Suppose that a game of chance is played with a pair of fair 6-sided dice ( with

ID: 3360321 • Letter: S

Question

Suppose that a game of chance is played with a pair of fair 6-sided dice ( with the sides numbered from 1 to 6). In the game, you can pick any number from 1 to 6 and the two dice are "rolled" in a cage. If $1 is bet and exactly one of the numbers that you picked is rolled you win $1, and if both of the dice are the number that you picked you win $7 (in each of those cases you also get your initial $1 back). If none of your number winds up being rolled you lose your $1 bet. Suppose that you play this game 6 times and pick the same number each time.

a.) What is the probability that doubles of YOUR number (both dices come up your number) does not occur in the 6 rolls?

b.) What is your TOTAL expected win or loss in both the dollar amount to the nearest $0.01 AND whether it is a win or loss.

Explanation / Answer

there are six face of a dice 1,2,3,4,5,6

P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=1/6

if two dice are rolled then sample space will be

(1,1)

(1,2)

(1,3)

(1,4)

(1,5)

(1,6)

(2,1)

(2,2)

(2,3)

(2,4)

(2,5)

(2,6)

(3,1)

(3,2)

(3,3)

(3,4)

(3,5)

(3,6)

(4,1)

(4,2)

(4,3)

(4,4)

(4,5)

(4,6)

(5,1)

(5,2)

(5,3)

(5,4)

(5,5)

(5,6)

(6,1)

(6,2)

(6,3)

(6,4)

(6,5)

(6,6)

(a) P(a desired double number )=1/36

now we use binomial distribution with parameter n=6 and p=1/36 and for

Binomial distribution ,P(X=r)=nCrpr(1-p)n-r

P(X=0)=0.8445 ( using ms-excel=binomdist(0,6,1/36,0))

(b) P( of a one desired bumber on either of dice)=11/36

P(of a double desired number on the dice)=1/36

so if W=winning then

W=1 with p(w)=11/36

W=-1 with p(w)=25/36

W=7 with p(w)=1/36

E(W)=W*p(w)=1*(11/36)+(-1)*(25/36)+7*(1/36)=-7/36=-0.2

there is expected loss i.e. $ 0.2