've been thinking about it for some time now and was unable to find a closed formula nor find it mentioned anywhere. So any more info would be appreciated.
The best I could do was a simulation. Interestingly it seems that when the coin is fair, the probability of $A$ winning is proportional to $a_0$ but when the coin is unfair, the graph looks more peculiar:
Here, the horizontal axis represents $a_0$ and the vertical one represents $P(A wins given that the coin is fair)$
In the next graph, the horizontal axis also represents $a_0$ and the vertical one represents $P(A wins given that the coin ends up Heads 1 in 3 times)$.
The code for the simulation can be found here.
probability economics
asked Aug 1 at 21:31
Peter Jankuliak
1454
asked Aug 1 at 21:31
Peter Jankuliak
1454
asked Aug 1 at 21:31
Peter Jankuliak
1454
asked Aug 1 at 21:31
Peter Jankuliak
1454
1454
As far as I can tell the only question you have is a request for more information. I think you'll need to be more specific. The idea behind your questions is intriguing, though.
– Theoretical Economist
Aug 1 at 21:37
Mainly whether there exist a closed formula, If only for some special cases (e.g. it's easy if $a_0 = b_0$). Or whether the problem has a name by which I could google more. Also, the area under the curve seems to be equal to $P(coin ends up Heads)$ which is interesting to me.
– Peter Jankuliak
Aug 1 at 21:51
The way you describe it, the player with the smaller amount is “all in†and this game should terminate the first time that player loses their bet.
– Laars Helenius
Aug 1 at 22:19
1
@PeterJankuliak the fact that "if the coin is fair, your odds of winning are proportional to your starting income" is a consequence of what is known as martingale theory (in particular, a consequence of optional stopping theorem). As for the second one, this looks quite interesting, though one might be able to do what is known as a doob decomposition and recover a similar result; the jumps look quite strange though)
– E-A
Aug 1 at 23:02
1
You will find a copy of the same graph in an answer to another question and some similar graphs in a third
– Henry
Aug 1 at 23:13
 |Â
show 1 more comment
As far as I can tell the only question you have is a request for more information. I think you'll need to be more specific. The idea behind your questions is intriguing, though.
– Theoretical Economist
Aug 1 at 21:37
Mainly whether there exist a closed formula, If only for some special cases (e.g. it's easy if $a_0 = b_0$). Or whether the problem has a name by which I could google more. Also, the area under the curve seems to be equal to $P(coin ends up Heads)$ which is interesting to me.
– Peter Jankuliak
Aug 1 at 21:51
The way you describe it, the player with the smaller amount is “all in†and this game should terminate the first time that player loses their bet.
– Laars Helenius
Aug 1 at 22:19
1
@PeterJankuliak the fact that "if the coin is fair, your odds of winning are proportional to your starting income" is a consequence of what is known as martingale theory (in particular, a consequence of optional stopping theorem). As for the second one, this looks quite interesting, though one might be able to do what is known as a doob decomposition and recover a similar result; the jumps look quite strange though)
– E-A
Aug 1 at 23:02
1
You will find a copy of the same graph in an answer to another question and some similar graphs in a third
– Henry
Aug 1 at 23:13
As far as I can tell the only question you have is a request for more information. I think you'll need to be more specific. The idea behind your questions is intriguing, though.
– Theoretical Economist
Aug 1 at 21:37
As far as I can tell the only question you have is a request for more information. I think you'll need to be more specific. The idea behind your questions is intriguing, though.
– Theoretical Economist
Aug 1 at 21:37
Mainly whether there exist a closed formula, If only for some special cases (e.g. it's easy if $a_0 = b_0$). Or whether the problem has a name by which I could google more. Also, the area under the curve seems to be equal to $P(coin ends up Heads)$ which is interesting to me.
– Peter Jankuliak
Aug 1 at 21:51
Mainly whether there exist a closed formula, If only for some special cases (e.g. it's easy if $a_0 = b_0$). Or whether the problem has a name by which I could google more. Also, the area under the curve seems to be equal to $P(coin ends up Heads)$ which is interesting to me.
– Peter Jankuliak
Aug 1 at 21:51
The way you describe it, the player with the smaller amount is “all in†and this game should terminate the first time that player loses their bet.
– Laars Helenius
Aug 1 at 22:19
The way you describe it, the player with the smaller amount is “all in†and this game should terminate the first time that player loses their bet.
