Question

1) In the example of a single player betting $ 1 on any number between 1 and 6 on the roll of a die and getting a prize of $ 4 if the die shows that number,

(a) Assume that the player has an unlimited amount of money. What is the expected number of bets until the player gets the $4 prize for the first time?

(b) Assume that the player has four $ 1 coins. What is the expected number of bets until the player either gets the $ 4 prize for the first time or gets bankrupt?

Answer 1

a)here as probability of getting the number that was bet =1/6 (as 6 equal probable outcome on die)

hecnce from geometric distribution: expected number of bets until the player gets the $4 prize for the first time =1/p=1/(1/6)=6

b)P(X=1)=P(gets desired number on 1st roll)=1/6

P(X=2)=P(not on firt and gets on 2nd) =(5/6)*(1/6)=5/36

P(X=3)=P(not on first two and gets on 3rd) =(5/6)²*(1/6)=25/216

P(X=4)=1-P(X=1)-P(X=2)-P(X=3)=1-(1/6)-(5/36)-(25/216)=125/216

hence expected bets =E(X)=xP(x)

=1*(1/6)+2*(5/36)+3*(25/216)+4*(125/216)=671/216=3.11