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Algebra Difficulty 2.8 Junior Find the answer

A person starting with 64andmaking 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:
(A) a loss of \textbf{(A)}\text{ a loss of } 27 \qquad \textbf{(B)}\text{ a gain of }27(C) a loss 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}$

Multiple choice: answer with the letter of the option you want.

Solution

If the person wins the bet, the person has 32\frac{3}{2} of the previous amount. If the person loses the bet, the person only has 12\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 64323232121212=2764 \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.Theansweris. The answer is \boxed{\textbf{(C)}}$.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.