# 1963 AHSME Problems/Problem 36

## Problem

A person starting with $64$ and making $6$ bets, wins three times and loses three times, the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is: $\textbf{(A)}\text{ a loss of } 27 \qquad \textbf{(B)}\text{ a gain of } 27 \qquad \textbf{(C)}\text{ a loss of } 37 \qquad \\ \textbf{(D)}\text{ neither a gain nor a loss}\qquad \\ \textbf{(E)}\text{ a gain or a loss depending upon the order in which the wins and losses occur}$

## Solution

If the person wins the bet, the person has $\frac{3}{2}$ of the previous amount. If the person loses the bet, the person only has $\frac{1}{2}$ of the previous amount.

Because of the Commutative Property, the order of multiplying the multipliers does not matter. Thus, the person walks away with $64 \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = 27$ dollars, so the person loses $37$. The answer is $\boxed{\textbf{(C)}}$.

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