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- 11/30/12
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X: # Heads, X~BIN(5,1/2), E(X)=5*1/2=5/2=2.5, x = 0,1,..,5. We know the expectation of a discrete distribution is calculated by summing up the probabilities over all possible x. But here in this specific setting P(HHHHH) is not possible and P(HHHHT) is also not possible so they must be subtracted. But you also have to take into account P(HHHH).
So the expected value of this game is 2.5 - 5*(1/2)^5 - 4*(1/2)^5 + 4*(1/2)^4 = 2.5 - 1/32 = 2.46875
So the expected value of this game is 2.5 - 5*(1/2)^5 - 4*(1/2)^5 + 4*(1/2)^4 = 2.5 - 1/32 = 2.46875