weighted sum of independent indicator random variables I am trying to prove the following claim:  

Let $v \in \mathbb R_{\ge 0}^n $ such that $||v||_1 = 1$ (i.e. $\sum_{i=1}^n
 v_i = 1$) and let $w \in \{0, 1\}^n $ be a random vector so that
  each $w_i$ is $1$ with probability $ \frac{1}{3} $ and $0$ with
  probability $\frac{2}{3}$ with all choices being independent.
  Then $\mathrm {Pr}(w \cdot v \ge 1/3) \ge \frac{1}{3}$

I know by linearity of expectation that $E[w \cdot v] = 1/3$, so it seems very intuitive.
The result is obvious for $v=(1, 0, 0, \ldots ,0)$ but I am not sure how to prove it for the general case.
 A: Think what happens if you take three such random vectors $w^1,w^2,w^3$, each supported on a different part of your probability space.
Since their sum is always 1 in every coordinate, at least one of them must satisfy : $<w^i,v>\ \geq\  \frac{1}{3}$
EDIT:
As some people misunderstood what I meant I am writing here a full proof:
Let $X_1,X_2,...,X_n$ be uniform Rvs on $[0,1]$. We sample $w^1,w^2,w^3$ as follows:
Declare the i'th coordinate of $w^1$ to be $1$ iff $X_i\leq\frac{1}{3}$, the i'th coordinate of $w^2$ is $1$ iff $\frac{1}{3}<X_i\leq\frac{2}{3}$ and the i'th coordinate of $w^3$ is $1$ otherwise.
Clearly the distribution of all three of them is as defined in the question, and $w^1+w^2+w^3=(1,1,...,1)$ always. Therefore we have that $\mathbb{P}(<w^1+w^2+w^3,v>\ =\ 1)=1$. 
Since the inner product is linear, and the fact that a sum of three number is $1$ implies that one of them is at least $\frac{1}{3}$ we have
$\mathbb{P}(\exists i.\ <w^i,v>\ \geq \frac{1}{3})=1$.
From the union bound this implies that at least one of them has probability at least $\frac{1}{3}$ to be at least $\frac{1}{3}$, but since they have the same distribution it is true for all three of them.
