3. Consider two games at a village fair, Game A and Game B. Let the gain in doll

Question

3. Consider two games at a village fair, Game A and Game B. Let the gain in dollars in one play of Game A be denoted by the random variable X, and let the gain in dollars in one play of Game B be denoted by the random variable (A negative gain means that you receive less money back from the game than you paid to take part.) The probability distributions of Xand Y are given below. -5 -2 1 4 10 Probability 0.33 0.27 0.20 0.13 0.07 Y -5 -2 1 4 25 Probability 0.37 0.32 0.16 0.10 0.05 You may assume that the result of any play of either of the games is independent of the result of any other play of either of the games. (a) In any one play of Game A, what is the probability that your gain will be positive? In any one play of Game B, what is the probability that your gain will be positive?

Explanation / Answer

a) P(positive gain in game A) = P(X=1) + P(X=4) + P(X=10) = 0.20 + 0.13 + 0.07 = 0.4
P(positive gain in game B) = P(Y=1) + P(Y=4) + P(Y=25) = 0.16 + 0.10 + 0.05 = 0.31

b) For game B

E[Y] = Y*P(Y) = -5*0.37 + -2*0.32 + 1*0.16 + 4*0.10 + 25*.05 = $-0.68

= sqrt((Y-E[Y])2*P(Y))
= sqrt((-5+0.68)2*0.37 + (-2+0.68)2*0.32 + (1+0.68)2*0.16 + (4+0.68)2*0.10 + (25+0.68)2*0.05
=$6.56

c) No, the expected value of Game B is less negative than game A and is a safer bet in the long run.

The probability doesnt take into account the amount of gain, but only the probability. Expected value takes into account the gain value also and therefore is a better indicator.

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