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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. $6$

8 Answers

Best answer
95 votes
95 votes

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: $A$

edited by
112 votes
112 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.$  

edited by
4 votes
4 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.
edited by
4 votes
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

Answer:

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