GATE Overflow Test Series | Probability | Test 1 | Question: 27

Question

An unbiased coin is tossed repeatedly and its outcome is recorded. What is the expected number of trials to get $\text{HT}$ consecutively?

Comments

https://math.stackexchange.com/questions/521130/expected-value-of-flips-until-ht-consecutively

Similar question: https://gateoverflow.in/3778/Gate-it-2005-question-32

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https://www.codechef.com/wiki/tutorial-expectation

Go through this link, you will find similar kind of problems, you can have better understanding. 

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@jaswanth431 Thanks for sharing :) This is a really beautiful application of recurrence in a probabilistic world, Saved tons of calculations of infinite series.

Answer 1

Expected number of tosses to get a Head is $E(H) = 1 + 0.5 \times E(H) \implies E(H) = 2.$

Similarly, $E(T) = 2.$

Now, $E(HT) =1+ 0.5E(T) +0.5E(HT) $

$\implies 0.5E(HT) = 2 \implies E(HT) = 4.$

Comments

@Arjun sir, i am not able to understand these kind of question, can you please share any good reference?

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Using infinite series;

 $2*P(HT) + 3*(P(THT)+P(HHT))+4*(P(HHHT)+P(TTHT)+P(THHT))…..$

$ S = 2*(\frac{1}{2^2}) + 3*(2*\frac{1}{2^3}) + 4*(3*\frac{1}{2^4}) + 5*(4*\frac{1}{2^5})….. $

$ \frac{S}{2} = 2*(\frac{1}{2^3}) + 3*(2*\frac{1}{2^4}) + 4*(3*\frac{1}{2^5}) + 5*(4*\frac{1}{2^6})…. $.

$ S-\frac{S}{2} = 2*(\frac{1}{2^2}) + 2*(2*\frac{1}{2^3}) + 2*(3*\frac{1}{2^4}) + 2*(4*\frac{1}{2^5})…. $.

$ \frac{S}{2^2} = 2*(\frac{1}{2^3}) + 2*(2*\frac{1}{2^4}) + 2*(3*\frac{1}{2^5}) + 2*(4*\frac{1}{2^6})…. $.

${ \frac{S}{2}}  – {\frac{S}{2^2}} =  2*(\frac{1}{2^2}) + 2*(\frac{1}{2^3}) + 2*(\frac{1}{2^4}) + 2*(\frac{1}{2^5})…. $
 

$ \frac{S}{4} = 2 * (\frac{\frac{1}{2^2}}{\frac{1}{2}}) $

$ S = 4 $

Answer 2

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Answer 3

As everyone is trying: let me share my approach too : 

E[HT] = (1/4)*E[TT] + (1/4)*E[TH] + (1/4)*E[HH] + (1/4)*2

E[HH] = E[T] + 2

E[TH] = E[T] + 2

E[TT] = E[HT] + 2

Now, the only issue is to find E[T] : 

E[T] = (1/2) * (E[T] + 1) + (1/4) * 1

which implies → E[T] = 2

Putting these values in the above expression we get E[HT] = 4

Comments

Can you explain how you got this below equation or what the below equation means?

$E[HT] = \frac{1}{4}*E[TT] + \frac{1}{4}*E[TH] + \frac{1}{4}*E[HH] + \frac{1}{4} * 2$

From your solution, $E[HT]$ means expected number of trails to get $HT$ consecutively.

Similarly, $E[TT], E[HH], E[TH]$ should mean expected number of trails to get $TT$ consecutively, $HH$ consecutively and $TH$ consecutively, respectively.

On separately solving we get $E[TT] = E[HH] = 6$ and $E[TH] =4$

Putting these values in your equation we end up with $E[HT] = 4.5$