While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:

Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function $f \colon [0,1] \to [0,1]$ such that $z$ is minimal.

$ b(v) = \int_0^v f(x)\ \mathrm{d}x \\ z=\int_0^1{ \max \left( \begin{array}{l} (1 + 2a)λ, \\ (2 + 2a)x - λ, \\ 2 a b(x) + 1, \\ (2 + 2a)x + b(1) - 2b(x) \\ \end{array} \right) } \mathrm{d}x $

That is the complete problem, but a partial solution would already help me a lot, for example:

• A solution only for certain values of $a$
• Finding values of $z$ without knowing $λ$ and $f$
• Only one pair $(λ, f)$ for an $a$ instead of all the pairs
• Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
• Is it possible that all values of $a$ have at least one pair in which the $f$ has $\{0, 1\}$ as its range?

I'll give you some info on what this formula is about. The goal of the project I'm working on is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows: In the beginning both players are required to bet a certain amount, the ante, variable $a$ in the formula. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:

• Check > Check
• Check > Raise > Fold
• Check > Raise > Call
• Raise > Fold
• Raise > Call

After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a raise followed by a call occurred. The goal for each player is to have as much expected profit as possible.

When player 1 uses the best response to the strategy of player 2, the expected value of the game will be $z - 1 - a$. A positive value indicates profit for player 1, a negative value profit for player 2. If player 1 checks, then player 2 will check with probability $f(c)$ and raise otherwise, where $c$ is the card player 2 has. If player 1 raises, then player 2 will call if his card is at least $λ$.

If I have the solution to this problem and didn't make any mistakes in making the formula, then I have the optimal strategy for player 2 and the expected value in the Nash equilibrium. And then I only have to do something similar for player 1.

Any ideas on minimizing $z$?

While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:

Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function $f \colon [0,1] \to [0,1]$ such that $z$ is minimal.

$ b(v) = \int_0^v f(x)\ \mathrm{d}x \\ z=\int_0^1{ \max \left( \begin{array}{l} (1 + 2a)λ, \\ (2 + 2a)x - λ, \\ 2 a b(x) + 1, \\ (2 + 2a)x + b(1) - 2b(x) \\ \end{array} \right) } \mathrm{d}x $

That is the complete problem, but a partial solution would already help me a lot, for example:

• A solution only for certain values of $a$
• Finding values of $z$ without knowing $λ$ and $f$
• Only one pair $(λ, f)$ for an $a$ instead of all the pairs
• Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
• Is it possible that all values of $a$ have at least one pair in which the $f$ has $\{0, 1\}$ as its range?

I'll give you some info on what this formula is about. The goal of the project I'm working on is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows: In the beginning both players are required to bet a certain amount, the ante, variable $a$ in the formula. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:

• Check > Check
• Check > Bet > Fold
• Check > Bet > Call
• Bet > Fold
• Bet > Call

After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a bet followed by a call occurred. The goal for each player is to have as much expected profit as possible.

When player 1 uses the best response to the strategy of player 2, the expected value of the game will be $z - 1 - a$. A positive value indicates profit for player 1, a negative value profit for player 2. If player 1 checks, then player 2 will check with probability $f(c)$ and bet otherwise, where $c$ is the card player 2 has. If player 1 bets, then player 2 will call if his card is at least $λ$.

If I have the solution to this problem and didn't make any mistakes in making the formula, then I have the optimal strategy for player 2 and the expected value in the Nash equilibrium. And then I only have to do something similar for player 1.

Any ideas on minimizing $z$?

While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:

Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function $f \colon [0,1] \to [0,1]$ such that $z$ is minimal.

$ b(v) = \int_0^v f(x)\ \mathrm{d}x \\ z=\int_0^1{ \max \left( \begin{array}{l} (1 + 2a)λ, \\ (2 + 2a)x - λ, \\ 2 a b(x) + 1, \\ (2 + 2a)x + b(1) - 2b(x) \\ \end{array} \right) } \mathrm{d}x $

That is the complete problem, but a partial solution would already help me a lot, for example:

• A solution only for certain values of $a$
• Finding values of $z$ without knowing $λ$ and $f$
• Only one pair $(λ, f)$ for an $a$ instead of all the pairs
• Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
• Is it possible that all values of $a$ have at least one pair in which the $f$ has $\{0, 1\}$ as its range?

To give you any idea on why you are thinking about this problem right now I will tell you something about the project I'm working on. The goal of the project is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows: In the beginning both players are required to bet a certain amount, the ante. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:

• Check > Check
• Check > Raise > Fold
• Check > Raise > Call
• Raise > Fold
• Raise > Call

After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a raise followed by a call occurred. The goal for each player is to have an expected value that is as high as possible.

I won't give you the relation of the poker game to the above formula, since then I would have to explain a lot more. But if I have the solution to the problem I'm 3/4 on the way to have a Nash equilibrium, if I didn't make any mistakes.

Any ideas on minimizing $z$?

While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:

Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function $f \colon [0,1] \to [0,1]$ such that $z$ is minimal.

$ b(v) = \int_0^v f(x)\ \mathrm{d}x \\ z=\int_0^1{ \max \left( \begin{array}{l} (1 + 2a)λ, \\ (2 + 2a)x - λ, \\ 2 a b(x) + 1, \\ (2 + 2a)x + b(1) - 2b(x) \\ \end{array} \right) } \mathrm{d}x $

That is the complete problem, but a partial solution would already help me a lot, for example:

• A solution only for certain values of $a$
• Finding values of $z$ without knowing $λ$ and $f$
• Only one pair $(λ, f)$ for an $a$ instead of all the pairs
• Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
• Is it possible that all values of $a$ have at least one pair in which the $f$ has $\{0, 1\}$ as its range?

I'll give you some info on what this formula is about. The goal of the project I'm working on is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows: In the beginning both players are required to bet a certain amount, the ante, variable $a$ in the formula. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:

• Check > Check
• Check > Raise > Fold
• Check > Raise > Call
• Raise > Fold
• Raise > Call

After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a raise followed by a call occurred. The goal for each player is to have as much expected profit as possible.

When player 1 uses the best response to the strategy of player 2, the expected value of the game will be $z - 1 - a$. A positive value indicates profit for player 1, a negative value profit for player 2. If player 1 checks, then player 2 will check with probability $f(c)$ and raise otherwise, where $c$ is the card player 2 has. If player 1 raises, then player 2 will call if his card is at least $λ$.

If I have the solution to this problem and didn't make any mistakes in making the formula, then I have the optimal strategy for player 2 and the expected value in the Nash equilibrium. And then I only have to do something similar for player 1.

Any ideas on minimizing $z$?