Consider a casino game that an individual (Joe) wants to play. It costs him N do

Question

Consider a casino game that an individual (Joe) wants to play. It costs him N dollars each time to play. He loves this game and wants to continue playing until he is either broke or he breaks the bank (wins all the money). The probability of winning is p; the probability of losing is q. These are fixed probability values every time the game is played.

Joe brought $M to the casino. Every time one plays you either lose the entrance fee ($N) or you win and are paid back D dollars.

(a) How much money does Joe expect to have after playing n times? Derive a formula for how much money he has.

(b) Suppose Joe starts with $100, p=0.3, q=0.7, N=$5, and D=$20. Is it likely that Joe will break the bank?

(c) If the answer to (b) is no, how many times is it likely that Joe can play this game before he is broke?

Explanation / Answer

a)Let the random variable 'W' be our net gain/lose in one play of the game.

Joe wins 'D'$ with probability 'p'

Joe lose 'N'$ with probability 'q'

Therefore,expected net gain/lose is (in one play) is,

E(W)=pD-qN

So expected net gain/lose for 'n' times of play is,

E(nW)=nE(W)=n(pD-qN)

Therefore expected money that joe will have is given by subtracting expected net gain/lose for 'n' times of play from total money that Joe had ('M'$),Thus we get,

T=M-n(pD-qN)____(Ans) (where T is just a notation)

b)Given,

p=0.3 , q=0.7 , M=100 , N=5 , D=20.

Substituiting the given values in the above derived equation we get we get,

T=100-(0.3*20-0.7*5)*n=100-(6-3.5)n=100-2.5n

Thus it is not likely that Joe will break the bank since '2.5n' is a positive value and it is getting subtracted from total money 100$

c)To find how many times is it likely that Joe can play this game before he is broke,we equate expected money that joe will have to '0' and solve for n,we get,

T=0

100-(0.3*20-0.7*5)*n=0

n=100/2.5=40____(Ans)

So Joe can play the game '40' times before he is broke.

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