in Probability
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60 votes

An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is

  1. $3$
  2. $4$
  3. $5$
  4. $6$
in Probability


Ok! Last thing: We write $\frac{1}{4}*(k’+2)$ in equation $(1)$ so that:  “to get one T, expected no. of tosses is $k’$ else $2$ turns are wasted”. Is my reasoning correct @ankitgupta.1729 Sir?


yes. You have divided the cases like $HT,HH,T$. Now for $HH$, your motive is to get $HT$ but you got H at the end in HH already, so you only have to get “first” tail. So, you have to introduce one new variable, say, $k’$  in your equation and $+2$ shows you have wasted $2$ coin flips in $HH$. 

I have missed the word “first” while describing $k’.$ So,

$k’$ is expected number of tosses to get “first” tail.

All these things come from the Law of Total Expectation.

Thanks :)

7 Answers

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This can be solved using recurrence relation as follows:
Let X be the expected value of number of tosses until the outcomes of 2 tosses are same.
We win when outcomes are HH,TT and we lose when outcomes are HT,TH.


So Prob(win) = 1/2 and Prob(lose) = 1/2

In the case of winning after first coin is tossed it needs one more toss to win while in the case of lose, after tossing first time we need to toss X more times as we could not win.
This gives us a Recurrence relation as follows:

X = 1/2∗(1+1) + 1/2∗(1+X)
2X = 2+1+X
X = 3
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This is going to be a long one. Let’s start.

Definition – $\bf{X}$ :- This is a random variable which denotes the number of tosses required to get two consecutive tosses with the same outcome.

$P(X=1) = 0$ . This is because we need atleast two tosses to get two consecutive tosses.

$(P(X=2)=$ {$H,H$} or {$T,T$} = $\frac{1}{2}$

Notice that for any sequence of coin tosses, if the last two outcomes are {$T,T$} or {$H,H$} then $\exists$ only one valid sequence for the preceding part of it. Hence for every value of $X$, there are only two valid sequences.

For example if ($X=5$) and the last two outcomes are the same, then the possibilities are : $\color{red}{xxx}TT$ and $\color{red}{xxx}HH$

There is only possibility for $\color{red}{xxx}$ for each of the outcomes, which are $\color{red}{HTH}TT$ and $\color{red}{THT}HH$ respectively. Otherwise, we’d get a smaller sequence with same outcomes which refutes our definition of $X$.

Therefore $P(X=n)=\frac{2}{2^n}=\frac{1}{2^{n-1}}$ …. $\bf{(i)} $ for $ n \geq 2 $

We know, $\bf{E[X]}= \sum_{x} x \times p(x)$

Let us denote $\bf{E[X]}$ by S.

$ S= P(X=1) \times 1 + P(X=2) \times 2+ …… $to $\infty$

$S= 0+ 2 \times \frac{1}{2^1} + 3 \times \frac{1}{2^2} + 4 \times \frac{1}{2^3}… $ to $\infty$ – $\bf(iii)$

$2S= 0+ 2 \times \frac{1}{2^1} \times 2 + 3 \times \frac{1}{2^2} \times 2 + 4 \times \frac{1}{2^3} \times 2… $ to $\infty$ -$(\bf{iv})$

Subtracting $\bf{(iv)}$ from $\bf({iii})$ we have :

$S= 2 + \frac{1}{2}+ (3-2)\frac{1}{2^2}+ (4-3)\frac{1}{2^3}+ \frac{1}{2^4} ….$ to $\infty$

$S= 2 + \frac{\frac{1}{2}}{1-\frac{1}{2}}=2+1=3$

Hence, the expected value $\bf{E[X]}$=3.

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