Probability of getting an even sum in one roll is 1/2.. Thus, expected number of rolls required to get even sum is inverse of that i.e. 2.
Alternatively, Going by basics... Let P(x) be probability of getting Even sum in x rolls. P(1) = 1/2....(Even) P(2) = (1/2) * (1/2) (Odd + Odd) P(3) = (1/2) * (1/2) * (1/2) (Odd + Even + Odd) P(4) = (1/2) * (1/2) * (1/2) * (1/2) (Odd + Even + Even + Odd) & So on.... Expected number is given by Summation( X*P(x) )...i.e. 1*(1/2) + 2*(1/4) + 3*(1/8) + 4*(1/16) + ....(theoretically infinite terms) Let this sum be S. Thus, S = (1/2) + (2/4) + (3/8) + (4/16) + (5/32) +.... S / 2 = (1/4) + (2/8) + (3/16) + (4/32) +.... S - (S/2) = (1/2) + (2-1)/4 + (3-2)/8 + (4-3)/16 + (5-4)/32 + .... Thus, S/2 = (1/2) + 1/4 + 1/8 + 1/16 + 1/32 + ....(theoretically infinite terms) R.H.S. is GP which evaluates to 1. Thus, S = 2. Thus, expected number of rolls required to get even sum is 2. -- 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.
