Three people play a game in which one person loses and two persons win each game. the loser must double the amount of money that each player has at the time. the three players agree to play 3 games. at the end of the three games, each player has lost one game and each person has $8.00. what was the original stake of each player?

At the beginning of the third round, the player who is going to lose has to have 8 dollars to pay his loses and 8 dollars to complete  the game. Therefore the ending is

A           B          C

4           4          16

C gives 4 to A and 4 to B So everyone has 8 dollars in the end.

Let us suppose that A has lost the second round.

C has to have 8 dollars at the start of round 2 and B has to have two dollars. In order to wind up with 4 dollars A has to have 10 dollars + 4 dollars to start with. So A needs 14 dollars.

A           B          C

14          2          8              Beginning of round 2

4            2+2     8 + 8        

4             4         16            End of Round 2

Round One. This time B will loose

If B is going to lose, He needs to pay off A and C and their totals at the beginning of round 2 have to be cut in 1/2

A           B          C

7           13           4    B needs 11 + 2 for himself

14          2            8

This is actually a lot of Mathmate's solution. I was too worried about the number of ways this could have been done to have gotten it  right.  

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.