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!)
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