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?

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Ishrat Jahan
asked
in Probability
Nov 3, 2014

17,518 views
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

- $3$
- $4$
- $5$
- $6$

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?

2

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.

2

0 votes

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

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

0 votes

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.

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.