Question

A gambler bets repeatedly $1 on red at the roulette (there are 18 red slots and 38 slots in all). He wins $1 if red comes up loses $1 otherwise. What is the probability that he will be ahead. (a) After 100 bets? (b) After 1,000 bets?

Answer 1

here expected win in one game E(X)= =1*(18/38)-1*(20/38)=-0.0526

and E(X²)==1²*(18/38)+(-1)²*(20/38)=1

therefore std deviation =sqrt(E(X²)-(E(X))²)=0.9986

a)

for 100 bets expected value =-0.526*100=-5.263

and std deviation =0.9986*(100)^1/2 =9.986

therefore from normal approximation probability that he will be ahead

=P(X>0)=P(Z>(0-(-5.263))/9.986)=P(Z>0.53)=0.2981

b)

for 1000 bets expected value =-0.526*1000=-52.63

and std deviation =0.9986*(1000)^1/2 =31.579

therefore from normal approximation probability that he will be ahead

=P(X>0)=P(Z>(0-(-52.63))/31.579)=P(Z>1.67)=0.0475