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.