let expected time is T , if we get even(probability 1/2) on first throw so T become 1 If dont then you are at same place where you started so T = (1/2) (1) + (1/2) ( 1+T) => T =2
On Sun, Aug 7, 2011 at 6:05 PM, rajul jain <rajuljain...@gmail.com> wrote: > If we get even number (probability 1/2) in first throw , so the expected > run is 1 > If first throw give ODD (probability 1/2) ,the run needed become 1 + run > needed until second odd number .This last waiting time is 2 . > Thus expected run become (1/2) (1) + (1/2)(1+2) = 2 > > On Sat, Aug 6, 2011 at 8:04 PM, shady <sinv...@gmail.com> wrote: > >> Hi, >> >> A fair dice is rolled. Each time the value is noted and running sum is >> maintained. What is the expected number of runs needed so that the sum is >> even ? >> Can anyone tell how to solve this problem ? as well as other related >> problems of such sort.... >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> 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 post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
