@Shady: There is a 1/2 probability of getting an even number on the
first roll. In the 1/2 of the cases where the first roll is odd, roll
again and note that there is a 1/2 chance of getting an odd number.
Etc. The probability of needing n rolls is 1/2 to the nth power.
The series is 1 + 2 * (1/2) + 3 * (1/4) + 4 * (1/8) + ....
Call the infinite sum S:
S = 1 + 2 * (1/2) + 3 * (1/4) + 4 * (1/8) + ....
Then 1/2 S = 1 * (1/2) + 2 * (1/4) + 3 * (1/8) + ..... Subtract
the two series, getting
1/2 S = 1 + 1/2 + 1/4 + 1/8 + ....
So 1/2 S = 2, and S = 4.
The expected number of rolls needed to that the sum is even is 4.
Dave
On Aug 6, 9:34 am, 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....
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