[Solution] Let $M_{0}$ be the initial amount. If he wins the first bet, his amount will be
$$
M_{1}=M_{0}+\frac{1}{2} M_{0}=\frac{3}{2} M_{0} .
$$

If he wins the second bet again, his amount will be
$$
M_{2}=M_{1}+\frac{1}{2} M_{1}=\frac{3}{2} M_{1}=\left(\frac{3}{2}\right)^{2} M_{0} \text {. }
$$
$\qquad$
From the above analysis, we can see that each win increases his amount to $\frac{3}{2}$ of the original. Similarly, each loss reduces his amount to $\frac{1}{2}$ of the original. Since he wins three times and loses three times in six attempts, regardless of the order, his final amount will be
$$
\left(\frac{3}{2}\right)^{3}\left(\frac{1}{2}\right)^{3} M_{0}=\frac{27}{64} M_{0}
$$

Thus, his loss is
$$
M_{0}-\frac{27}{64} M_{0}=\frac{37}{64} M_{0} .
$$

Given that $M_{0}=64$, after six attempts, he loses 37 yuan.
Therefore, the correct choice is $(\mathrm{C})$.
$36 \cdot 59$ A person initially has 64 yuan, and bets 6 times, winning 3 times and losing 3 times, with the order of wins and losses being arbitrary. The probability of winning and losing is equal. If the betting amount is half of the remaining money each time, then the final result is
(A) Lost 27 yuan.
(B) Won 27 yuan.
(C) Lost 37 yuan.
(D) Neither won nor lost.
(E) The result depends on the order of wins and losses.
(14th American High School Mathematics Examination, 1963)