# prob dice question

• May 6th 2006, 12:45 PM
amitbern
prob dice question
you throw 1 dice until you get K times the number '6'
X = the number of your throws.
what is P(X) ?
• May 6th 2006, 01:37 PM
rgep
The probability that N throws contains (K-1) '6's somewhere in the first (N-1) throws, with a '6' on the N-th throw is $\binom{N-1}{K-1} \left(\frac56\right)^{(N-1)-(K-1)} \left(\frac16\right)^{K-1} \frac16$.
• May 6th 2006, 08:54 PM
amitbern
how can i prove that ∑ p(X)=1 ?
• May 6th 2006, 10:41 PM
CaptainBlack
Quote:

Originally Posted by amitbern
how can i prove that ∑ p(X)=1 ?

I'm not quite sure what you want to prove here. This sum has to be 1
since with probability 1 you will eventually reach K sixes. So the sum
of the probabilities of the number of throws needed for this is 1.

RonL
• May 6th 2006, 10:54 PM
amitbern
I know that it's supposed to be 1,
but how can i prove it mathematically (Newton’s binomial )?