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Ram has a fair coin, i.e., a toss of the coin results in either head or tail and each event happens with probability exactly half $(1/2)$. He repeatedly tosses the coin until he gets heads in two consecutive tosses. The expected number of coin tosses that Ram does is.

  1. $2$
  2. $4$
  3. $6$
  4. $8$
  5. None of the above
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19 votes
19 votes

Answer is (C)

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

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

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

  • AP terms can be handled by integration or differentiation
  • Fibonacci generating function is $= \begin{align} \frac{1}{1-x-x^2} \end{align}$
  • probability on each branch is $x = \frac{1}{2}$

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

Similar Kind of Question as a Reference

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29 votes
29 votes

Let x be the Expected Number of tosses.

If we get a tail immediately (Probability $\frac{1}{2}$ ) then the Expected number = (x+1)

If we get a head then a tail (Probability $\frac{1}{4}$ ) then the Expected number = (x+2)

If first 2 tosses are head then the Expected number  =2

Thus 

x = $\frac{1}{2}$ (x+1) + $\frac{1}{4}$ (x+2) + $\frac{1}{4}$ 2 

$\frac{1}{4}$ x = $\frac{3}{2}$ x

x=6

Hence (c) 6 is the Answer.

2 votes
2 votes

The expected number of coin flips required to obtain $n$ consecutive heads  $= E_n = 2(2^n-1)$

here $n=2$ $\implies E_2 = 2(2^2-1) = 2*3=6$

So option $C.$ is correct.


For details :-

https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads

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0 votes
If you simply want the number of tosses for consecutive heads or tails, there is a shortcut formula ie En = 2(2^n-1) and here in our case n is 2(as we need two consecutive tosses).
Answer:

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