In a certain game of chance, a square board with area 1 is colored with sectors of either red or blue. A player, who cannot see the board, must specify a point on the board by giving an x-coordinate and a y-coordinate. The player wins the game if the specified point is in a blue sector. The game can be arranged with any number of red sectors, and the red sectors are designed so that

[[math]] R_i = \left(\frac{9}{20} \right)^i [[/math]]

where [math]R_i[/math] is the area of the [math]i^{\textrm{th}}[/math] red sector.

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Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

Suppose there are N red sectors. Let w be the probability of a player winning the game. Then, w = the probability of a player missing all the red sectors and

[[math]] w = 1 - \left [ \frac{9}{20} + \left (\frac{9}{20} \right)^2 + \ldots + \left( \frac{9}{29}\right)^N\right ] \frac{1-\left( \frac{9}{20}\right)^N}{1-\frac{9}{20}} = \frac{2}{11} + \frac{9}{11} \left ( \frac{9}{20}\right)^N [[/math]]

Using the geometric series formula,

[[math]] w = 1 - \frac{\frac{9}{20} - \left( \frac{9}{20}\right)^{N+1}}{1- \frac{9}{20}} = 1- \frac{9}{20} \frac{1- \left( \frac{9}{20} \right)^{N}}{1- \frac{9}{20}} = \frac{2}{11} + \frac{9}{11} \left ( \frac{9}{20}\right)^{N} [[/math]]

Thus we need

[[math]] \begin{align*} 0.2 \gt w = \frac{2}{11} + \frac{9}{11} \left ( \frac{9}{20}\right)^N \\ 2.2 \gt 2 + 9 \left ( \frac{9}{20} \right ) ^N \\ 0.2 \gt 9 \left ( \frac{9}{20} \right )^N \\ \frac{2}{90} \gt \left (\frac{9}{20} \right)^N \\ \left ( \frac{20}{9}\right)^N \gt 45 \\ N \gt \frac{\ln(45)}{\ln(20/9)} \approx 4.767 \end{align*} [[/math]]

Thus [math]N[/math] must be the first integer greater than 4.767, or 5.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.