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

17,505 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

86 votes

Best answer

Probability on each branch $= x = \frac{1}{2}$

2nd toss onwards, each toss layer gives us two success. (i.e. HH event or TT event )

$$\begin{align} E &= \sum k.p(k) \\ &= 2.(2x^2) + 3.(2x^3) + 4.(2x^4) + 5.(2x^5) + ... \\ &= 2.\left [ 2x^{2} + 3x^{3} + 4x^{4} + 5x^{5} + .... \right ] \\ &= 2.\left [ \frac{x}{(1-x)^2 } - x \right ] \\ \\ &\text{ putting x = } \frac{1}{2} \ \ \text{ ;}\\ \\ &= 2.\left [ \frac{\frac{1}{2}}{\left(\frac{1}{2}\right )^2} - \frac{1}{2} \right ] \\ &=3 \end{align}$$

A very similar QS :

An unbiased coin is tossed repeatedly and outcomes are recorded. What is the expected no of toss to get HT ( one head and one tail consecutively) ?

Probability in each branch $=0.5$. I double circled the satisfying toss events.

While observing the diagram I noticed that, from 2nd toss onward our required event starts showing up. Additionally,

1. in the $\text{2nd}$ toss (or the 3rd level) we have one satisfying case.

2. in the $\text{3rd}$ toss (or the 4th level) we have two satisfying case.

3. in the $\text{4th}$ toss (or the 5th level) we have three satisfying case.

4. in the $\text{5th}$ toss (or the 6th level) we have four satisfying case.

5. etc.

i.e. in the $\text{kth}$ toss we would have $(k-1)$ satisfying case.

So,

$$\begin{align} E(x) &= \sum_{k=2}^{\infty } k.P(k)\\\ &= \sum_{k=2}^{\infty } k.\left \{ (k-1)*(0.5)^k \right \}\\ &= \sum_{k=2}^{\infty } \left \{ (k^2-k)*(0.5)^k \right \}\\ \end{align}$$

Using geometric series identity : https://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series

$$\begin{align} \sum_{k=2}^{\infty}k(k-1)x^{k-2} = \frac{2}{(1-x)^3}\ \ \text{for } |x| < 1 \\ \end{align}$$

In our case : $x = 0.5$ So,

$$\begin{align} E &= \sum_{k=2}^{\infty}k(k-1)x^{k} = x^2\sum_{k=2}^{\infty}k(k-1)x^{k-2} = x^2.\frac{2}{(1-x)^3} \\ \end{align}$$

putting $x = \frac{1}{2}$ ; we get $E = 4$

More example:

For consecutive two heads ; HH

By drawing the tree diagram we can find the following series :

$$\begin{align*} E &= \sum{k.P(k) } \\ &=2.(1.x^2) + 3.(1.x^3) + 4.(2.x^4) + 5.(3.x^5)+6.(5.x^6)+7.(8.x^7)+.....\infty\\ \end{align*}$$

Above series is a nice combination of AP , generating function and Fibonacci numbers !!!!

- AP terms can be handled by integration or differentiation
- Fibanacci Generating function is = $\begin{align*} \frac{1}{1-x-x^2} \end{align*}$

$$\begin{align} &\Rightarrow \frac{E}{x} =2.(1.x^1) + 3.(1.x^2) + 4.(2.x^3) + 5.(3.x^4)+6.(5.x^5)+7.(8.x^6)+.....\infty\\ &\Rightarrow \int \frac{E}{x} .dx = 1.x^2+1.x^3+2.x^4+3.x^5+5.x^6+.....\infty \\ &\Rightarrow \int \frac{E}{x} .dx = x^2.\left ( 1.x^0+1.x^1+2.x^2+3.x^3+5.x^4+.....\infty \right ) \\ &\Rightarrow \int \frac{E}{x} .dx = \frac{x^2}{1-x-x^2} \\ &\Rightarrow \frac{E}{x} = \frac{\mathrm{d}}{\mathrm{d} x}\left [ \frac{x^2}{1-x-x^2} \right ] \\ &\Rightarrow \frac{E}{x} = \frac{2x(1-x-x^2)+(1+2x)x^2}{(1-x-x^2)^2} \\ &\Rightarrow E = x.\left \{ \frac{2x(1-x-x^2)+(1+2x)x^2}{(1-x-x^2)^2} \right \} \\ &\Rightarrow E = \frac{1}{2}.\left \{ \frac{2.\frac{1}{2}(1-\frac{1}{2}-\frac{1}{4})+(1+2.\frac{1}{2}).\frac{1}{4}}{(1-\frac{1}{2}-\frac{1}{4})^2} \right \} \\ &\Rightarrow E = 6 \\ \end{align}$$

Infact 2nd QS on HT can also be done in the above way using integration.

Correct Answer: $A$

0

@dd When we have three toss, we don't get two consecutive head or tail in 2nd toss. Then in 3rd toss the outcomes possible are HTT, HTH, THT, THH and remaining four outcomes will no be possible because we did not toss coin if we get two consecutive head(HH) or tail(TT) in second toss . In this 2 cases are preferable, then the probability =2/4=½ but you have taken probability 2$x^{3}$ =2/8.