– Laars Helenius
Aug 1 at 22:19
1
1
@PeterJankuliak the fact that "if the coin is fair, your odds of winning are proportional to your starting income" is a consequence of what is known as martingale theory (in particular, a consequence of optional stopping theorem). As for the second one, this looks quite interesting, though one might be able to do what is known as a doob decomposition and recover a similar result; the jumps look quite strange though)
– E-A
Aug 1 at 23:02
@PeterJankuliak the fact that "if the coin is fair, your odds of winning are proportional to your starting income" is a consequence of what is known as martingale theory (in particular, a consequence of optional stopping theorem). As for the second one, this looks quite interesting, though one might be able to do what is known as a doob decomposition and recover a similar result; the jumps look quite strange though)
– E-A
Aug 1 at 23:02
1
1
You will find a copy of the same graph in an answer to another question and some similar graphs in a third
– Henry
Aug 1 at 23:13
You will find a copy of the same graph in an answer to another question and some similar graphs in a third
– Henry
Aug 1 at 23:13
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
This actually has been analyzed, at least from the point of view of the player who has the worst of things. The game you describe is equivalent to the following. Aplayer plays a game in a casino. His probability of winning is $p<frac12.$ When he wins a bet, he wins the amount of the bet. He starts withs a bankroll $< 1$ and attempts to raise it to $1$. The idea is that he needs $1$ for some important purpose, and if he doesn't have it, he might as well have nothing.
So, on each bet, he bets either everything he has, if his bankroll is less than or equal to $1/2$ or just enough to raise his bank roll to 1, if he has more than $1/2,$ so if he has $.8,$ he bets $.2.$ Of course, he stops when he goes broke, or when he has raised his bankroll to $1.$
This was shown to be the best strategy in "How to Gamble if You Must," by Dubins and Savage. Let $f(x)$ be the probability of winning if the current bankroll is $x$. Then $$
f(x)=casesp+(1-p)f(2x-1),&$xgeqfrac12$\
pf(2x),&$x<frac12$
$$
Also of course, $f(0)=0, f(1)=1.$ It's easy to compute the values at the dyadic rationals. $f(.5)=p, f(.25)=p/2, f(.75)=(1+p)/2,$ etc.
I saw an analysis of $f$ somewhere years ago, and I can't remember where. It was in a popular science book, which only asserted the results, and I had to prove them for myself. If I recall correctly, $f$ is continuous and strictly increasing (no surprise there) but wildly non-differentiable. Of course, since it's monotonic, it must be differentiable almost everywhere, but what I think I remember is that it is not differentiable on any interval, and that the graph is not rectifiable.
Since any $0<x<1$ can be approximated arbitrarily closely by dyadic rationals, and $f$ can be computed exactly at dyadic rationals, $f(x)$ can be computed to any desired precision without simulation. Now that I think of it, there is a formula that computes $f(x)$ from the binary fraction for $x$. In that sense, there is a closed form formula.
The formula for $f$ is valid whether or nor $p<frac12$ but I don't know how much of the analysis holds up in that case.
answered Aug 1 at 23:21
saulspatz
10.5k21324
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This actually has been analyzed, at least from the point of view of the player who has the worst of things. The game you describe is equivalent to the following. Aplayer plays a game in a casino. His probability of winning is $p<frac12.$ When he wins a bet, he wins the amount of the bet. He starts withs a bankroll $< 1$ and attempts to raise it to $1$. The idea is that he needs $1$ for some important purpose, and if he doesn't have it, he might as well have nothing.
So, on each bet, he bets either everything he has, if his bankroll is less than or equal to $1/2$ or just enough to raise his bank roll to 1, if he has more than $1/2,$ so if he has $.8,$ he bets $.2.$ Of course, he stops when he goes broke, or when he has raised his bankroll to $1.$
This was shown to be the best strategy in "How to Gamble if You Must," by Dubins and Savage. Let $f(x)$ be the probability of winning if the current bankroll is $x$. Then $$
f(x)=casesp+(1-p)f(2x-1),&$xgeqfrac12$\
pf(2x),&$x<frac12$
$$
Also of course, $f(0)=0, f(1)=1.$ It's easy to compute the values at the dyadic rationals. $f(.5)=p, f(.25)=p/2, f(.75)=(1+p)/2,$ etc.
