# Help! I need the expected value for a tricky situation!

 P: 330 See http://en.wikipedia.org/wiki/Stirlin...he_second_kind How many ways are there to have exactly 10 different numbers out of the 60? C(256,10) * S(60,10) * 10! So the average is $\frac{1}{256^60} \sum_{i=1}^60 i * C(256,i) * S(60,i) * 10!$
 P: 327 Let $$X_i = 1$$ if number i is drawn at least once, = 0 otherwise, for i = 1, 2, 3, ..., 256. Then $$P(X_i = 1) = 1 - (1 - 1/256)^{60}$$ and $$E(X_i) = 1 - (1 - 1/256)^{60}$$ So the expected number of distinct numbers drawn is $$E(\sum_{i=1}^{256} X_i) = \sum_{i=1}^{256} E(X_i) = 256 \; [1 - (1 - 1/256)^{60}]$$ Here we have made use of the theorem that E(X+Y) = E(X) + E(Y). It's important to realize that this theorem holds even if X and Y are not independent, which is good for us in this case because the X_i's are not independent.
 Quote by awkward Let $$X_i = 1$$ if number i is drawn at least once, = 0 otherwise, for i = 1, 2, 3, ..., 256. Then $$P(X_i = 1) = 1 - (1 - 1/256)^{60}$$ and $$E(X_i) = 1 - (1 - 1/256)^{60}$$ So the expected number of distinct numbers drawn is $$E(\sum_{i=1}^{256} X_i) = \sum_{i=1}^{256} E(X_i) = 256 \; [1 - (1 - 1/256)^{60}]$$ Here we have made use of the theorem that E(X+Y) = E(X) + E(Y). It's important to realize that this theorem holds even if X and Y are not independent, which is good for us in this case because the X_i's are not independent.