A certain Las Vegas casino offers one fair game. Players get to sit in comfortab

Question

A certain Las Vegas casino offers one fair game. Players get to sit in comfortable over- stuffed chairs overlooking the casino's food court and repeatedly put a $1.00 chip in the slot on the chair's arm. On each turn, a player will win $1.00 or lose $1.00, each with probability 1/2. (By the way, the casino makes money from these players by wafting enticing aromas from the food court toward them as an encouragement to overeat.) Suppose that you show up with $500 on a given day and you play this game 500 times. What's the probability that you'll be $25.00 ahead at the end of the day?

Explanation / Answer

here expected win on one game E(X) =(1/2)*(1)+(1/2)*(-1) =0

and E(X2) =(1/2)*(1)2+(1/2)*(-1) 2 =1

therefore std deviation=(E|(X2)-(E(X))2)1/2 =(1-02)1/2 =1

hence expected win from 500 games =500*0=0

and std deviation =1*(500)1/2 =22.3607

therefore P(X>25)=P(Z>(25-0)/22.3607)=P(Z>1.1180)=0.1318

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