Flipping random coins from a bag - equivalent to a single coin? My first and I think naive question here.
I am trying to model a certain business, and the simplest model I am willing to test is:
1. there is a bag of differently biased coins.
2. every step, a single coin is chosen with equal probability.
3. the chosen coin is flipped and returned to the bag.
The business goal is to predict the rate of heads in future trials (probably using Bayesian inference).
This got me thinking - am I not over-complicating things? Isn't this process (observationally) equivalent to a single biased coin?
Thanks in advance!
 A: Yes, this process is equivalent to flipping a single biased coin. 
In practice, there is often room to change the procedure, in which case there may be differences. For example, if after you choose a coin, you get to observe multiple flips from that coin rather than just one, then the distribution of biases matters. A bag with a $0\%$ coin and a $100\%$ coin will never produce the sequence HT (with the altered procedure), whereas a bag with a $50\%$ coin may.
A: This is a standard example illustrating the notion that conditional independence does not imply unconditional independence. If the experiment consists of choosing a coin at random and then tossing it $n > 1$ times, then the results of the $n$ tosses are conditionally independent observations of a $\text{Bernoulli}(p)$ random variable where $p$ is the parameter for the coin chosen, and thus the total number of Heads is (conditionally)
a $\text{Binomial}(n,p)$ random variable. But, the $n$ tosses are not unconditionally independent.
More explicitly, with two coins with $P(\text{Heads}) = p_i$ and $n=2$, the conditional probabilities of a Head on each trial is $p_i$ and the two tosses are conditionally
independent. But, the law of total probability says that 
$$P(\text{Heads}) = \frac{p_1+p_2}{2}$$ on each toss, whereas for the two trials,
it also says that
$$P(2 ~\text{Heads on}~2~\text{tosses}) = \frac{p_1^2+p_2^2}{2} 
\neq \left(\frac{p_1+p_2}{2}\right)^2$$
showing that the two tosses are not unconditionally independent.
In short, if the experiment consists of choosing a fresh coin for
each toss, then the trials are independent and the result is
equivalent to tossing a coin with probability $\bar{p}$ where
$\bar{p}$ is the population mean, but this is not the case
if the experiment consists of choosing a coin and tossing it
several times. If the OP is receiving data about "big batches
of trials, not individual ones" it is important to look carefully
at the conditions of the experiment that resulted in the big
batches of data.
