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Consider the following experiment.

Step 1. Flip a fair coin twice.

Step 2. If the outcomes are (TAILS, HEADS) then output $Y$ and stop.

Step 3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output $N$ and stop.

Step 4. If the outcomes are (TAILS, TAILS), then go to Step $1.$

The probability that the output of the experiment is $Y$ is (up to two decimal places)

A much easier way for this would be:

Let $W = \text{We get a Y}$.
Let $D = \text{The game is decided after the first two tosses}$

The crucial observation here is that if we don't finish after the first two tosses, we are facing the same game again.

Therefore, $P(W/D^c) = P(W) \implies P(W/D) = P(W)$.

This gives us that $P(W) = P(W \cap D)/P(D) = \frac{1/4}{1/4 + 2/4} = 0.33$
Sample Space = $\{Y,N\}$

Therefore, $P(Y)+P(N)=1$

Number of favourable cases for $P(Y)=1$

Number of favourable cases for $P(N)=2$

Total cases = $3$ (NOT 4)

So, $P(Y)=1/3=0.33$
is this a Markov chain?

Answer is ${0.33}$

$1^{st}$ time it is ${0.25}\left(\dfrac{1}{4}\right),$ when tail tail comes, entire process gets repeated, so next time probability of $Y$ to happen is ${0.25}\times {0.25}\left(\dfrac{1}{4}\times \dfrac{1}{4}\right),$ likewise it goes on as infinite GP

Sum of infinite GP $= \dfrac{a}{(1-r)}$

here, $a= \dfrac{1}{4}$ and $r =\dfrac {1}{4}$

so answer becomes $\dfrac{1}{3}$ i.e ${0.33}$
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@Sreyas, nice explanation,but can someone explain why the following solution isn't the answer..
q=1-p(not getting Y)=3/4

so,all together....put it as....pq+p^2q+p^3q+...... +p^nq =1/4
reshown
Why the failure of getting something other than $Y$ in the second flip is not considered for computing the probability of getting Y in next flip.
edited
let $P$ be the probability that $Y$ is printed.

one-fourth of the time we have $Y$ is printed and $\dfrac{1}{4}^{th}$ of the time the experiment is repeated all again and we have the same chance for printing $Y$ again. which can be written as,

$P=\dfrac{1}{4}+\dfrac{1}{4} P$

$\dfrac{3}{4} P = \dfrac{1}{4}$

$P=\dfrac{1}{3} = .33$
edited

Alternate approach:

If we try to draw the probability tree it shall be something like this:

Now if we see the probability tree, a certain structure is getting repeated recursively.. Now if we assume that this structure which repeats, gives us our required probability $p$ of getting $\text{print(Y)}$ in its structure, then we can calculate as follows:

Which I guess highly simplifies the calculation to just few lines...

wrong

@swami_9 Are you really dumb or just acting like one?

@JAINchiNMay First time it is $1/4$, 2nd time it is $1/4*1/4$ and so on. Sir has just written $0.25 = 1/4$.

That’s why $a= 1/4 \ and \ r= 1/4$

$P(TH)=\dfrac{1}{4}$

$P(HH + HT)=\dfrac{1}{2}$

now if $TT$ comes then toss again,

So, $P(TTTH)=\dfrac{1}{16}$ and so on.... $P(TH+TTTH+\ldots) = \dfrac{1}{4} + \dfrac{1}{16}+\ldots= \dfrac{1}{3}$

P(getting the output Y) = P(TH) +P(TTTH) + P(TTTTTH) + P(TTTTTTTH) +  P(TTTTTTTTTH)+  P(TTTTTTTTTTTH)...........

P(getting the output Y) = 1/4 + 1/42 +1/43 +1/44 +1/45 +1/46+................ = (1/4) / ( 1- 1/4) =1/3

The correct answer is 0.33 .
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### 1 comment

I also applied the same approach... Easy to understand.

You can correct me if I am wrong.

### 1 comment

bro you adding p(TH) case in the first if we get (TH) we simply stop as mentioned in question the how its possible please explain if am wroong