Let X be the random variable for this question. P(X=x) = 1/3 for all x. When X=x, we have that it took x times for us to retrieve the correct key.Thus, we have the infinite sum 1/3 + 22/31/3 + 3*(2/3)^21/3 + ...Factoring out 1/3 we get 1/3(1+2(2/3)+3*(2/3)^2+...)The infinite sum 1 + 2x^2 + 3x^2 can be sequentially written as 1 + x + x^2 + ... + x*(1+x+x^2+...) + x^2*(1+x+x^2+..) + ....which is equivalent to (1-x)^-2. Thus the expectation value is 1/3*(1-2/3)^-2 = 3.
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