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I will be assuming that money starts at 1 for each player and players can bid any real number. I'll also assume that prizes are real numbers.

First, it should be noted that in each position with prizes $p_1 , \dots , p_n$ and accumulated scores $s_A, s_B$, if one player wins with money $m$ then he can win with any amount $m'>m$ and if that player loses with money $m$ he also loses with $m' < m$.

Let's assume that there is no partition of the prizes into two classes which yield the same sum; the above argument shows that there is a "singular value" which is the separator of the classes of winning and losing position for a player (such existence is guaranteed in $\mathbb{R}$).

There is a practical rule to find out what that value is in general, but let's start with simple positions: if player $A$ has to win $n$ times in a row in order to win, it is easy to see that the winning condition is $\frac{1}{n} \cdot m_A \geq m_B$.

To analyse a general position, given that the win-the-prize option has a winning condition of the form $\chi_W \cdot m_A \geq m_B$ and the lose-the-prize option has a winning condition of the form $\chi_L \cdot m_A \geq m_B$, the position the player $A$ is in has a winning condition of $\frac{1+\chi_L}{1+ \frac{1}{\chi_W}} m_A \geq m_B$.

To see why this happens, just consider the associated system of inequalities:

$$\chi_W \cdot (m_A - p) \geq m_B$$

$$\chi_L \cdot m_A \geq m_B - p$$

then deduce

$$m_A - \frac{ m_B}{\chi_W} \geq p \geq m_B - \chi_L \cdot m_A$$

Taking equalities yields the relation stated above.

This way, starting form the end of the auction tree you can go backward and find out all the $\chi$ for each position, until you get to the first one. Note that exacxtly meeting the inequality does not guarantee a win, it depends on who is going first on that auction (i.e. who won the last auction).

Prizes: 10 6 4 8 5

Minimal winning sequences: 10 8, 10 4 5, 10 6 4, 10 6 5, 6 8 5, 6 4 8, 4 8 5.

Results: In the starting position, $\chi=1$. After the first auction, the losing and winning $\chi$ are respectively $\frac{4}{7}$ and $\frac{7}{4}$; after the second auction the $\chi$ are $\frac{3}{1}$ if you won both 10 and 6, $\frac{4}{3}$ if you won 10 and lost 6, $\frac{3}{4}$ if you lost 10 and won 6, $\frac{1}{3}$ if you lost both.

Note that in the general case where there could be ties, the $\chi$s may not be one the inverse of the other, resulting in a $\chi>1$ for the starting position, which means the first player has no way to win (but still has a way to draw). In fact, I think that every time there is a partition in two subsets so that each yields the same sum, it is not possible for the first player to win (he can just draw).

2 edited body

I will be assuming that money starts at 1 for each player and players can bid any real number. I'll also assume that prizes are real numbers.

First, it should be noted that in each position with prizes $p_1 , \dots , p_n$ and accumulated scores $s_A, s_B$, if one player wins with money $m$ then he can win with any amount $m'>m$ and if that player loses with money $m$ he also loses with $m' < m$.

Let's assume that there is no partition of the prizes into two classes which yield the same sum; the above argument shows that there is a "singular value" which is the separator of the classes of winning and losing position for a player (such existence is guaranteed in $\mathbb{R}$).

There is a practical rule to find out what that value is in general, but let's start with simple positions: if player $A$ has to win $n$ times in a row in order to win, it is easy to see that the winning condition is $\frac{1}{n} \cdot m_A \geq m_B$.

To analyse a general position, given that the win-the-prize option has a winning condition of the form $\chi_W \cdot m_A \geq m_B$ and the lose-the-prize option has a winning condition of the form $\chi_L \cdot m_A \geq m_B$, the position the player $A$ is in has a winning condition of $\frac{1+\chi_L}{1+ \frac{1}{\chi_W}} m_A \geq m_B$.

To see why this happens, just consider the associated system of inequalities:

$$\chi_W \cdot (m_A - p) \geq m_B$$

$$\chi_L \cdot m_A \geq m_B - p$$

then deduce

$$m_A - \frac{ m_B}{\chi_W} \geq p \geq m_B - \chi_L \cdot m_A$$

Taking equalities yields the relation stated above.

This way, starting form the end of the auction tree you can go backward and find out all the $\chi$ for each position, until you get to the first one. Note that exacxtly meeting the inequality does not guarantee a win, it depends on who is going first on that auction (i.e. who won the last auction).

Prizes: 10 6 4 8 5

Minimal winning sequences: 10 8, 10 4 5, 10 6 4, 10 6 5, 8 6 8 5, 8 6 4 8, 4 8 54.

Results: In the starting position, $\chi=1$. After the first auction, the losing and winning $\chi$ are respectively $\frac{4}{7}$ and $\frac{7}{4}$; after the second auction the $\chi$ are $\frac{3}{1}$ if you won both 10 and 6, $\frac{4}{3}$ if you won 10 and lost 6, $\frac{3}{4}$ if you lost 10 and won 6, $\frac{1}{3}$ if you lost both.

Note that in the general case where there could be ties, the $\chi$s may not be one the inverse of the other, resulting in a $\chi>1$ for the starting position, which means the first player has no way to win (but still has a way to draw).

1

I will be assuming that money starts at 1 for each player and players can bid any real number. I'll also assume that prizes are real numbers.

First, it should be noted that in each position with prizes $p_1 , \dots , p_n$ and accumulated scores $s_A, s_B$, if one player wins with money $m$ then he can win with any amount $m'>m$ and if that player loses with money $m$ he also loses with $m' < m$.

Let's assume that there is no partition of the prizes into two classes which yield the same sum; the above argument shows that there is a "singular value" which is the separator of the classes of winning and losing position for a player (such existence is guaranteed in $\mathbb{R}$).

There is a practical rule to find out what that value is in general, but let's start with simple positions: if player $A$ has to win $n$ times in a row in order to win, it is easy to see that the winning condition is $\frac{1}{n} \cdot m_A \geq m_B$.

To analyse a general position, given that the win-the-prize option has a winning condition of the form $\chi_W \cdot m_A \geq m_B$ and the lose-the-prize option has a winning condition of the form $\chi_L \cdot m_A \geq m_B$, the position the player $A$ is in has a winning condition of $\frac{1+\chi_L}{1+ \frac{1}{\chi_W}} m_A \geq m_B$.

To see why this happens, just consider the associated system of inequalities:

$$\chi_W \cdot (m_A - p) \geq m_B$$

$$\chi_L \cdot m_A \geq m_B - p$$

then deduce

$$m_A - \frac{ m_B}{\chi_W} \geq p \geq m_B - \chi_L \cdot m_A$$

Taking equalities yields the relation stated above.

This way, starting form the end of the auction tree you can go backward and find out all the $\chi$ for each position, until you get to the first one. Note that exacxtly meeting the inequality does not guarantee a win, it depends on who is going first on that auction (i.e. who won the last auction).

Results: In the starting position, $\chi=1$. After the first auction, the losing and winning $\chi$ are respectively $\frac{4}{7}$ and $\frac{7}{4}$; after the second auction the $\chi$ are $\frac{3}{1}$ if you won both 10 and 6, $\frac{4}{3}$ if you won 10 and lost 6, $\frac{3}{4}$ if you lost 10 and won 6, $\frac{1}{3}$ if you lost both.
Note that in the general case where there could be ties, the $\chi$s may not be one the inverse of the other, resulting in a $\chi>1$ for the starting position, which means the first player has no way to win (but still has a way to draw).