# 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: $P(X)=1$; $X_{odd} = X_{even}$; $P(X_{even})=\frac{n(X_{even})}{n(X)}=\frac{1}{2}$

• Because $X_{odd}\ne X_{even}$, that's why. – Parcly Taxel Oct 25 '16 at 10:34
• What do you mean with $X_{odd}=X_{even}$? – drhab Oct 25 '16 at 10:34
• The question should be "... of getting the first head at a ... ". And then the probability of odd is bigger, since you start with an odd number. – Jimmy R. Oct 25 '16 at 10:34

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.

• Let's take a discrete case then. $X \in {1,2,3,4,5,6,7,8,9,10}$. Here $X_{odd} = X_{even}$. Using official approach $P(X_{even}) = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024} = \approx \frac{1}{3}$. Why does $P(X_{even}) = \frac{n(X_{even})}{n(X)} = \frac{5}{10} = \frac{1}{2}$ yield a different result? – A.L. Verminburger Oct 25 '16 at 12:45
• No $X_{\rm odd}\ne X_{\rm even}$, even in the discrete case. The possible results $X = i$ for $i= 1, \ldots, 10$ do not have the same probiability, so you must not use the formula $n(A)/n({\rm total})$. – martini Oct 25 '16 at 13:25

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)