Events with constant probabilities regardless of $p$ for Bernoulli random variables Let $\{X_i \}_{i \in \mathbb{N}}$ be independent Bernoulli random variables with parameter for $0 < p < 1$. Find a single
event $A$ that has probability $1/2$ regardless of the value of $p$ and a second event $B$ that always has
probability $1/3$. Use the $\sigma - $algebra axioms to show that $A$ and $B$ are actually events.
I'm unsure how to proceed with this question.  Do I need to look at all the possible events such as the probabilities of the intersections of all the random variables?  I realized that $\frac{E[X]}{2pj}= \frac{1}{2}$ for all values of $ p $ where $j$ is the number of random variables and $X$ is the sum of the $ j $ random variables, but I'm uncertain whether this is relevant.  Any suggestions for how to approach this problem would be appreciated.
 A: Suggestion: Treat the results of $\{X_n\}_{n=1}^{\infty}$ as the base two expansion of a number between $0$ and $1$, e.g. $$X_0 = 1, X_1 = 0, X_2 = 1 \rightarrow 0.101_2 = 1 \times \frac{1}{2} + 0 \times \frac{1}{4} + 1 \times \frac{1}{8} = \frac{5}{8}.$$
Each additional digit splits the interval $[0,1]$ into twice as many intervals, and you should be able to find a "leading segment" (i.e. an interval of form $[0,\alpha]$) where the probability of the corresponding real number falling into that segment is close to $\frac{1}{2}$. (Of course the value of $\alpha$ will depend on $p$.) As you look at more and more $X_n$ the intervals get finer, and you should be able to pick successive values of $\alpha$ so the probability gets closer and closer to $\frac{1}{2}$.
(I have left this really hand-wavy and haven't introduced any of the notation that a serious answer should have to provide so that you can figure that stuff out. Also, I'm pretty sure this would work, but I might have missed some issues with this method. Also also, this argument looks to be a bit of a pain to formalize, so if there's a slick easy answer I've certainly missed it here. Good Luck.)

Reply to comment:
Let's take $n=2$. The "leading segment" candidates are $[0,0/4]$, $[0,1/4]$, $[0,2/4]$, $[0,3/4]$, or $[0,4/4]$. As $i$ goes from 0 to 4,  $P(\text{mapped number lying in }[0,i/4])$ goes from 0 to 1, so at some point the probability goes from being less than 1/2 to greater than 1/2. Maybe instead of looking at a single value, we think of a "bracket" of two values, let's say it's when growing from $[0,2/4]$ to $[0,3/4]$. Now add another digit, or as you say "further expansion", we can consider $P(\text{mapped number lying in }[0,5/8])$. Depending on whether this is greater or less than 1/2 our bracket shrinks to either "between 1/2 and 5/8" or "between 5/8 and 3/4". If you've ever done a proof by bisection, or programmed a bisection search, this should look familiar.
