Probability of getting a head on even No of tosses: why it is 1/3 and not 1/2? $X \in \mathbb{R}$ is the number of coin tosses it takes to get the first head.
Question: find the probability of getting the first head at an even number of tosses.
Official solution:
$P(X_{even}) = \frac{1}{2^{2}} + \frac{1}{2^{4}} + ... = \frac{1}{3}$ (sum of a geometric progression).
Why is this not correct solution:
$P(X)=1$; $X_{odd} = X_{even}$; $P(X_{even})=\frac{n(X_{even})}{n(X)}=\frac{1}{2}$
 A: Because $X_{\rm odd} \ne X_{\rm even}$. The most probable result is to get a head with the first toss (we have $P(X = 1) = \frac 12$), and $1$ is odd.
A: there is 0.5 chance that you will immediately get a head on toss 1 - but that is not the only way you can fail, you can also get a head on 3,5,7... and fail
after two opening tails (probability 1/4), the game is back where it started - the probability of winning is prob of getting head on toss 2 (if there is one) plus the probability of winning from start multiplied by probability we go to a third toss
P(head on even toss) = P(first head on toss 2) + P(game goes to a third toss) 
P(head on even toss)= (1/2)(1/2) + (1/2)(1/2)P(head on even toss)
(3/4)P(head on even toss) = 1/4
P(head on even toss) = (1/3)
A: Another way of looking it is as follows: suppose that you throw pairs of coins, one for odd, one for even, the odd one first, until at least one of them comes up heads. Then, when your experiment has ended, the outcome of the last two coins is either HH, HT, or TH. In two out of those three, the odd one gets the first heads (HH, HT); and in one of three, the even coin does (TH).
