Independent die roll

#1
Hi i have this problem, where a fair die is rolled until the same number occurs 2 successive times. Let X be the trial on which the repeat occurs, X=10 find f(x)=p(X=x)

For example, 1,2,3,4,5,6,1,5,3,3

I'm thinking the answer is as follows:
f(x)=P(X=x)=1/n for i=1,2,….,n
Ans: X=10=n=1/10

am i correct or absolutely incorrect heh ?
 

Dason

Ambassador to the humans
#2
consider the sequence 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

It hasn't had the same number 2 success times but goes much longer than what you have listed as the max (10)
 

BGM

TS Contributor
#9
Do you understand the rationale behind? How do asterisk come up with that answer?

If yes you should not in doubt.
 
#10
I dont understand what you mean "if yes you should not in doubt" but the number after the asterisk is the probability of rolling any number on a die, 1/6 since there are 6 numbers on a die.

My question is why are we using 5/6 ? this is the probability that any OTHER different number is rolled right
 
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#11
1/6 is the probability of matching the previous roll. (stop rolling)
5/6 is the probability of not matching the previous roll. (continue rolling)

This is basically a negative binomial distribution, except on the first roll probability of stopping = 0 and the probability of continuing is 1.

\({k + r -1}\choose{k} \)\( * (1 - p)^r * p^k\)

where

r= number of failures = 1
k \(\in\) {1,2,3 ... } = number of successes
p = probability of success = 5/6
(1 - p) = probability of failure = 1/6

By convention we usually call the stopping event (rolling the same number twice in a row) a failure. It may seem more natural to call this a success, but it just means the stopping event.

Because r=1 the first term simplifies to \({k + r -1}\choose{k}\) = \({k}\choose{k}\) = 1

So so for P(X = 10) we need

SSSSSSSSSF = \(1 * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{5}{6} * \frac{1}{6}\)
 
#13
I think you made a mistake wouldnt p(x=1)=1/6 and p(x=2)=(5/6)*(1/6) and p(x=3)=(5/6)^2 * (1/6) and so on so that its P(X = x) = (5/6)(x-1) × (1/6) or is this just in the case where if you are looking to roll a specific number for example a 5 instead of a repeat which is whats asked in the question ?
 
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#14
I don't see how p(x=1) could equal 1/6.

How could you roll the same number twice on the first roll? Unless you are not counting the first roll... then it is just semantics.