Solution to Spring 2016 POTW #2 (Weird? You Bet!)

See here for the original post of the puzzle.

Trial and error may work here, but I don't recommend it. Instead:

• Suppose Adam starts with $A and Brendan with $B
• Adam loses first game, pays out to double Brendan's money.
• New results: Adam has $(A-B)$ and Brendan has $(2B).
• Brendan loses second game, pays out to double Adam's money.
• New results: Adam has $(2(A-B))=$(2A-2B) and Brendan has $(2B-(A-B)) = $(3B-A).
• By assumption: 120 = 2A-2B = 3B-A.
• Solve this system of equations to get A=150 and B=90.

Playing it out, just to check:

• Start: Adam has $150, Brendan has $90.
• Adam loses. New results: Adam has $60, Brendan has $180.
• Brendan loses. New results: Adam has $120, Brendan has $120.

(Here's a bonus challenge for you readers: Is there anything special about $120? Could this puzzle be solved for any ending amount? And what if there were 3 players and each one lost a round? Try to generalize!)

Spring 2016 POTW #2: Weird? You Bet!

Submissions due by midnight on Sunday, March 27, 2016.

Two players, Adam and Brendan, play high-stakes cribbage. They both entered with a certain amount of money, and they decided that the loser of a game should double his opponent's money. (For example, if Adam (with $10) lost to Brendan (with $6), then Adam would pay out $6 to double Brendan's money to $12, leaving Adam with just $4.)

Adam lost the first game, and then Brendan lost the second game, with each loser paying out according to the rule described above. After these two games and payouts, they miraculously discovered that both of them had exactly $120.

How much money must each player have started with, before the two games, to end up in this situation?

(Use the submission box below to submit your answer. No need to explain: a correct answer will suffice for 10 points.)