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I want to solve following question on probability distribution:

A coin is tossed repeatedly and probability that a head appears at any toss is $p$ where $0<p<1$. What is the expected length of initial run ?


A run of length $r$ is $r$ consecutive toss that are all the same. (e.g. $HTTHH$ has a run of length $1$ and two runs of length $2$).

Let us assume, we need $r$ tosses until first head appears, then the probability distribution function $$m(r)=(1-p)^{r-1}p.$$ If $t$ be the time of first head, then expectation of head $H$ is $$E(t)=1 \cdot p+(1-p)p+\cdots.$$ If $r=1$, then always the length of run is $1$.

If $r=2$, then we have $HH,TT,TH,HT$, so there are two runs ($HH, TT$) of length $2$ half of the time and two runs $(HT,TH$) of length $1$ half of the time.

But I am confused with the term "initial run" in my question. What does mean it ?

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    $\begingroup$ If the first toss is heads, the initial run refers to the number of heads obtained before the first tail appears; if the first toss is tails, the initial run refers to the number of tails obtained before the first head appears. $\endgroup$ Jan 7, 2023 at 3:16
  • $\begingroup$ @N.F.Taussig, thanks. so what would be the expected length ? Is it $p(1-p)^{r-1}$ at $rth$ toss ? $\endgroup$
    – MAS
    Jan 7, 2023 at 3:24

2 Answers 2

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"But I am confused with the term "initial run" in my question. What does mean it ?"

In e.g. $HHTHTT$ the initial run is $HH$, in e.g. $TTTHHT$ the initial run is $TTT$.

I hope that these examples are enough to make clear what is meant by "initial run".


Problems of this sort can often be solved without looking too much at distributions.

Let $\mu$ denote the expectation of the length of the first run.

Let $\mu_H$ denote the expectation of the length of the first run under condition that the first toss results in heads.

Let $\mu_T$ denote the expectation of the length of the first run under condition that the first toss results in tails.

Defining $q:=1-p$ we find:$$\mu_H=p(1+\mu_H)+q1=1+p\mu_H\text{ and }\mu_T=q(1+\mu_T)+p1=1+q\mu_T$$leading to:$$\mu_H=q^{-1}\text{ and }\mu_T=p^{-1}$$ Then finally we find:$$\mu=p\mu_H+q\mu_T=pq^{-1}+qp^{-1}$$Of course you can make it an expression in $p$ only by substituting $q=1-p$.

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  • $\begingroup$ Thanks. But what is the distribution function, in this case ? I wrote one. Is it correct? $\endgroup$
    – MAS
    Jan 8, 2023 at 7:32
  • $\begingroup$ Can you explain how did you find $\mu_H=p(1+\mu_H)+q\cdot 1$ ? $\endgroup$
    – MAS
    Jan 8, 2023 at 7:48
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    $\begingroup$ $P(N=n)=P(HH\cdots HT)+P(TT\cdots TH)=p^nq+q^np$ for positive integer $n$. Both strings have length $n+1$. Later I will try to react on your second comment. $\endgroup$
    – drhab
    Jan 8, 2023 at 7:51
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    $\begingroup$ Let $N$ denote the length of the initial run and let $H,HH,HT$ denote the events of starting with one head, two heads, head followed by tail respectively. Then: $$\mu_{H}:=\mathbb{E}\left[N\mid H\right]=P\left(H\right)\mathbb{E}\left[N\mid HH\right]+P\left(T\right)\mathbb{E}\left[N|HT\right]=p\mathbb{E}\left[N\mid HH\right]+q\mathbb{E}\left[N|HT\right]$$ This with $\mathbb{E}\left[N\mid HH\right]=1+\mathbb{E}\left[N\mid H\right]=1+\mu_{H}$ and $\mathbb{E}\left[N\mid HT\right]=1$. If this is outside your scope then just use the distribution given in my former comment to find $\mathbb EN$. $\endgroup$
    – drhab
    Jan 8, 2023 at 14:44
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$E$(length of run)$ = (1 + $E(length until we see tails given first is head)$)*p+ (1 + $E(length until we see heads given first is tails) $)*(1-p)$

$E = p*(1 + 1/(1-p)) + (1-p)*(1 + 1/p) $ [1/p => expected number of tosses until we see a head and likewise for tails also]

$= p(2-p)/(1-p) + (1-p)(p+1)/p = (2p^2 - p^3 + 1 - p^2 - p + p^3)/(p(1-p))$

Hence, $E = (p^2 - p + 1)/(p*(1-p))$ tosses

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