TITLE: A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd?
QUESTION [6 upvotes]: Q. A fair coin is tossed untill head appears for the first time. What is probability that the number of tosses required is odd?

My work:
suppose that head comes in first toss so probability of getting head in the first toss $=\dfrac{1}{2}$
suppose that first & second tosses show tails & third toss shows head so  probability of getting head in the third toss  $=(1-\dfrac12)(1-\dfrac12)\dfrac{1}{2}$ $=\dfrac1{2^3}$
suppose that first 4 tosses show tails & fifth toss shows head so  probability of getting head in the fifth toss  $=(1-\dfrac12)^4\dfrac{1}{2}$ $=\dfrac1{2^5}$
suppose that first 6 tosses show tails & seventh toss shows head so  probability of getting head in the fifth toss  $=(1-\dfrac12)^6\dfrac{1}{2}$ $=\dfrac1{2^7}$
…………….
and so on
But I am not able to find the final probability of getting head first time so that the number of tosses required is odd.  what should do I next to it? please help me.

REPLY [2 votes]: Here's another way to solve the problem by considering pairs of tosses at a time instead of single tosses.
Let $\sigma$ be an arbitrary infinite sequence of heads and tails. $\sigma$ uses 1-based indexing.
$$ \text{e.g.}\;\;\; \sigma = HTHTHTHTHTTTTTHHHH\cdots $$
Imagine grouping the elements of $\sigma$ into pairs.
$$ \sigma = HT,HT,HT,HT,HT,TT,TT,HH,HH\cdots $$
Let's imagine we have three states, $S$, $E$, and $O$.

$S$ is the start state, we haven't seen a head yet.
$E$ is the state marking that we saw a head at an even index first.
$O$ is the state marking that we saw a head at an odd index first.

$E$ and $O$ are both absorbing states. Once we enter one of those states, we are never going to leave it.
Our state at the beginning of our process is always $S$ because, initially, we haven't observed any tosses at all of our coin.
Next let's consider what happens when we read our first toss-pair from $\sigma$.
There are twice as many ways to transition from $S$ to $O$ than there are to transition from $S$ to $E$.
     TT
S  ----->   S

     TH
S  ----->   E

     HT
S  ----->   O

     HH
S  ----->   O

As the number of pairs processed approaches infinity, the probability that the current state is $S$ approaches zero.
However, the ratio of the probability that that the current state is $O$ is always twice the probability that the current state is $E$.
Therefore, the limiting probability that the state is $O$ is $2/3$