Section on: Discrete Probability, Expected Value and Variances Suppose that in g
ID: 3231268 • Letter: S
Question
Section on: Discrete Probability, Expected Value and Variances
Suppose that in game of chance, you bet an initial amount of money, and every time you play, the money is doubled with probability p and halved with probability 1 - p What value of p gives you an expected value of your money equal to its value before the play? With this value of p, start the game with $100 and play the game for 5 plays by (using a random number generator app), listing the value of your money after each play Do three more 5-play games each starting with $100 and list the value after each play for each game.Explanation / Answer
5. (a) Let x be the initial of money and the expected value of money.
The money is doubled (2x) with probability p and halved (x/2) with probability 1 - p.
The Probability distribution for X is
E (X) = xipi
x = (2x) (p) + (x/2) (1 - p)
x = 2x [p + (1 - p)/4]
x / 2x = p + 1/4 - p/4
1/2 = (4p + 1 - p) / 4
4/2 = 3p +1
2 = 3p +1
3p = 2 - 1
p = 1/3
p = 1/3 gives the expected value of money equal to the value before the play.
(b) First 5-play game
E (X) = xipi = (50) (2/3) + (100) (1/3) + (200) (1/3) + (100) (2/3) + (200) (1/3) = $ 266.67
Expected value of money at the end of the first game = $ 266.67
Second 5-play game
E (X) = xipi = (50) (2/3) + (100) (1/3) + (200) (1/3) + (100) (2/3) + (50) (2/3) = $ 333.33
Expected value of money at the end of the second game = $ 333.33
Third 5-play game
E (X) = xipi = (200) (1/3) + (100) (2/3) + (50) (2/3) + (25) (2/3) + (12.5) (2/3) = $ 191.67
Expected value of money at the end of the third game = $ 191.67
Fourth 5-play game
E (X) = xipi = (50) (2/3) + (100) (1/3) + (50) (2/3) + (25) (2/3) + (50) (1/3) = $ 133.33
Expected value of money at the end of the fourth game = $ 133.33
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