Does the distribution of a randomly baised coin approach a fair coin distribtion if the sample size is relatively large? I appolgize if the answer is trivial, as I wanted to confirm my suspicions about the question, and that the answer is indeed trival.
Given:


*

*The bias, $B$, of an unfair coin is defined as ${B∈ℝ:0≤B≤1}$, where $B = P(Heads)$

*Likewise, $P(Tails) = 1-P(Heads) = 1-P(B)$

*The possible values of $B$ are evenly distrubuted within $(0,1)$

*Bias $B$i is the bias on the ith coin flip, with  $B$i+1 being determined prior to the $i+1$th coin flip. The particular values of these variables are not known.

*$n$ is the total number of coin flips


If $n$ is relatively large, I would think that the distribution of every coin flip would approach that of a fair coin, as the expected vaule for the trend of a particular bias $B$i would be one flip. So, I would assume all the biases would eventually "cancel" each other out, causing the disturbtion to apporach that of a fair coin. 
 A: In fact, your coin is indistinguishable from a fair coin.
The probability that the $i$th toss is heads equals the probability that a uniform random number $X_i\in(0,1)$ is $<B_i$. With $X_i$ and $B_i$ independent, it follows by symmetry that this probability is $\frac12$. (And of course different tosses are independent provided the $B_i$ are also independent, which you probably intend to have).
A: You seem to be saying each coin $i$ has a probability of heads of $B_i$ where the different $B_i$s are uniformly and independently distributed on $[0,1]$
In that case, by symmetry, the probability that a particular coin shows heads is $\frac12$.  
So the law of large numbers says that the proportion showing heads $X_n$ of a sample size $n$ converges in probability and almost surely to $\frac12$ as the sample size increases.  $\mathbb E[X]=\frac12$ and $\text{Var}(X_n)=\frac1{4\sqrt{n}}$, and the central limit theorem shows $\sqrt{n}(X_n -\frac12)$ converges in distribution to the normal distribution $\mathcal N(0,\frac14)$
