Probability of a "run" of coin tosses.

Question

Given probability of tails is p, so heads is 1-p. Define X as a random variable for the length of a run (X=5 is either TTTTTH or HHHHHT). Find pmf.

So I think a run of just heads or just tails is pretty straightforward. A run of tails being, $P(X=x)=p(1-p)^x$ and a run of heads being $P(X=x)=(1-p)p^x$. I'm just not really sure how I would combine the two. At first I considered the pmf being $p(1-p)^x+(1-p)p^x$, But $\sum_{x=0}^\infty p(1-p)^x+(1-p)p^x=2$, so it's not a valid pmf. Any help on this one?

The random variable $X$ has not been fully described. Do you mean the length of the run we get starting with the first toss? So a string that starts with $TTHHHHHTTTH$ would give $X=2$? — André Nicolas, Oct 07 '13 at 21:26
Sorry, it's started by the first trial. Yes, what you described would be equivalent to $TTH$, so $X=2$. — Probability Student, Oct 07 '13 at 21:40
A somewhat similar question is here: http://math.stackexchange.com/questions/417762/probability-of-20-consecutive-success-in-100-runs/420724#420724 — awkward, Oct 07 '13 at 22:06

score 2 · Answer 1 · answered Oct 07 '13 at 21:24

2

$\textbf{Hint:}$ It is not possible to get a run of length 0.

answered Oct 07 '13 at 21:24

Arthur

5,524

Is it as simple as changing it to; $$\sum_{x=1}^\infty p(1-p)^x+(1-p)p^x$$ ? – Probability Student Oct 07 '13 at 21:35
Yes, for $x \geq 1$ you have correctly found $P(X = x) = p(1−p)^x+(1−p)p^x$, and we also have $P(X = 0) = 0$. – Arthur Oct 07 '13 at 21:39

Probability of a "run" of coin tosses.

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