Calculating Probability of results of 3 12-sided dice (*not* finding probabilities of sums) I have a complicated bit of probability calculating I don't have the math for:
There are three dice with 12 sides each. 
Only one of the dice will be used as the "result" die. 
There are three different ways to determine which die to use: 
1: "Omit the lowest die, Omit the highest die, the remaining die is the result." 
2: "The result is the lowest number rolled." 
3: "The result is the highest number rolled." 
So the result is always a number 1-12. 
I need to know the probability of each possible result for each of the three methods. (or a formula I could use to figure it out myself...) 
Help?
 A: You’re asking about quantities that are known as order statistics. There’s a brief introduction to the subject on Wikipedia that includes an explicit formula for discrete random variables, which is what we’re dealing with here.
The $k$th-order statistic of a set of random variables $X_1,\dots,X_n$, denoted by $X_{(k)}$, is basically the $k$th-lowest value among them. In particular, $X_{(1)}$ is the minimum value and $X_{(n)}$ the maximum.  
For this problem, the random variables $X_k$ are the values rolled on the individual dice. Since you’re rolling only three dice (and assuming that you keep the “middle” value even when it’s equal to the highest or lowest), your case 1 is the order statistic $X_{(2)}$. The random variables representing the die rolls are independent and identically distributed, so we can use the formula from the Wikipedia page:
$$\Pr(X_{(k)}=x)=\sum_{j=0}^{n-k}\binom n j\left((1-F(x))^j(F(x))^{n-j}-(1-F(x)+f(x))^j(F(x)-f(x))^{n-j}\right)$$
where $f(x)$ is the probability mass function—the probability that a die roll equals $x$—and $F(x)$ is the cumulative distribution function—the probability that a die roll is less than or equal to $x$. Assuming that the three dice are fair, $f(x)=\frac1{12}$ and $F(x)=\frac x{12}$, so we have
$$\begin{align}\Pr(X_{(k)}=x) &= \sum_{j=0}^{3-k}\binom3j\left(\left(1-\frac x{12}\right)^j\left(\frac x{12}\right)^{3-j}-\left(1-\frac x{12}+\frac1{12}\right)^j\left(\frac x{12}-\frac1{12}\right)^{3-j}\right) \\
&= \frac1{12^3}\sum_{j=0}^{n-k}\binom3j\left((12-x)^jx^{3-j}-(13-x)^j(x-1)^{3-j}\right).\tag{*}\end{align}$$
Taking $X_{\text{max}}=X_{(3)}$ first since there’s only one term in the sum, equation (*) becomes
$$\Pr(X_{\text{max}}=x)=\frac1{12^3}\binom30\left(x^3-(x-1)^3\right)={x^3-(x-1)^3\over12^3}={3x^2-3x+1\over12^3}.\tag1$$
To compute the probabilities of the minimum value, we could work through the hairy sum, but by symmetry $\Pr(X_{\text{min}}=x)=\Pr(X_{\text{max}}=13-x)$, so we can simply substitute $13-x$ into equation (1) and simplify to get $$\Pr(X_{\text{min}}=x) = {(13-x)^3-(12-x)^3\over12^3} = {3x^2-75x+469\over12^3}.\tag2$$ The calculation for the middle value ($k=2$) is a bit messy, but with a bit of work (or by feeding it to a CAS program) we end up with $$\Pr(X_{(2)}=x)={-6x^2+78x-38\over12^3}.\tag3$$ (It’s tempting to guess that $\Pr(X_{(2)}=x)=1-\Pr(X_{(1)}=x)-\Pr(X_{(3)}=x)$, but this doesn’t account for the cases in which the middle value is also an extreme value.) This peaks at $x=6.5$ and is symmetric about this value, which is what we’d expect. As further sanity check, we can compute $\Pr(X_{\text{max}}=1)=\Pr(X_{\text{min}}=12)=\frac1{12^3}$, which checks out because the only way for either to happen is to roll a triple one or triple twelve, respectively.  
A: I'm not allowed to comment on other people's answers yet, but this is meant as an addition to what @kccu said, and a hint for method 1.
Say $Y$ is the result of one die roll.
You can easily compute $P(Y=k)$, $P(Y\le k)$ and $P(Y\ge k)$.
From that it is not too difficult to deduce the $P(X=k)$ for each method.
Assuming each roll is independant from one another,
with method 2 you have 
$$
P(X=k)
= {3\choose 1} P(Y=k) P(Y\ge k)^2
=3P(Y=k) P(Y\ge k)^2
$$
Method 3 is nearly identical, and as for method 1
you just need the proper combination of $P(Y=k)$, $P(Y\le k)$ and $P(Y\ge k)$.
A: I recommend using counting methods. Let's assume the dice are distinguishable, say one is red, one is yellow, and one is blue. Then there are $12^3$ possible outcomes for a single die roll. 
Let's look at method 2: the result is the lowest number rolled. Let's say $X$ is the result using this method. We want $P(X=1)$, $P(X=2)$, ..., $P(X=12)$. To find $P(X=k)$, we need to know how many outcomes of the three dice satisfy the lowest number rolled is $k$. For instance, if $k=2$, then the following outcomes work:
$(2,2,2),(2,2,3),...,(2,2,12),(2,3,2),(2,3,3),...,(2,3,12),...,(2,12,12),...,(12,12,2)$
There's way too many to just count in the list, so we need to be systematic about it. We know there is at least one $2$. If there is only one $2$, the other dice show more than $2$. There are $3$ choices for which die shows $2$, and $10$ choices for what the other dice show, so $3\cdot 10 \cdot 10=300$ outcomes. If there are two $2$s showing, there are three choice for the die which does not show $2$, and $10$ choices for what it does show, so $3\cdot 10=30$ choices. Obviously there is only one way for all three dice to show $2$. So we find there are $300+30+1=331$ outcomes in which the lowest die shows $2$. This gives $P(X=k)=\frac{331}{12^3}$. 
That was a sort of tedious calculation, so we can try to think of ways to make it easier. For instance, we could instead find $P(X \geq k)$. For $k=2$, this means we want the lowest die to be at least $2$. This means there are $11$ choices for what each die shows, so $P(X \geq 2)=\frac{11^3}{12^3}$. We know $P(X \geq 1)=1$, so $P(X=1)=P(X \geq 1)-P(X \geq 2)=1-\frac{11^3}{12^3}=\frac{397}{12^3}$. Similarly, $P(X=2)=P(X \geq 2)-P(X \geq 3)=\frac{11^3}{12^3}-\frac{10^3}{12^3}=\frac{331}{12^3}$, the same as what we found above. 
For method 3, it will probably be easier to find $P(X \leq k)$. For method 1, you probably just have to count the number of outcomes with $X=k$; I can't immediately see an easier way.
