The expectation of a Random Process

So, in step t, assuming a sufficiently large number for t, do we have:

N/16 groups of 4 blue balls
And, since the scenario seems to favor red balls. (If only 2 blue balls, all turn to red.)
3N/16 groups of 4 red balls.
So, if N=256, for example, our expectation would be Nb(t)=64

Funny how the supposed answers to probability questions have a high probability of being incorrect. Fingers crossed:

This question can't be answered unless and until you specify the relationship between Nb and Nr. I assume you meant to note that Nb = Nr

E [ Nb (1) ] = N*5/16 >>>>>>>>>>>>>>>>>>> E [ Nr (1) ] = N*11/16

E [ Nb (2) ] = N*5/16*5/16

E [ Nb (t) ] = N*(5/16)^t

The expectation for the number of black balls after trial t using N balls is:

N*(5/16)^t

And wtf do you mean 'unless they turn red'?

What's the difference between a duck?

Are you saying that for any given subset, the minority balls will turn the color of the majority balls? What about two blue, two red?

Lack of specificity will ruin the result.
I had to guess that initial Nb=Nr and 2 blue, 2 red would mean all turn red. But it was just guess.

Of course, if we specify that Nb(0) = Na(0), and we also specify that the balls *stay the same color* in the event of a 2-2 tie within a group, then this becomes a very easy problem.

The answer in that specific case would be E [Na (t)]=E [Nb (t)] = N/2, by symmetry considerations. There is no dependence on time in that specific case.

I plan to revisit the more general case in a few weeks when I have more time free.

For a given Nb and Nr, you could use Markov chains to determine probabilities for each possible distribution. Once you have the probabilities, the expected value is simple to compute. It will take some work to figure out the transition probabilities.

If a 2/2 split turns into 4 red, then at t=1 both colors will have a multiple of 4 balls. That means your matrix for the Markov chain requires states only for multiples of 4 (fewer states and fewer transition probabilities to calculate).

If Nb and Nr aren't multiples of 4 at the start, you will still have to calculate the probabilities for going from the t=0 distribution to the t=1 distributions that are all multiples of 4.

I manually worked out the matrices that would be used beyond t=1 (if not starting with multiples of 4 for each color) for Nb+Nr = 4, 8, 12, 16. The row and column headings are the number of blue balls.

They are:

It's still a fool's errand without fully defined terms.

I'm going with the assumption that "unless" was intended to be "otherwise."
As for the initial relationship between Nb and Nr, that does not need to be specified. The ideal general solution would be specified in terms of Nb and Nr.
In any case, the technique applies, but possibly with different matrices.

True that, assumptions on those are only way to proceed.
Interesting approach to solution btw.