Two players A and B play a marble game. Each player has both a red and blue marble. They present one marble to each other. If both present red, A wins $3. If both present blue, A wins $1. If the colors of the two marbles do not match, B wins $2. Is it better to be A, or B, or does it matter?

Respuesta :

Answer:

Step-by-step explanation:

A has one red and blue marble and B has one red and blue marble.

Hence selecting one marble is equally likely with prob = 0.5

Since A and B are independent the joint event would be product of probabilities.

Let A be the amount A wins.

If each selects one, the sample space would be

             (R,R)  (R,B)  (B,R) (B,B)

Prob     0.25  0.25  0.25  0.25

A              3       -2     -2       1

E(A)      0.75   -0.5    -0.5   0.25   =    0

The game is a fair game with equal expected values for A and B.

It does not matter whether to be A or B

Both players have the same chances of winning, so it does not matter whether you are player A or B.

Since two players A and B play a marble game, and each player has both a red and blue marble, and they present one marble to each other, and if both present red, A wins $ 3, while if both present blue, A wins $ 1, and if the colors of the two marbles do not match, B wins $ 2, to determine if it is better to be A, or B, or does it matter, the following calculation must be performed:

  • The probability of each of the outcomes must be calculated.
  • Blue and blue = 0.25
  • Red and red = 0.25
  • No match = 0.50

  • Then, the probability of the results must be multiplied by the amount that each player could win.
  • Red and red = 0.25 x 3 = 0.75 (A)
  • Blue and blue = 0.25 x 1 = 0.25 (A)
  • No match = 0.50 x 2 = 1 (B)

Therefore, since 0.75 + 0.25 is equal to 1, both players have the same chances of winning, so it does not matter whether you are player A or B.

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