Consider two games. In game A, each time we play we win Rs. 2 with probability 2/3 and
lose Rs.1 with probability 1/3. In game B, each time we play we win Rs.1002 with probability
2/3 and lose Rs. 2001 with probability 1/3. What is the expected winnings in both games?
What is the variance? Using Chebyshev’s inequality, compute an upper bound on the
probability that after playing 10 rounds of each game the winnings deviate by more than ±10
from the expected value.
Answers
Given : In game A, each time we play we win Rs. 2 with probability 2/3 and lose Rs.1 with probability 1/3. In game B, each time we play we win Rs.1002 with probability 2/3 and lose Rs. 2001 with probability 1/3.
To Find : What is the expected winnings in both games?
Solution:
Game A
Winning Rs 2 Probability = 2/3
Loose Rs 1 Probability = 1/3
Game B
Winning Rs 1002 Probability = 2/3
Loose Rs 2001 Probability = 1/3
A B
W W (2/3)2 + (2/3)1002 = 1004/3
W L (2/3)2 - (1/3)(2001) = -1997/3
L W -(1/3)1 + (2/3)1002 = 2003/3
L L -(1/3)1 - (1/3)(2001) = - 2002/3
Expected winning = 1004/3 - 1997/3 + 2003/3 - 2002/3
= -992/3
expected winnings in both games = -992/3
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