A game is played with two fair dice. The first die has sides numbered 1 through

Question

A game is played with two fair dice. The first die has sides numbered 1 through 6, while the second die has two sides numbered 1, two sides numbered 2, and one side each numbered 3 and 4. In a game, a fair coin is tossed. If the coin lands showing heads, die one is rolled. If the coin lands showing tails, die two is rolled. Let X = number observed when the game is played. (a) What is the pmf of X? (Evaluate px (1), px (2) etc.). (b) Calculate the expected value and standard deviation of X. (c) You pay $1 to play a game and win $2 if X lessthanorequalto 2. If X lessthanorequalto 3, you win nothing. Calculate the expected winnings in the game. Does this game favor you as player or the "house" (casino)? Explain your reasoning.

Explanation / Answer

(a) In the case of first die say D1, probabilirty of getting any number from 1 to 6, p1(x) = 1/6 for x = 1,2,3,4,5,6

In the case of secoon die say D2. probability of getting 1, p2(1) = 2/6, p2(2) = 2/6, p2(3) = 1/6 and p2(4) = 1/6

Since a fair coin is thrown to decide which die to throw, probability throwing die 1. p(D1) = 1/2 and p(D2) = 1/2

Hence probability of getting1 in the game, p(1) = p(D1) X p1(1) + p(D2) X p2(1) = 0.5 X 1/6 + 0.5 X 2/6 = 0.25

Similarly p(2) = 0.5 X 1/6 + 0.5 X 2/6 = 0.25

p(3) = 0.5 X 1/6 + 0.5 X 1/6 = 0.5 X 2/6 = 0.1667

p(4) = 0.5 X 1/6 + 0.5 X 1/6 = 0.5 X 2/6 = 0.1667

p(5) = 0.5 X 1/6 + 0.5 X 0 = 0.5/6 = 0.0833

p(6) = 0.5 X 1/6 + 0.5 X 0 = 0.5/6 = 0.0833

Thus pmf of X is given by p(1) = 0.25, p(2) = 0.25, p(3) = 0.1667, p(4) = 0.1667, p(5) = 0.0833 and p(6) = 0.0833

(b) expected value of x = 1 X 0.25 + 2 X 0.25 + 3 X 0.1667 + 4 X 0.1667 + 5 X 0.0833 + 6 X 0.0833 = 2.8332

variance of x = (1 X 0.25 + 4 X 0.25 + 9 X 0.1667 + 16 X 0.1667 + 25 X 0.0833 + 36 X 0.0833) - 2.83322

= 10.4988 - 8.0270 = 2.4718

Hence standard deviation of x = sqrt (variance) = sqrt(2.4718) = 1.5722

(c) Probability of X <= 2 = p(1) + p(2) = 0.25 + 0.25 = 0.5. Probability of X(>=3) = 0.5

Winnings if X <= 2 = 2 - 1 = 1 $

  Winnings if X >= 3 = 0 - 1 = -1 $

Hence expected winnings = 1 X 0.5 + -1X0.5 = 0.5 - 0.5 = 0 $.

Thus the game does not favour either the player or the casino, as both have equal chance of winning or losing the same amount (1$) and expected winnings for both are 0 $.

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