Question

Three people play a game in which there are always two winners and one loser. The loser gives each winner an amount equal to what the winner already has. After three games each has lost once and each has 24$. With how much money did each begin?

Answer 1

39$, 21$ and 12$

Explanation:

Let's denote the losers of the first, second and third game by A, B and C, respectively.

After each game, the winner's holdings double. The net holdings stay constant at 72$.

Before the last game, the holdings of A and B must have been 12$ each (since after the last game, which they had both won, they had 24$ each), while C had `24\times 3-12-12 = 48`$

Before the second game, the holdings of A and C are 6$ and 24$, respectively. That of B is `72-6-24=42`

At the beginning B and C must have had 21$ and 12$, A must have had `72-21-12=39`$.