# probabilty

• Oct 13th 2012, 12:29 AM
nitin1
probabilty
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. thanks
• Oct 13th 2012, 05:46 AM
Nappy
Re: probabilty
The expected value of the dice face will me 3.5= (1+2+3+4+5+6)/6 . So you should move up in multiples of 3.5 until you hit an even number:

1st roll expected value = 3.5

2nd roll expected value = 7(3.5+3.5)

3rd roll expeacted value = 10.5

4th roll = 14 = 3.5x4= An even number

So the expected amount of runs is 4.
• Oct 13th 2012, 08:37 AM
awkward
Re: probabilty
Quote:

Originally Posted by nitin1
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. thanks

Let's say the expected number of rolls is E. With probability 1/2, you roll an even number on your first roll and the number of rolls required is 1. Also with probability 1/2, you roll an odd number on your first roll; you are then right back where you started, except that you have rolled the die once, so the expected number of rolls is 1+E. Therefore

$\displaystyle E = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot (1 + E)$

Solve this equation for E.

(It's also possible to find an infinite series for the expected number of rolls and sum the series, but you should get the same result either way.)

P.S. The singular of "dice" is "die". No one ever said English makes sense.