doubt

[maahi]

Suppose you keep rolling four dice simultaneously until at least one of them
shows a 6. What is the expected number of \rounds" (each round consisting
of a simultaneous roll of four dice) you have to play?

[anonymouse]

Hi Maahi,

This is how I did it. Let p be the probability of success given by p = 1 - (5/6)^4.

Then, the probability that the nth trial has the first success is given by p(1-p)^n-1.

Then E(X) is given by the sum (n, 1, inf) np(1-p)^n-1.

I really didn't know how to solve this, so I used a widget online to get the answer, which is approximately 1.93.

Of course we can't do this in the exam, so I tried to look online for another way to solve the question, and found out that the question describes a geometric distribution (I don't think I've encountered it before), for which the expected value is simply 1/p.

So, the answer is 1296/691, which is approximately 1.93.


tl;dr: The question describes a geometric distribution of the number of trials needed to get a success (at least one 6). So, E(X) = 1/p = 1.93, where p = 1 - (5/6)^4.

[pbansal]

i v also done it in same manner....its 1296/671=1.93

[MI]

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.

[pbansal]

@MI
THIS FOLLOWS GEOMETRIC DISTRIBUTION WITH P=1-(5/6)^4
I.E 1-NONE
AND GEOMETREIC DIST.. WITH X>1 FOLLOWS E(x)=1/P
YOU CAN HAVE MORE OF SUCH QUESTIONS IN ANY STATS BOOK

[MI]

Thanks :D

[maahi]

thanks so much for answering my query

THIS FOLLOWS GEOMETRIC DISTRIBUTION WITH P=1-(5/6)^4  
can u explain d intuition of how to get this value of p ??/?

[anonymouse]

@ maahi

It's because P(getting at least one six) = 1 - P(getting no six)

and P(getting no six) = (5/6)^4

[maahi]

i don't know if its right but i understand this that getting no six in one trial is 5/6 and hence in 4 being independent it is 5/6 ^4 and total prob is 1 . taht is why we subtract it ??

[maahi]

q37 dse 2007 http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/30_jun_2007_option_a.pdf

 i get option b as d ans. i m going somewhere wrong . please help .

[anonymouse]

You have understood it correctly. Also, the solution of Q.37 has been posted previously.

[maahi]

thank you . i keep on taking x as the savings .