TIFR CSE 2011 | Part A | Question: 6

Question

Assume that you are flipping a fair coin, i.e. probability of heads or tails is equal. Then the expected number of coin flips required to obtain two consecutive heads for the first time is.

1. $4$
2. $3$
3. $6$
4. $10$
5. $5$

Answer 1

Answer is (C)

Let the expected number of coin flips be $X$. The case analysis goes as follows:

a. If the first flip is a tails, then we have wasted one flip. The probability of this event is $\frac{1}{2}$ and the total number of flips required is $X+1.$
b. If the first flip is a heads and second flip is a tails, then we have wasted two flips. The probability of this event is $\frac{1}{4}$ and the total number of flips required is $X+2.$ as the same scenario as beginning is there even after 2 tosses.
c. If the first flip is a heads and second flip is also heads, then we are done. The probability of this event is $\frac{1}{4}$ and the total number of flips required is $2$. 

Adding, the equation that we get is -
$X = \frac{1}{2}(X+1) + \frac{1}{4} (X+2) + \frac{1}{4}2$

Solving, we get $X = 6$.

Thus, the expected number of coin flips for getting two consecutive heads is 6.

Comments

Don't we will waste two flips in the case of TT?

K=1/4(K+2)+ 1/4(K+1)+ 1/4(2) + 1/4(k+2)

     $TT\parallel TH\parallel HH\parallel HT\parallel $

K = (3K+7)/4

K=7

CORRECT ME IF SOMEWHERE WRONG.

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@Divyanshu Shukla whenever we get first T in first flip then we wasted 1 flip and will require (X+1) flip  to get answer ==> this will also cover the case of getting TT.

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@Avdhesh Singh Rana how we will solve if we want HT ?? i am getting answer as 5 but correct answer is 4.

 

Answer 2

Check the first answer from William Chen,

on -> https://www.quora.com/What-is-the-expected-number-of-coin-flips-until-you-get-two-heads-in-a-row

link found by @Anurag Pandey !

Nice answer !

Comments

Okay got it. Just want to verify my approach.

That series really doesnot go anywhere.

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yes that series never ends and also we not get a proper pattern in that series. so tey for recurrence method

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I have one doubt, the question mentions obtaining two consecutive heads for the first time. So when,

X=2, p(2) = HH = 1/2

X= 3, p(x) = now here we are tossing coins for three 3 times, so that second consecutive head appears on the 3rd toss, because if we get 2nd head on 2nd toss, we are not going to toss it for third time and that case is already covered in X=2. So

p(3) = THH= 1/8

But this approach isn't fetching me the correct answer.

Can someone point out the mistake in this approach?

Answer 3

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.

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For details :-

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

Answer 4

$2^{N+1}-2$

PUT N=2

6 IS THE ANSWER

SOURCE: GFG