prob dice question

**amitbern** · May 6th 2006, 12:45 PM

you throw 1 dice until you get K times the number '6'
X = the number of your throws.
what is P(X) ?

**rgep** · May 6th 2006, 01:37 PM

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 $\binom{N-1}{K-1} \left(\frac56\right)^{(N-1)-(K-1)} \left(\frac16\right)^{K-1} \frac16$ .

**amitbern** · May 6th 2006, 08:54 PM

how can i prove that ∑ p(X)=1 ?

**CaptainBlack** · May 6th 2006, 10:41 PM

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

**amitbern** · May 6th 2006, 10:54 PM

I know that it's supposed to be 1,
but how can i prove it mathematically (Newton’s binomial )?

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