Question

(6(4 pts) A player (Joe) goes to a casino and plays a fair game. The player may wager any amount of money. There is a 0.5 probability of winning. If the player wins, then the player get twice the amount of the bet in winnings. If the player loses, the player gets nothing. Think of betting on a coin toss. If you win you double your money, if you lose you lose your money. This is a "fair" game because the expected winnings is the same as the cost of the bet. There are no fair games in casinos, but for simplicity, imagine that there is. Joe hears about a "fool proof" method to "beat the casino." He will enter the casino with a large amount of money, His first bet will be for Sl. If he wins he gets $1 of profit (S2-SI). If he loses his next bet is for $2. If he wins he gets $4 back. Since he has spent $3 (S1 one the first bet and $2 on the second), he again has $1 profit. If he loses the second bet, he then bets $4 on the third bet. If he wins this bet he will again have $1 profit ($8 - $1. $2 -S4). He plans to continue this process until he wins, so he feels he is guaranteed of always having $1 of profit. We will consider one round of Joe playing until he wins (or until he loses everything) as our experiment Express probabilities as a fraction. () Joe enters the casino with $31. This is enough to place bets. The probability of Joe winning eventually and getting the S1 profit is The probability of Joe losing all his money is V 3.6 Joe's expected profit is (show work) Low: .20 + L :(-00)=0 (ii) Joe enters the casino with $511. This is enough to place 9 bets. The probability of Joe winning eventually and getting the $1 profit is The probability of Joe losing all his money is Joe's expected profit is (show work) bets. (iii) Joe enters the casino with $1,048,575. This is enough to place 20 The probability of Joe winning eventually and getting the S1 profit is The probability of Joe losing all his money is Joe's expected profit is (show work) bets. (iii) Joe enters the casino with infinite money. This is enough to place The probability of Joe winning eventually and getting the $1 profit is The probability of Joe losing all his money is Joe's expected profit is_ Joe will increase his wealth by 0 %

Answer 1

John places bets in the manner 1, 2, 4, 8, …2^n-1, where 1+ 2 + 4 + 8, …+ 2^n-1 = 2ⁿ – 1. Total amount that he places on bets is 2ⁿ – 1. We can use this result in all the parts of question.

Let us first calculate Expected gain in n bets,

There are two cases,

(a) He does not lose all the games (probability P) and his gain is $ 1, here P = 1 – (1/2)ⁿ

(b) He lost all the games (probability Q) and his gain is – (2ⁿ – 1), Q = (1/2)ⁿ

Expected Value E(x) = 1xP – (2ⁿ – 1)Q

Expected Value E(x) = 1[1 – (1/2)ⁿ] – (2ⁿ – 1)[(1/2)ⁿ] = 0.

(i) John places bets in the manner 1, 2, 4, 8, …2^n-1, where 1+ 2 + 4 + 8, …+ 2^n-1 = 2ⁿ – 1, With $ 31, Joe can place 5 bets.

Probability of winning last bet and getting $1 profit = LLLLW = (1/2)⁴ (1/2) = 1/32.

Probability of losing all his money = probability of losing all his games = (1/2)⁵ = 1/32.

Expected gain E(x) = 1x(31/32) – (31)(1/32) = 0

(ii) John places bets in the manner 1, 2, 4, 8, …2^n-1, where 1+ 2 + 4 + 8, …+ 2^n-1 = 2ⁿ – 1, With $ 511, Joe can place 9 bets as 2⁹ – 1 = 511.

Probability of winning last bet and getting $1 profit = (1/2)⁸ (1/2) = 1/512.

Probability of losing all his money = probability of losing all his games = (1/2)⁹ = 1/512.

Expected gain E(x) = 1x(511/512) – (511)(1/512) = 0

(iii) John places bets in the manner 1, 2, 4, 8, …2^n-1, where 1+ 2 + 4 + 8, …+ 2^n-1 = 2ⁿ – 1, With $ 1048575, Joe can place 20 bets as 2²⁰ – 1 = 1048575.

Probability of winning last bet and getting $1 profit = (1/2)¹⁹ (1/2) = 1/1048576.

Probability of losing all his money = probability of losing all his games = (1/2)²⁰ = 1/1048576.

Expected gain E(x) = 1x(1048575/1048576) – (1048575)(1/1048576) = 0

(iv) With infinite money, he can place infinite bets.

Probability of winning last bet and getting $1 profit = (1/2)^n-1 (1/2) = (1/2)ⁿ = 0, where n is infinite.

Probability of losing all his money = probability of losing all his games = (1/2)ⁿ = 0 as n is infinite.

Expected gain E(x) = 1[1 – (1/2)ⁿ] – (2ⁿ – 1)[(1/2)ⁿ] = 0. This answer is 0 and independent of n, even if we try for infinite times.