Gambling Big

Probability Level 3

Alice and Bob are playing a game with 100 100 rounds (there are no ties). With each successive round, the money Alice and Bob are betting increases. Namely, for round n n , with the first round being round 1 1 , the amount of money the loser will need to pay the winner is 2 n − 1 2^{n-1} dollars. By the end of the game, Alice has won a profit of $ 2017 \$2017 . What is the sum of every round n n she has won?


The answer is 145.

Zain Majumder by Zain Majumder , 19, USA

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1 solution

Zain Majumder
Dec 16, 2017

Let set A A contain all the rounds Alice won and set B B contain the rounds Bob won (Alice lost). Then,

∑ n ∈ A 2 n − 1 − ∑ n ∈ B 2 n − 1 = 2017 \sum\limits_{n \in A} 2^{n-1} - \sum\limits_{n \in B} 2^{n-1} = 2017

∑ n ∈ A 2 n − 1 + ∑ n ∈ B 2 n − 1 = ∑ n = 1 100 2 n − 1 = 2 100 − 1 \sum\limits_{n \in A} 2^{n-1} + \sum\limits_{n \in B} 2^{n-1} = \sum\limits_{n = 1}^{100} 2^{n-1} = 2^{100}-1

Adding these two equations and dividing by two leaves ∑ n ∈ A 2 n − 1 = 2 100 + 2016 2 = 2 99 + 1008 = 2 99 + 2 9 + 2 8 + 2 7 + 2 6 + 2 5 + 2 4 . \sum\limits_{n \in A} 2^{n-1} = \frac{2^{100}+2016}{2} = 2^{99}+1008 = 2^{99}+2^9+2^8+2^7+2^6+2^5+2^4.

Therefore, Alice won rounds 5 5 , 6 6 , 7 7 , 8 8 , 9 9 , 10 10 , and 100 100 , the sum of which is 145 . \boxed{145}.

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