Your confusing the notation. In this e-mail E[X] represented the expected number of numbers that land in the correct position after X hits.
In the previous e-mail, E[X] represented the number of hits necessary to sort X unsorted numbers. So the difference between the 2 e-mails is that the E[X] had a different meaning in each. But in both cases he was correct. Imagine a game where you have equal chances of winning and losing. When you win I give you £2, when I win nothing happens. The expectation of each game is then that I will give you £1, even though that outcome is impossible in a single game. On Saturday, May 14, 2011, Eagle <[email protected]> wrote: > @Pedro, > Are you contradicting yourself or the contest analysis, in the > following statement? >> E[X] = p(X=0) * 0 + p(X=2) * 2 = 0.5 * 0 + 0.5 * 2 = 1 > Until now, you were saying E[2] = 2. Now, you are saying it is 1. > Check your last mail to me, you have given following- >> E[n=3] = 1 + (1/6 * 0 + 0 * E[n=1] + 1/2 * E[n=2] + 1/3 * E[n=3]) = 1 + >> (1/2 * E[n=2] + 1/3 * E[n=3]) > The dimension (units) on the LHS & RHS of the equation are not > matching. I have read the links about > expected value that you have given. The problem is, I am also having > the same feeling (please do not > get offended) that you are not using the concept properly. > Thanks again for your patience to continue the discussion. > > Eagle > > > > On May 14, 8:27 pm, Pedro Osório <[email protected]> wrote: >> You are saying that it is impossible to put "exactly one element into the >> right position" when N=2, right? I agree, obviously, we have 2 elements in >> the right position, or 0 elements in the right position. >> >> What you don't seem to understand is that this doesn't contradict the >> statement "The expected number of elements that are put into position is >> exactly 1". The reason why you don't understand it, is because you don't >> understand the concept of expected value (I won't link it again, since you >> refuse to read and try to understand what it means). >> >> What baffles me is that for the case N=2, it is trivial to understand that, >> even though "having exactly one element in the right position" is >> impossible, the expected value of elements put into the right position is >> exactly 1: >> >> Initial: 2 1 >> >> Possible outcomes: >> 1 2 - 0.5 probability - 2 elements in the right position >> 2 1 - 0.5 probability - 0 elements in the right position >> >> X - elements that go into the right position after one hit for the array [2 >> 1] >> >> E[X] = sum(p(X=x) * x) = p(X=0) * 0 + p(X=1) * 1 + p(X=2) * 2 + ... = >> >> All elements with p(X=x)=0 will be eliminated so only x=0 and x=2 matter (as >> you have repeatedly said): >> >> E[X] = p(X=0) * 0 + p(X=2) * 2 = 0.5 * 0 + 0.5 * 2 = 1 >> >> Do you understand now that p(X=E[X]) can be 0? > > -- > You received this message because you are subscribed to the Google Groups > "google-codejam" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-code?hl=en. > > -- Paul Smith [email protected] -- You received this message because you are subscribed to the Google Groups "google-codejam" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-code?hl=en.
