Question

as a project for math class, two students devised a game in which 3 black marbles and 2 red marbles are put into a bag. first the players must decide who is playing black marbles and who is playing red marbles. then each player takes a turn at drawing a marble, noting the color, replacing the marble in the bag, and then drawing a second marble and noting the color before returning it to the bag. the point scheme for the game is detailed in the table below.

point values for marble game

black marble points	red marble points
different colors: -1 point	different colors: -1 point
both red: 0 points	both black: 0 points

if seth is challenged to a game by a classmate, which statement below is correct in all aspects in helping him make the correct choice?

• since ( e(\text{black}) approx 0.24 ) and ( e(\text{red}) approx 0.16 ), seth should choose to play black marbles.
• ( e(\text{red}) ) will be twice that of ( e(\text{black}) ), so seth should choose to play red marbles.
• since ( e(\text{red}) = e(\text{black}) ), it is a fair game, so it doesnt matter which color seth chooses.
• both options will lose points because there are two ways to lose points and only one way to gain points. he should...

Explanation:

Step1: Calculate probabilities for black player

There are 3 black (B) and 2 red (R) marbles, total 5. Probability of B: \( P(B)=\frac{3}{5} \), R: \( P(R)=\frac{2}{5} \).

• Both black: \( P(BB)=(\frac{3}{5})^2=\frac{9}{25} \), points: +2.
• Different: \( P(BR)+P(RB)=2\times\frac{3}{5}\times\frac{2}{5}=\frac{12}{25} \), points: -1.
• Both red: \( P(RR)=(\frac{2}{5})^2=\frac{4}{25} \), points: 0.

Expected value \( E(\text{black}) = 2\times\frac{9}{25}+(-1)\times\frac{12}{25}+0\times\frac{4}{25}=\frac{18 - 12}{25}=\frac{6}{25}=0.24 \).

Step2: Calculate probabilities for red player

• Both red: \( P(RR)=\frac{4}{25} \), points: +4.
• Different: \( P(RB)+P(BR)=\frac{12}{25} \), points: -1.
• Both black: \( P(BB)=\frac{9}{25} \), points: 0.

Expected value \( E(\text{red}) = 4\times\frac{4}{25}+(-1)\times\frac{12}{25}+0\times\frac{9}{25}=\frac{16 - 12}{25}=\frac{4}{25}=0.16 \).

Step3: Analyze options

• First option: \( E(\text{black})\approx0.24 \), \( E(\text{red})\approx0.16 \), so choose black. Correct.
• Second: \( E(\text{red}) \) is not twice \( E(\text{black}) \). Wrong.
• Third: \( E(\text{black}) \) and \( E(\text{red}) \) are positive, not equal. Wrong.
• Fourth: Expected values are positive, not losing. Wrong.

Answer:

Since \( E(\text{black}) \approx 0.24 \) and \( E(\text{red}) \approx 0.16 \), Seth should choose to play black marbles. (The correct option is the first one: "Since \( E(\text{black}) \approx 0.24 \) and \( E(\text{red}) \approx 0.16 \), Seth should choose to play black marbles.")