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A coin tossed 3 times

1) if output at  last two toss is $HT$ then stop tossing

2) if output of last two toss  is $TT$ then toss again and go to step 1 or step 3

3) if output of last two toss is $TH$ or $HH$ then toss again and goto step 2

What will be probability, that at last the coin will stop tossing?
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Let us assume E and E' be two events where

E = { (HHT) , (THT) }

and E' = { (HTT), (TTT), (HTH), (TTH), (HHH), (THH) }

Then P(E) = 2/8 = 1/4  and  P(E') = 6/8 = 3/4

Now, according to the given condition in the problem at each chance we are tossing the coin 3 times , if the event E occurs then we will not get another chance to toss  and if event E' occurs then we will get another chance to toss.

Now, the probability that I got only 1 chance to toss is 1/4 ( the event E occurred on the 1st chance)

The probability that I got 2 chances to toss is 3/4 * 1/4 ( the event E' occurred on the 1st chance and event E occurred on the 2nd chance)

So, if we go on like this we can find the probability that I got k chances to toss is (3/4)k * (1/4) (where the event E' occurred on first k-1 times after that event E occurred). 

So, the probability that tossing would terminate within k chances is

1/4 + 1/4 * 3/4 + 1/4 * (3/4)2 + ..................... + 1/4 * (3/4)k

Now, in the problem it is not given that at most how many chances we can take. So, the number of chances taken by me can run up to infinity.

So, the probability that tossing would terminate is

1/4 + 1/4 * 3/4 + 1/4 * (3/4)2 + ...........................(Up to infinity)

Now, the above expression is the sum of infinite G.P series

We know that the sum of an infinite G.P. series is a/ (1-r) where a= the first term of infinite G.P. and r = common ratio

So, the sum of the above expression is (1/4) / (1- 3/4 ) = 1

So, the probability that the tossing would terminate is 1.

Hence our tossing will definitely terminate it cannot go forever.

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