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To find out the expected no of flips of a coin to get a TT, i want to find it out using a series of probability multiplied with their values. In a similar question using sum of series the expected values of no of flips to get a HT was found {"Expected Value of Flips Until HT Consecutively"}, I tried this, lets say at nth flip we get a T (with (n-1) th flip being tail ). Say m be the no of flips before a TT comes, here m = n-2, in these m no of flips the m th flip must be H (since if its T, this and (n-1) th flip would make a TT ) and the no of tails in these m throws can be between {0,…,$\lfloor {\frac{n}{2}} \rfloor $ } , however i couldn’t calculate the possibility of the different strings of length m, how to calculate it

Cloverr
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First we need to check that the expectation is finite, for the subsequent part to work. To do so we compare with a game where we stop only when we get "TT" where the first "T" is on an odd flip. Clearly if we stop, we would have stopped in the original game, possibly earlier. In the new game, the expected number of flips is $\sum_{k=1}^\infty (\frac{3}{4})^k \frac{1}{4} (2k)$, which is clearly finite by any test of your choosing (such as explicit evaluation or integral test). Thus the original game also has finite expected number of flips.

Let $x$ be the expected number of flips needed if the previous flip was "T".

Let $y$ be the expected number of flips needed if the previous flip was "F".

Then it is easy to find the relations between $x,y$ since the coin has no memory. For example, $x = 1 + \frac{1}{2} (0) + \frac{1}{2} (y)$ because after a "T", after one more flip, either the game ends or the game continues in the state with previous flip "F", each with probability one half. I leave you to find the other relation.

Now the original game can be considered as starting in the state with previous flip "F", so the answer is $y$.

user21820
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  • Ty for the answer, i wasn't expecting an answer of this form although, but yes this method solves it very nicely! – Cloverr Apr 10 '15 at 08:51
  • @Nilanjan: The direct computation would require Fibonacci numbers since the number of $n$-digit binary strings without consecutive "1"s is $F_{n+1}$. I did not try but I think the summation can still be computed and simplified to the answer, but in general such problems cannot be solved without using this technique. But it is always important to check that the expectation is finite so that we can obtain the simultaneous equations. – user21820 Apr 10 '15 at 09:00
  • Yes other than yours most of the other solutions ignore calculating if the expectation is finite, now that you have provided a way to approach this using the fibonacci numbers i would definitely try to solve using that :) , Ty again! – Cloverr Apr 10 '15 at 09:09