Re: doubt

Posted by MI on May 28, 2013; 11:30am
URL: http://discussion-forum.276.s1.nabble.com/doubt-tp7581374p7581381.html

probability that game gets finished in 1 chance P(1)

= (1/6)^4  x [  1+  5 x 4C1 + 25 x 4C2 + 125 x 4C3 ]

= (1/6)^4 x [670]

probability that game gets finished in 1 chance P(2)

= (1/6)^4 x [670] x (5/6)^4

Here a= (1/6)^4 x [670]    r= (5/6)^4

So expected value

= 1 p(1) + 2 p(2) +................

= 1 x a + 2 x a x r + 3 x a x (r)^2 +.................



Expected value = a / (square of (1-r))

=  1.92

approximately 2


 I am expecting some smart way to do this....but not able to get it. Pleas share if someone finds out.