Stuck on probability/statistics question EDIT: Post as been edited to address relevant questions raised in comments.
I'm new to the site and I'm stuck on a probability question.  I don't think it's trivial, certainly not to me, as I am relying on a single probability and statistics class I took 20 years ago...  The problem is this:


*

*Imagine there are 10 white squares (the number of white squares is not terribly important other than it has to be 5 or greater).

*The probability of turning a white square black is 25% per attempt.

*The goal is to maximize the number of black squares you get.

*You have 10 "moves" to do so.

*Attempting to turn a white square black uses one move.

*Moving from one square to another also uses one move.

*You send in your "moves" in batches of 10 and, therefore, you cannot adjust subsequent moves regardless of the results of previous moves.

*Attempting to convert a square that you had previously successfully converted from white to black does not revert it back to white. It just means you wasted a move trying to convert an already black square.


How would I even start figuring out the best way to maximize the number of black squares I get (i.e., how many attempts per square before moving to the next square)?
Thank you very much!
 A: You didn’t specify a geometry, so I’ll assume that we can always move to a fresh square.
You say you want to maximize the number of black squares, but that’s not a meaningful objective since this is a random variable. I’ll assume that you in fact want to maximize the expected number of black squares.
Your terminology is a bit confusing, since you’re using “move” with two different meanings; so I’ll call the moves “turns” instead.
It certainly makes sense to try each square once before moving on, and it makes no sense to move on the last turn. So we should never move more than $4$ times.
Given a number of moves from $0$ to $4$, it’s clear that we should distribute the attempts over the squares as equally as possible.
That yields $5$ inequivalent strategies: $(10)$, $(5,4)$, $(3,3,2)$, $(2,2,2,1)$ and $(2,1,1,1,1)$, with expected numbers of black squares
\begin{eqnarray}
1-\left(\frac34\right)^{10}=\frac{989527}{1048576}&\approx&0.94369\;,\\
1-\left(\frac34\right)^5+1-\left(\frac34\right)^4=\frac{1481}{1024}&\approx&1.44629\;,\\
2\left(1-\left(\frac34\right)^3\right)+1-\left(\frac34\right)^2=\frac{51}{32}&=&1.59375\;,\\
3\left(1-\left(\frac34\right)^2\right)+1-\left(\frac34\right)^1=\frac{25}{16}&=&1.5625\;,\\
1-\left(\frac34\right)^2+4\left(1-\left(\frac34\right)^1\right)=\frac{23}{16}&=&1.4375\;,\\
\end{eqnarray}
respectively, so the optimal strategy is to move twice, trying two squares thrice and one square twice, with an expected number $\frac{51}{32}$ of black squares.
