Problem with a probability measure Suppose that $\Omega=\{w_{1},\dots,w_{n}\}$ and define a probability measure $\mathbb{P}$ with the condition that the probability of $w_{j+1}$ is the double of the probability of $w_{j}$. Let $A_{k}=\{w_{1},\dots,w_{k}\}$. Compute $\mathbb{P}(A_{k})$.
Could someone give me a hint? 
 A: $1\\=\mathbb{P}(\Omega)\\=\mathbb{P}(\{\omega_1,\dots,\omega_n\})\\=\mathbb{P}(\cup_{i=1}^n\{\omega_i\})\\=\sum_{i=1}^n\mathbb{P}(\{\omega_i\})\\=\sum_{i=1}^n\mathbb{P}(\{\omega_1\})\cdot 2^{i-1}\\=\mathbb{P}(\{\omega_1\})\sum_{i=1}^n2^{i-1}\\=\mathbb{P}(\{\omega_1\})\cdot (2^n-1)$
Thus $\mathbb{P}(\{\omega_1\})=\frac{1}{2^n-1}.$ Now use induction and the fact that $\omega_{j+1}$ has double the probability of $\omega_{j}$, then compute $\mathbb{P}(A_k)$.
A: I believe you can use the normalization axiom of your probability rule in conjunction with the given recursive definition. To do this, set the first value, $w_0$ equal to a constant, $k$, then each successive probability is given by $\mathbb{P}(w_1)=2C$, $\mathbb{P}(w_2)=4C$ and so on following $\mathbb{P}(w_i)=2^iC$. We know the total probability has to be one to have a valid probability rule, so if we sum over all the events, we should get one. 
Thus, $\sum_{i=0}^{N-1}2^{i+1}k = 1$ 
Using the formula for a finite geometric series,
$1=C2(\frac{1-2^{N}}{1-2})$, $C=\frac{1-2}{2-2^{N+1}}$.
So for $\mathbb{P}(w_k)=2^kC=2^k(\frac{1-2}{2-2^{N+1}})$
