Finding three unknowns

Question

Three gamblers agree that the loser of a game has to double the amount of money each of the two others has.
For example, if they start with $100, $30, and $20, and the first one loses, then the first one has to pay $30 to
the second one, and $20 to the third one. They play three games, and each gambler loses once. At the end,
each gambler has $80. How much did each gambler start out with?

Answer 1

Call the starting amounts of the gamblers x, y, and z. The loser of each game will lose an amount equal to the sum of his competitors' totals. The non-losers of each game will have their totals doubled.

 

It doesn't matter which order they're in, so let's just say x loses the first game. After game 1, then, the gamblers have the following money remaining:

 

x --> x-y-z

y --> 2y

z --> 2z

 

Let's say y loses the second game. After game 2, then, the gamblers have:

 

x-y-z --> 2(x-y-z)

2y --> 2y-(x-y-z)-2z

2z --> 4z

 

Finally, z loses the third game. After game 3, the gamblers have:

 

2(x-y-z) --> 4(x-y-z)

2y-(x-y-z)-2z --> 2(2y-(x-y-z)-2z)

4z --> 4z-2(x-y-z)-(2y-(x-y-z)-2z)

 

Those three ugly equations all equal $80. Let's simplify them:

 

4(x-y-z) = 80

x-y-z = 20

 

2(2y-(x-y-z)-2z) = 80

2y-(x-y-z)-2z = 40

2y-x+y+z-2z = 40

3y-x-z = 40

-x+3y-z = 40

 

4z-2(x-y-z)-(2y-(x-y-z)-2z) = 80

4z-2x+2y+2z-(2y-x+y+z-2z) = 80

6z-2x+2y-2y+x-y-z+2z = 80

7z-x-y = 80

-x-y+7z = 80

 

 

 

That gives you a system of three equations:

 

x - y - z = 20

-x + 3y - z = 40

-x - y + 7z = 80

 

To solve this system, let's solve the second equation for z:

 

-x + 3y - z = 40

-x + 3y = 40 + z

-x + 3y - 40 = z

 

Plug that into the first equation and into the third equation:

 

x - y - (-x + 3y - 40) = 20

x - y + x - 3y + 40 = 20

2x - 4y + 40 = 20

2x - 4y = -20

x - 2y = -10

 

-x - y + 7(-x + 3y - 40) = 80

-x - y - 7x + 21y - 280 = 80

-8x + 20y = 360

-2x + 5y = 90

 

 

 

Now we've got a system of two equations:

 

x - 2y = -10

-2x + 5y = 90

 

You can solve this system by solving the first equation for x, then plugging that into the second equation:

 

x = 2y - 10

 

-2(2y - 10) + 5y = 90

-4y + 20 + 5y = 90

y + 20 = 90

y = 70

 

Now, plug y = 70 back into x = 2y - 10 to solve for x:

 

x = 2(70) - 10

x = 140 - 10

x = 130

 

Finally, plug x = 130 and y = 70 into one of the equations with three variables and solve for z:

 

x - y - z = 20

130 - 70 - z = 20

60 - z = 20

60 = 20 + z

40 = z

 

 

 

So, the first gambler who lost started out with $130, the second gambler who lost started out with $70, and the third gambler who lost started out with $40.

 

 

 

Let's check our work by playing out the scenario with these numbers. x loses the first game:

 

130 --> 130 - 70 - 40 = 60 - 40 = 20

70 --> 2(70) = 140

40 --> 2(40) = 80

 

y loses the second game:

 

20 --> 2(20) = 40

140 --> 140 - 20 - 80 = 120 - 80 = 40

80 --> 2(80) = 160

 

z loses the third game:

 

40 --> 2(40) = 80

40 --> 2(40) = 80

160 --> 160 - 40 - 40 = 120 - 40 = 80

 

That checks out!