If we expand it.. E(t) = E(t1) + E(t2) + E(t3) + ... + E(tn); here I am able to derive E(t1) as N/1 using the expression E(t1) = 1/N + ((N-1)/N)(E(t1) + 1); but how do I proceed? How do I derive the E(t2) and so on??
What do these values mean?? Does it mean like E(t2) is the no. of expected throws to get value 2?? Any help on this? On Sunday, 17 June 2012 00:09:13 UTC+5:30, amitesh wrote: > > This problem is similar to Coupan collector problem. > http://en.wikipedia.org/wiki/Coupon_collector%27s_problem > > In your case the answer is > > [image: For N-Dice ; \newline \sum_{i=1}^{N} N/i \newline for\; N =~2 ; > \newline \sum_{i=1}^{2} 2/i = 2/1 + 2/2 = 3 \newline] > > > Hope it helps! > > > -- > Amitesh > > > > > On Sat, Jun 16, 2012 at 5:18 PM, Gaurav Popli <[email protected]>wrote: > >> What is the expected number of throws of his die while it has N sides >> so that each number is rolled at least once? >> e.g >> for n=2 ans 3.00 >> n=12 ans is 37.24... >> i refrd to expectation tutuorial at >> http://www.codechef.com/wiki/tutorial-expectation but still couldnt >> get the logic... >> >> any help? >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" 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/algogeeks?hl=en. >> >> > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/xLsfC_Gc8z0J. 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/algogeeks?hl=en.
