Roll $m$ dice and re-roll up to $n$ of them where the result is less than $t$. What is the probability that a single die will have the value $v$? I'm trying to combine some dice probabilities into one. 
If I roll $m$ dice, and after check each die individually against some threshold $t$. I can then re-roll up to $n$ of them if the result of the die is below the threshold $t$. What is the probability that a die from this set will have a value of $v$?
Let's assume:


*

*All dice are fair and have 6 sides

*$m \gt 0$

*$n \le m$

*$t \le 6$

*$v \le 6$
Initially starting with $m = 2$, $n = 1$ and $t = 4$, I'm happy with generating the probabilities of each die separately:
Die without re-roll:
$\mathbb{P}(v\in \{1\dots6\}) = \frac{1}{6}$
Die with re-roll:
$\mathbb{P}(v\in \{1\dots3\}) = \frac{1}{12}$
$\mathbb{P}(v\in \{4\dots6\}) = \frac{1}{4}$
My question really is how do I combine them? Do I just take a weighted average of each result? 
$\mathbb{P}(v\in \{1\dots3\}) = \frac{1}{8}$
$\mathbb{P}(v\in \{4\dots6\}) = \frac{5}{24}$
I feel that may be a little too straight forward?
P.S. First post on here, I hope my LaTeX is good enough.
 A: Here there are three ways a die can have a value less that $4$:


*

*initially both dice were less than $4$ (probability $1/4$) and you rerolled the other one (probability $1/2$);

*initially both dice were less than $4$ and you rerolled this one (probability $1/8$) but it was still less than $4$ (probability $1/2$);

*initially only this die was less than $4$ (probability $1/4$) and when you rerolled it it still was (probability $1/2$).


Thus the probability of a particular die being less than $4$ at the end is $1/8+1/16+1/8=5/16$. If it is at least $4$ then clearly all values $4,5,6$ have the same probability which is $7/48$. If you choose randomly which die to reroll when there is a choice, all the smaller numbers will have probability $5/48$, but if you always chose the smaller number (say) that would change.
You can generalise this but it gets messier. The probability of rerolling a given die is given by $\sum_{k=0}^m\binom mk\big(\frac{t-1}{6}\big)^k\big(\frac{7-t}{6}\big)^{m-k}\cdot\frac{\max(k,n)}{m}$, which we obtain by conditioning on the number $k$ of dice originally below $t$, so the probability of being less than $t$ at the end is given by
$$\frac{t-1}{6}-\sum_{k=0}^m\binom mk\Big(\frac{t-1}{6}\Big)^k\Big(\frac{7-t}{6}\Big)^{m-k}\cdot\frac{\max(k,n)}{m}\cdot\frac{t-1}{6}$$
