What is the probability that the expectation value is the actual value in this experiment

Solution 1:

$20$ percent of the time, I've picked up a blue ball. From here, I can say that the probability of me obtaining a blue ball is $0.2$.

Yes, based on the empirical/frequentist interpretation of probability.

However, we know from the definition of probability that the number of blue balls in the bag is just the total number of balls multiplied by the probability of getting a single blue ball.

The last line should read “multiplied by the probability of getting a blue ball in a single draw” instead.

Note that here, we're relying on the classical (equal-possibility) interpretation of probability.

Doing this in the above example, I'd get $20$ blue balls in the bag. This is nothing but the expectation value of the number of blue balls in my bag.

Yes: you empirically obtained (estimated) the expectation of a Binomial experiment, computed its probability, derived the expectation of the corresponding $100$-trial experiment to finally infer an estimate of the expected number of balls in your bag.

However, If I repeated the trial infinite times, and noticed that in exactly 20 percent of the times, I get a blue ball, can I say that the actual number of blue balls in the bag is equal to the expectation value of the number of blue balls ?

Yes, under the assumptions of classical probability, the actual number of blue balls in your bag equals its limiting expected value.

What would be the probability that there are actually $20$ blue balls in the bag ? I think this would take the form of some distribution, but I don't know what or how.

The number of blue balls in your bag is a random variable and indeed has a probability distribution. Using the Binomial distribution and the estimated probability 0.2, $$P(\text{bag has $20$ blue balls})={100\choose20}0.2^{20}\,0.8^{80}=9.93\%.$$

and not the actual value, I can say $P(b=\langle b\rangle)\lt 1$.

Based on the epistemic/subjective/Bayesian interpretation of probability:

• if you know that there are actually 20 blue balls, then $P(b=20)=1;$
• if you know that there is not actually 20 blue balls, then $P(b=20)=0;$
• $0<P(b=20)<1$ if and only if you don't know the actual number of blue balls (regardless of whether there are indeed $20$ blue balls).

Second question: What do you mean when you say to find the probability that $20$ balls are blue?

Does it ask us to find the probability that there are $20$ blue balls in the bag,

These are impossible to answer without more context. As opposed to $23$ balls being blue? As opposed to $20$ balls being red? What is the experiment (how many draws are there? are the balls replaced after each draw? Etc.), and what is the sample space? Etc.

or is it asking the probability that if we pick out $20$ balls at random, all of them would be blue ?

Possibly. You have just supplied some context; the scenario still needs to be further filled out before we can choose some probability interpretation and work out a reply.