Question

A person starting with $64$ cents 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 loss of $27$ ¢

a gain of $27$ ¢

a loss of $37$ ¢

neither a gain nor a loss

a gain or a loss depending upon the order in which the wins and losses occur

Answer: C

Concepts:exponent process simulation

Difficulty rating: 1470

Small Hint:

A win multiplies the current amount by $\frac{3}{2}$ , while a loss multiplies it by $\frac{1}{2}$

Big Hint:

Multiplication makes the order irrelevant

Solution:

After three wins and three losses, the amount, in cents, is $\begin{aligned} 64\left(\frac32\right)^3 \left(\frac12\right)^3 &=64\cdot\frac{27}{64}\\ &=27. \end{aligned}$ The loss is $64-27=37$ cents, regardless of order.

Thus, the correct answer is C.

1963 AMC 12 Problem 36

Answer: C

Difficulty rating: 1470

Small Hint:

Big Hint:

Solution:

Problem 36 in Other Years