Calculating chance of results when dropping lowest result of rolling 4 dice Say I want to roll 4 6 sided dice, then take the 4 results, drop the lowest result (or just 1 of the lowest values, if it is rolled more than once), and add the remaining 3 dice together to get the number.
(for those interested, this is the same as the rolling for abilities in D&D)
I've managed to make a python script that runs every possible combination to get percents for each result(for example, I know that the chance of rolling an 18 is aprox. 1.62%), but I am curious if there is a way to mathematically calculate it, without simulating each outcome, or counting them out.
I am specifically interested in the chances of rolling a 3 and an 18, but if there is a way to calculate each of the numbers, that would be even better.
Again, I'm not interested in the result, as I already have it. I am interested in the process in calculating it, if possible.
 A: You get $3$ only if your roll $[1,1,1,1]$.
Therefore, the probability of $3$ is $1/6^4$.

You get $18$ only if your roll any of the following:


*

*$[6,6,6,6]$, for which there is $1$ option

*$[6,6,6,X]$, for which there are $5$ options

*$[6,6,X,6]$, for which there are $5$ options

*$[6,X,6,6]$, for which there are $5$ options

*$[X,6,6,6]$, for which there are $5$ options


Therefore, the probability of $18$ is $(1+5+5+5+5)/6^4$.
A: To get $18$ you need three sixes.  You can get that either by rolling four sixes, which is $\frac 1{6^4}$ just like three ones or three sixes and one other number.  For three sixes, you have four ways to pick which die is the other number and five other numbers, so there are $20$ ways and the total probability is $\frac {21}{6^4}\approx 0.0162$ as you say.  The middle numbers are harder as there are more combinations to account for.  Even for $16$ you need to account for $664x$ and $655x$, where the $x$ is the die you throw out.
A: Say you want to compute the probability of getting a $3$.
If the result is $3$, then it is because you got $1$ on every dice, right? There can be no other way to get a $3$ as a result. So,
$$P(\text{result is $3$}) = P(\text{all dice roll $1$}) = \frac16\cdot\frac16\cdot\frac16\cdot\frac16 = \left(\frac16\right)^4$$
(I am assuming you know how to calculate the probability of a specific outcome of a single die);
How does one find $P(18)$? If you get $18$, three die rolled $6$ right? It does not matter what did the fourth die roll, because it will never be higher.
Lets say your dice are coloured, all with different colours. One red, one blue, one black and one white. Let's say the red was the one you ignored. The chances of getting $6$ in all other dice is
$$P(\text{white, black and blue rolled $6$}) = \left(\frac16\right)^3$$
But you chose a specific coloured die to be ignored. In how many ways could you make such a choice? In $4$ ways. So you should multiply that by $4$ right?
That would give $4\cdot\left(\frac16\right)^3 \approx 1.85\%$ which is more than you got. Why? Because you counted one specific event $4$ times: when all $4$ dice roll $6$. So you should subtract $3$ of those events, so as to take that into account only once:
$$4\cdot\left(\frac16\right)^3 - 3\cdot\left(\frac16\right) \approx 1.62\%$$ 
For other specific outcomes you think similarly. Either by counting every single way of that outcome happening, or by this method of starting by general things and then taking into account what was added or subtracted too many times.
