A simple dice game involves the player paying $1 to play. Two fair six-faced dic

Question

A simple dice game involves the player paying $1 to play. Two fair six-faced dice are
then rolled. The player receives a payout of $11 (net gain $10) with a “double six”, a
payout of $5 (net gain $4) with a sum of 11 or a payout of $2 (net gain $1) with a sum
of 10; otherwise there is no payout, and so the player loses the stake of $1.
(a) Complete the following table of the probability distribution of random variable
X, defined as the player’s net gain from a single game:

(b) Find E(X) and the standard deviation of X.
(c) If Aaron plays this game twice, what is the probability that he comes out
ahead (i.e. positive net gain)?
(d) If Barbara plays this game fifty times, find the mean and standard deviation of
her overall net gain.
(e) Use your answer to part (e) and a suitable approximation to calculate the
probability of coming out ahead after playing fifty games.

x p(x) xp(x) x2p(x) -1 1 4 10   

Explanation / Answer

(a)

E[X] = -0.25

VaR[X] = E[X2] - E[X]2 = 4.5833 -0.252 = 4.5208

SD[X] = 2.1262

(c) If he playes the game twice, he will have a net gain positive. It will happen when her wins either 4 or 10 once out of 2 chances

Pr(Positive gain) = 1 - Pr(Negative gain) - Pr(Zero gain)

= 1 - Pr($ -1, $ -1) - Pr($ 1, $ -1) - Pr($ -1 , $ 1)

= 1 - 5/6 * 5/6 - 3/36 * 5/6 - 5/6 * 3/36  

= 0.1667

(d) If Barbara playes fifty times this games

E[Overall Net gain] = -0.25 * 50 = -$ 12.50

VaR[ Overall Net Gain] = 50 * 4.5208 = 226.041667

SD[Overall net gain] = $ 15.035

(e) Now we can approximate the given distribution to normal as sample size is greater than 30.

so

Pr(gain > 0) = Pr(x > 0 ; - 12.50 ; 15.035)

Z = (0 + 12.50)/ 15.035 = 0.8314

Pr(gain > 0) = Pr(Z > 0.8314) = 1 - Pr(Z < 0.8314) = 0.7971

x p(x) xp(x) x2p(x) -1 0.8333 -0.8333 0.83333 1 0.0833 0.0833 0.08333 4 0.0556 0.2222 0.88889 10 0.0278 0.2778 2.77778

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