Transfer Game - A mix of percentage and fixed amounts

Let the amounts the four players have initially and after the game be $a_0$ , $b_0$ , $c_0$ , and $d_0$ ; and $a_1$ , $b_1$ , $c_1$ , and $d_1$ respectively. Then

$\small \begin{cases} b_1 = 0.8(0.2a_0 + b_0 + 10) - 20 & = 0.16a_0+0.8b_0-12 \\ c_1 = 0.8(0.04a_0 + 0.2b_0 + c_0 +22) - 30 & = 0.032a_0 + 0.16b_0+0.2c_0 -12.4 \\ d_1 = 0.8(0.008a_0 + 0.04b_0 + 0.2c_0 + d_0 + 34.4) - 40 & = 0.0064a_0+0.032b_0+0.16c_0+0.8d_0-12.48 \\ a_1 = 0.8a_0 - 10 + 0.2(0.008a_0 + 0.04b_0 + 0.2c_0 + d_0 + 34.4) + 40 & = 0.8016a_0+0.008b_0+0.04c_0+0.2d_0+36.88 \end{cases}$

Since

$\begin{array} {lll} b_1 = b_0 & \implies 0.2b_0 = 0.16a_0 - 12 & \implies b_0 = 0.8a_0 - 60 \\ c_1 = c_0 & \implies 0.2c_0 = 0.032a_0 + 0.16b_0 -12.4 & \implies c_0 = 0.8a_0 - 110 \\ d_1 = 1.5d_0 & \implies 0.7d_0 = 0.0064a_0+0.032b_0+0.16c_0-12.48 & \implies d_0 = \dfrac 8{35}a_0 - \dfrac {320}7 \\ a_1 = 0.9a_0 & \implies -0.0984a_0+0.008b_0+0.04c_0+0.2d_0+36.88 = 0 & \implies a_0 = 1600 \end{array}$

Therefore $c_0 = 0.8a_0 - 110 = \boxed{1170}$ .

Let the initial amounts of Players $A$ , $B$ , $C$ , and $D$ be $A_1$ , $B_1$ , $C_1$ , and $D_1$ , respectively.

When Player $A$ gives $20\%$ of what he has plus $\$10$ to Player $B$ :

• Player $A$ now has $A_2 = \frac{4}{5}A_1 - 10$

• Player $B$ now has $B_2 = B_1 + \frac{1}{5}A_1 + 10$

When Player $B$ gives $20\%$ of what he has plus $\$20$ to Player $C$ :

• Player $B$ now has $B_3 = \frac{4}{5}B_2 - 20 = \frac{4}{5}(B_1 + \frac{1}{5}A_1 + 10) - 20$

• Player $C$ now has $C_2 = C_1 + \frac{1}{5}B_2 + 20 = C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20$

When Player $C$ gives $20\%$ of what he has plus $\$30$ to Player $D$ :

• Player $C$ now has $C_3 = \frac{4}{5}C_2 - 30 = \frac{4}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) - 30$

• Player $D$ now has $D_2 = D_1 + \frac{1}{5}C_2 + 30 = D_1 + \frac{1}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) + 30$

When Player $D$ gives $20\%$ of what he has plus $\$40$ to Player $A$ :

• Player $D$ now has $D_3 = \frac{4}{5}D_2 - 40 = \frac{4}{5}(D_1 + \frac{1}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) + 30) - 40$

• Player $A$ now has $A_3 = A_2 + \frac{1}{5}D_2 + 40 = \frac{4}{5}A_1 - 10 + \frac{1}{5}(D_1 + \frac{1}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) + 30) + 40$ .

Since Player $B$ ended up with the same amount that he had originally:

• $B_1 = B_3 = \frac{4}{5}(B_1 + \frac{1}{5}A_1 + 10) - 20$ , which solves to $A_1 = \frac{5}{4}B_1 + 75$

Since Player $C$ ended up with the same amount that he had originally:

• $C_1 = C_3 = \frac{4}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) - 30$

• after substituting $A_1 = \frac{5}{4}B_1 + 75$ , this solves to $B_1 = C_1 + 50$

• substituting this into $A_1 = \frac{5}{4}B_1 + 75$ gives $A_1 = \frac{5}{4}C_1 + \frac{275}{2}$

Since Player $D$ gained $50\%$ of what he had originally:

• $\frac{3}{2}D_1 = D_3 = \frac{4}{5}(D_1 + \frac{1}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) + 30) - 40$

• after substituting $B_1 = C_1 + 50$ and $A_1 = \frac{5}{4}C_1 + \frac{275}{2}$ , this solves to $D_1 = \frac{2}{7}(C_1 -50)$

Since Player $A$ lost $10\%$ of what he had originally:

• $\frac{9}{10}A_1 = A_3 = \frac{4}{5}A_1 - 10 + \frac{1}{5}(D_1 + \frac{1}{5}(C_1 + \frac{1}{5}(B_1 + \frac{1}{5}A_1 + 10) + 20) + 30) + 40$

• after substituting $B_1 = C_1 + 50$ , $A_1 = \frac{5}{4}C_1 + \frac{275}{2}$ , and $D_1 = \frac{2}{7}(C_1 -50)$ , this solves to $C_1 = \boxed{\$1170}$ .

---

By the way,

• $A_1 = \frac{5}{4}C_1 + \frac{275}{2} = \frac{5}{4} \cdot 1170 + \frac{275}{2} = \$1600$

• $B_1 = C_1 + 50 = 1170 + 50 = \$1220$

• $D_1 = \frac{2}{7}(C_1 -50) = \frac{2}{7}(1170-50) = \$320$

Let the initial amounts with players $A, B, C , D$ be $x(0) = [x_1(0), x_2(0), x_3(0), x_4(0)]^T$ , then after the first move by player $A$ , the amounts will change to $x(1)$ , where,

$x(1) = M(1) x(0) + b(1)$

where $M(1) = \begin{bmatrix} 0.8 && 0, && 0 && 0 \\ 0.2 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$

and $b(1) = [-10, 10, 0, 0]^T$

Similarly, after the second move, we have

$x(2) = M(2) x(1) + b(2) = M(2) M(1) x(0) + M(2) b(1) + b(2)$

where $M(2) = \begin{bmatrix} 1 && 0, && 0 && 0 \\ 0 && 0.8 && 0 && 0 \\ 0 && 0.2 && 1 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix}$

and $b(2) = [0, -20, 20, 0]^T$

Continuing this pattern, one sees that,

$x(4) = M x(0) + b$

where

$M = M(4) M(3) M(2) M(1)$ and $b = b(4) + M(4) b(3) + M(4) M(3) b(2) + M(4) M(3) M(2) b(1)$

And we're also given that $x(4) = G x(0)$ , where,

$G = \begin{bmatrix} 0.9 && 0 && 0 && 0 \\ 0 && 1 && 0 && 0 \\ 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 1.5 \end{bmatrix}$

Therefore, $G x(0) = M x(0) + b$ , from which it follows that $x(0) = (G - M)^{-1} b$

With help of a small piece of code the above equations can programmed, and the results is that $x(0) = [1600, 1220, 1170, 320]^T$ . Thus, the answer is $\boxed{1170}$ .