A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is? A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is? How do I approach this problem?
 A: Add up the probabilities of the coin coming up heads for the first time on toss 1, 3, 5...
$p_o = 1/2 + 1/2^3 + 1/2^5 + ...$


*

*The $1/2$ term is pretty obvious, it's the probability of the first toss being heads.

*The $1/2^3$ term is the probability of getting heads for the first time on the third toss, or the sequence TTH. That sequence has a probability of $1/2 * 1/2 * 1/2$.

*The $1/2^5$ term is the probability of getting heads for the first time on the fifth toss, or the sequence TTTTH. That sequence has a probability of $1/2 * 1/2 * 1/2 * 1/2 * 1/2$.
Now we can rewrite the series above as
$p_o = 1/2 + 1/8 + 1/32 + ...$
This is a geometric series that sums to $2/3$. The easiest way to show this is with a visual example. Start with the series
$p = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ...$
This is a geometric series that sums to $1$.

If we sum just the even terms of that series, we can see that they sum to $1/3$.
$1/4 + 1/16 + 1/64 + 1/256 + ... = 1/3$

If you eliminate the even terms from the full sequence, you're left with just the odd terms, which must add up to $2/3$.
$p_o = 1/2 + 1/8 + 1/32 + ... = 2/3$
A: Think recursively - let $p_o$ be the probability of the first head on an odd toss, and let $p_e$ be the probability of the first head on an even toss. Now $p_o+p_e=1$, and we also have that $p_e$ equals the probability of first toss tails times $p_o$. Thus $p_e = 1/2\cdot p_o$; $p_o+1/2\cdot p_o = 1$; $p_o = 2/3$.
