A Puzzling Fortnight - Day 5

Never thought of actually solving the two inequalities. StRyKeR's solution is the right one.

Let w be the number of wins and T be the total number of games played right before Savior's win rate was strictly below 80%. That means

w/T < 4/5

and

(w+1)/(T+1) >= 4/5

First equation gives 5w < 4T, or 5w <= 4T - 1, since w and T are integers. Second gives

5w + 5 >= 4T + 4, or

5w + 1 >= 4T, or

5w >= 4T - 1.

Combined with the first equation, we see that we MUST have 5w = 4T - 1. In other words,

w = 4T/5 - 1/5 and

(w + 1)/(T + 1) = (4T/5 + 4/5)/(T + 1) = 4/5 = 80%, as desired. Basically, the first game that gives Savior a win rate of at least 80% is the game that puts him at EXACTLY 80%.

More generally, the trick works for any fractions p/q such that q - p = 1. In this case, q = 5 and p = 4.