Please correct me.

Please correct me.

0

102 votes

**Answer is (A)**

$E(X)= \sum X_{i} \times P_{i}$

Where X=no of tosses when you get successive HEAD/TAIL(only one is possible at a time though).

$P_{i}$=Probability that you get in $X_{i}$ tosses.

Now see solution:

You need atleast 2 tosses to get 2 heads/tails. Now see if you throw twice probability to get 2 heads/tails is $\dfrac{1}{2}$ out of $4$ outcomes $[HT,HH,TH,TT].$

Similarly if you get result in $3rd$ toss that means you did not get in $2nd$ toss so favourable cases for this can be $THH$ and $HTT$ only out of total $8$ outcomes. So probability is $\dfrac{2}{8}=\dfrac{1}{2^{2}}.$

To generalize ,you can see that in every case you will have only two favourable cases and $2^{n}$ sample space. So for n th throw probability is $\dfrac{1}{(2^{n-1})}.$

Now coming to $E(X)= 2\times \dfrac{1}{2} + 3\times \dfrac{1}{4} + 4\times \dfrac{1}{8}+\ldots \text{till infinity}.$

See this is combined AP-GP, So multiplying E(X) by $\dfrac{1}{2}$ and subtracting from E(X).

$E(X)=2\times \dfrac{1}{2}+ 3\times {1}{4} +4 \times \dfrac{1}{8}+\ldots$

${0.5}\times E(X)=2\times \dfrac{1}{4} +3\times \dfrac{1}{8}+\ldots$

subtracting, we get $\frac{1}{2} \times E(X)= 1+\frac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\ldots$

${0.5} \times E(X)= 1+ \left(\dfrac{1}{4}\right)\div {\left(1-{0.5}\right)}= 1+\dfrac{1}{2} =\dfrac{3}{2}$ $\left(\frac{a}{1-r}\right)$

$ E(x)= 3.$

**doubt** :

When I toss 3^{rd} time, that means I definitely didn’t get HH or TT last time..so I think the sample space for third time should not include** HHH ,HHT ,TTH ,TTT** that means

for 3^{rd} time favourable outcomes are **HTT, THH** only out of 2^3 – 4(removing not favourable) = 4

so 3^{rd} time probability of getting two successive outcomes should be 2 / 4….but you are taking 2 /8 ?

@Sandeep_Uniyal can you please explain?

2

4 votes

Let the expected number of coin flips be *X*.

The case analysis goes as follows:

a. If the first and second flips are HH or TT then we are done. The probability of this event is 2*(1/4) = 1/2

and the total number of flips required is *2*.

b.if the first and second flips are TH or HT then we wasted 2 flips. The probability of this event is 2*(1/4) = 1/2

and the total number of flips required is *X+2*.

Adding, the equation we get -

*X*=1/2(*2*)+1/2(*X*+2)

solving equation we get **X = 3**

**answer : 3**

@puja

HT or TH is a fail

case HT = prob of H =1/2 and prob of T=1..therefore 1/2...here 2 flips wasted (X+2)

case TH = prob of t =1/2 and prob of H=1..therefore 1/2...here also 2 flips wasted(X+2)

so X = (X+2)*1/2* (X+2) *1/2...but here shouldnt it be (X+2)*1/2 + (X+2)*1/2 ??

am not sure if my understanding is correct

HT or TH is a fail

case HT = prob of H =1/2 and prob of T=1..therefore 1/2...here 2 flips wasted (X+2)

case TH = prob of t =1/2 and prob of H=1..therefore 1/2...here also 2 flips wasted(X+2)

so X = (X+2)*1/2* (X+2) *1/2...but here shouldnt it be (X+2)*1/2 + (X+2)*1/2 ??

am not sure if my understanding is correct

1

3 votes

Number of expected toss to get a head (or tail), E(H) = E(T) = 2

let the expected number of tosses to get two successive same outcome is e.

e = 1/2*(1+E(H)) + 1/2*(1+E(T))

e = 1/2*(1+2) + 1/2 *(1+2)

e = 3

Therefore correct answer would be (A).

Answer is right, but procedure is wrong.

let the expected number of tosses to get two successive same outcome is e.

e = 1/2*(1+E(H)) + 1/2*(1+E(T))

e = 1/2*(1+2) + 1/2 *(1+2)

e = 3

Therefore correct answer would be (A).

Answer is right, but procedure is wrong.

Let the expected number of coin flips be x.

1. If first flip gives head and second gives tail then we have wasted 1 flip so need to do one more flip

1/2 * 1/2 * (x+1)

2. If first flip gives head and second gives tail then we are done. (required 2 flips) So.

2(1/2 * 1/2)

3. If first flip gives tail and second head then wastage = 1 flip. (Same as 1st case)

1/2 * 1/2 * (x+1)

4. If both gives tail (same as 2)

2 (1/2*1/2)

so

x = 2(1/2*1/2) + 1/2 *1/2 * (x+1) + 1/2 *1/2 *(x+1) + 2 * (1/2) * (1/2)

x= 1 + (x +1 )/ 2

x= 3

15