I saw an analysis of $f$ somewhere years ago, and I can't remember where. It was in a popular science book, which only asserted the results, and I had to prove them for myself. If I recall correctly, $f$ is continuous and strictly increasing (no surprise there) but wildly non-differentiable. Of course, since it's monotonic, it must be differentiable almost everywhere, but what I think I remember is that it is not differentiable on any interval, and that the graph is not rectifiable.
Since any $0<x<1$ can be approximated arbitrarily closely by dyadic rationals, and $f$ can be computed exactly at dyadic rationals, $f(x)$ can be computed to any desired precision without simulation. Now that I think of it, there is a formula that computes $f(x)$ from the binary fraction for $x$. In that sense, there is a closed form formula.
The formula for $f$ is valid whether or nor $p<frac12$ but I don't know how much of the analysis holds up in that case.
answered Aug 1 at 23:21
saulspatz
10.5k21324
add a comment |Â
up vote
4
down vote
accepted
This actually has been analyzed, at least from the point of view of the player who has the worst of things. The game you describe is equivalent to the following. Aplayer plays a game in a casino. His probability of winning is $p<frac12.$ When he wins a bet, he wins the amount of the bet. He starts withs a bankroll $< 1$ and attempts to raise it to $1$. The idea is that he needs $1$ for some important purpose, and if he doesn't have it, he might as well have nothing.
So, on each bet, he bets either everything he has, if his bankroll is less than or equal to $1/2$ or just enough to raise his bank roll to 1, if he has more than $1/2,$ so if he has $.8,$ he bets $.2.$ Of course, he stops when he goes broke, or when he has raised his bankroll to $1.$
This was shown to be the best strategy in "How to Gamble if You Must," by Dubins and Savage. Let $f(x)$ be the probability of winning if the current bankroll is $x$. Then $$
f(x)=casesp+(1-p)f(2x-1),&$xgeqfrac12$\
pf(2x),&$x<frac12$
$$
Also of course, $f(0)=0, f(1)=1.$ It's easy to compute the values at the dyadic rationals. $f(.5)=p, f(.25)=p/2, f(.75)=(1+p)/2,$ etc.
I saw an analysis of $f$ somewhere years ago, and I can't remember where. It was in a popular science book, which only asserted the results, and I had to prove them for myself. If I recall correctly, $f$ is continuous and strictly increasing (no surprise there) but wildly non-differentiable. Of course, since it's monotonic, it must be differentiable almost everywhere, but what I think I remember is that it is not differentiable on any interval, and that the graph is not rectifiable.
Since any $0<x<1$ can be approximated arbitrarily closely by dyadic rationals, and $f$ can be computed exactly at dyadic rationals, $f(x)$ can be computed to any desired precision without simulation. Now that I think of it, there is a formula that computes $f(x)$ from the binary fraction for $x$. In that sense, there is a closed form formula.
The formula for $f$ is valid whether or nor $p<frac12$ but I don't know how much of the analysis holds up in that case.
answered Aug 1 at 23:21
saulspatz
10.5k21324
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This actually has been analyzed, at least from the point of view of the player who has the worst of things. The game you describe is equivalent to the following. Aplayer plays a game in a casino. His probability of winning is $p<frac12.$ When he wins a bet, he wins the amount of the bet. He starts withs a bankroll $< 1$ and attempts to raise it to $1$. The idea is that he needs $1$ for some important purpose, and if he doesn't have it, he might as well have nothing.
So, on each bet, he bets either everything he has, if his bankroll is less than or equal to $1/2$ or just enough to raise his bank roll to 1, if he has more than $1/2,$ so if he has $.8,$ he bets $.2.$ Of course, he stops when he goes broke, or when he has raised his bankroll to $1.$
This was shown to be the best strategy in "How to Gamble if You Must," by Dubins and Savage. Let $f(x)$ be the probability of winning if the current bankroll is $x$. Then $$
f(x)=casesp+(1-p)f(2x-1),&$xgeqfrac12$\
pf(2x),&$x<frac12$
$$
Also of course, $f(0)=0, f(1)=1.$ It's easy to compute the values at the dyadic rationals. $f(.5)=p, f(.25)=p/2, f(.75)=(1+p)/2,$ etc.
