Let's say we take the simple question:
"What is the expected number of heads after 4 coin tosses?" In this case, the answer is straightforward. For $1≤i≤4$ let $X_{i}$ be a random variable; if coin $i$ results in head $X_{i}$ = 1 and $0$ otherwise. By linearity of expectation, we want $$E(∑X_{i})=∑E(X_{i})$$. But $E(X_{i})$ is just the probability that coin $i$ results in a head, so we arrive at $$E(∑X_{i})=∑P(X_{i})=4*1/2=2$$
But what if we add another condition to the question, namely the number of heads is greater than 2.
In this case, we can just forge ahead as follows: the number of heads must be 3 or 4. By Bayes' theorem the associated probabilities are $4/5$ and $1/5$, so the expected value of the number of heads is $$E=4/5 \cdot 3+1/5 \cdot 4=3.2$$
But I was wondering how to adapt the method of solution from the first part of the question without doing a direct computation as it might prove more useful if the number of heads is increased from, say just 4 to 50 or a hundred.
Any ideas?
I don't see any way to avoid doing the explicit math, just as was done in the posted question.
One idea is to try to use a scaling strategy. However, that doesn't seem to work here.
For example, the constraint of at least 3 heads eliminates the 11 smallest of the 16 possibilities.
However $~\dfrac{16}{5} \times 2 \neq 3.2,~$ so the $~\dfrac{16}{5}~$ scaling factor seems useless here.
Trying to remedy the scaling factor can be a challenge: Not all of the $~11~$ choices to be eliminated had an expected value of $~0,~$ so (arguably) some attempt must be made to determine the expected value of the choices that are eliminated.
However, by that approach you are often (but not always) chasing your tail. That is, you are forced to compute math similar to what you were trying to avoid.
Game isn't over yet.
Suppose that you have $~50~$ coin flips, and your expected number of heads is $~25.~$ Then, you might ask: what is the expected number of heads if you know that you had at least two heads?
I will lay out a strategy for that particular problem by asking a similar question: suppose that you had $~4~$ coin flips and your expected number of heads is $~2.~$
What is the expected number of heads, out of 4 coin flips, if you are given that you had at least one head?
$~1/16$-th of the time, you had $~0~$ heads, so the answer should be $~\displaystyle \frac{16}{15} \times 2.~$
Now, consider the moderately more cumbersome question:
What is the expected number of heads, out of 4 coin flips, if you are given that you had at least two heads?
To answer this, you are doing the same math that you wanted to avoid, but it is easier, because you only have two cases to deal with:
Let a group of 4 coin flips constitute a trial.
Assume that you have 16 trials, and that you got
you had 16 trials, got a total of 32 heads.
Average = total / number of trials.
Instead of having total = 32, trials = 16, you now have
total = (32 - 4), trials = (16 - 5), so the average is
28/11.
So, the expected number of heads in a single trial of 4 coin flips, given that you got at least two heads is $~\dfrac{28}{11}.$
So, in the latest question, the math is not avoided, but it is easier. Similarly, if you have 50 coin flips, and are asked for the expected number of heads given that you had (for example) at least 2 heads, then the above approach is arguably best.
Let a single trial represent flipping the coin 50 times.
In this latest question, you start with $~2^{50}~$ trials and a total number of heads = $~25 \times 2^{50}.$
Then, of the $~2^{50}~$ trials, $~\binom{50}{0}~$ of these trials produced 0 heads each, and $~\binom{50}{1}~$ of these trials produced 1 head each.
So, the new expectation is
$$\frac{\left[ ~25 \times 2^{50} ~\right] - 50}{2^{50} - 51}.$$