I saw an analysis of $f$ somewhere years ago, and I can't remember where. It was in a popular science book, which only asserted the results, and I had to prove them for myself. If I recall correctly, $f$ is continuous and strictly increasing (no surprise there) but wildly non-differentiable. Of course, since it's monotonic, it must be differentiable almost everywhere, but what I think I remember is that it is not differentiable on any interval, and that the graph is not rectifiable.
Since any $0<x<1$ can be approximated arbitrarily closely by dyadic rationals, and $f$ can be computed exactly at dyadic rationals, $f(x)$ can be computed to any desired precision without simulation. Now that I think of it, there is a formula that computes $f(x)$ from the binary fraction for $x$. In that sense, there is a closed form formula.
The formula for $f$ is valid whether or nor $p<frac12$ but I don't know how much of the analysis holds up in that case.
answered Aug 1 at 23:21
saulspatz
10.5k21324
This actually has been analyzed, at least from the point of view of the player who has the worst of things. The game you describe is equivalent to the following. Aplayer plays a game in a casino. His probability of winning is $p<frac12.$ When he wins a bet, he wins the amount of the bet. He starts withs a bankroll $< 1$ and attempts to raise it to $1$. The idea is that he needs $1$ for some important purpose, and if he doesn't have it, he might as well have nothing.
So, on each bet, he bets either everything he has, if his bankroll is less than or equal to $1/2$ or just enough to raise his bank roll to 1, if he has more than $1/2,$ so if he has $.8,$ he bets $.2.$ Of course, he stops when he goes broke, or when he has raised his bankroll to $1.$
This was shown to be the best strategy in "How to Gamble if You Must," by Dubins and Savage. Let $f(x)$ be the probability of winning if the current bankroll is $x$. Then $$
f(x)=casesp+(1-p)f(2x-1),&$xgeqfrac12$\
pf(2x),&$x<frac12$
$$
Also of course, $f(0)=0, f(1)=1.$ It's easy to compute the values at the dyadic rationals. $f(.5)=p, f(.25)=p/2, f(.75)=(1+p)/2,$ etc.
I saw an analysis of $f$ somewhere years ago, and I can't remember where. It was in a popular science book, which only asserted the results, and I had to prove them for myself. If I recall correctly, $f$ is continuous and strictly increasing (no surprise there) but wildly non-differentiable. Of course, since it's monotonic, it must be differentiable almost everywhere, but what I think I remember is that it is not differentiable on any interval, and that the graph is not rectifiable.
Since any $0<x<1$ can be approximated arbitrarily closely by dyadic rationals, and $f$ can be computed exactly at dyadic rationals, $f(x)$ can be computed to any desired precision without simulation. Now that I think of it, there is a formula that computes $f(x)$ from the binary fraction for $x$. In that sense, there is a closed form formula.
The formula for $f$ is valid whether or nor $p<frac12$ but I don't know how much of the analysis holds up in that case.
answered Aug 1 at 23:21
saulspatz
10.5k21324
answered Aug 1 at 23:21
saulspatz
10.5k21324
answered Aug 1 at 23:21
saulspatz
10.5k21324
answered Aug 1 at 23:21
saulspatz
10.5k21324
10.5k21324
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869540%2fplayers-a-and-b-repeatedly-flip-a-possibly-unfair-coin-until-one-loses-everyth%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
As far as I can tell the only question you have is a request for more information. I think you'll need to be more specific. The idea behind your questions is intriguing, though.
– Theoretical Economist
Aug 1 at 21:37
Mainly whether there exist a closed formula, If only for some special cases (e.g. it's easy if $a_0 = b_0$). Or whether the problem has a name by which I could google more. Also, the area under the curve seems to be equal to $P(coin ends up Heads)$ which is interesting to me.
– Peter Jankuliak
Aug 1 at 21:51
The way you describe it, the player with the smaller amount is “all in†and this game should terminate the first time that player loses their bet.
– Laars Helenius
Aug 1 at 22:19
1
@PeterJankuliak the fact that "if the coin is fair, your odds of winning are proportional to your starting income" is a consequence of what is known as martingale theory (in particular, a consequence of optional stopping theorem). As for the second one, this looks quite interesting, though one might be able to do what is known as a doob decomposition and recover a similar result; the jumps look quite strange though)
– E-A
Aug 1 at 23:02
1
You will find a copy of the same graph in an answer to another question and some similar graphs in a third
– Henry
Aug 1 at 23:13