I have a question about understanding the math to a previous question I had posed (answered successfully). Link at the bottom.
The question is how does it make sense that rolling a 2D6 Die consisting of 111 22 3 for sides has the math work out where the probability to roll TWO 1s is 25% chance and rolling a 1 and a 2 is 33% chance? what is the math here for a laman?
It is also interesting to me that only when rolling 2 die is there a 50% chance of getting 2 circles.
Here is the link (https://rpg.stackexchange.com/questions/207875/using-anydice-com-to-calculate-non-traditional-dice-combinations/207898?noredirect=1#comment570131_207898)
ABOVE is the original question. BELOW is more math about these dice.
I am not even sure if the below example is correct. I am trying to figure it out as i post this.
[![Here is prob for different combinations. 1= circle 2 =triangle and 3- diamond on the die][1]][1]
If you want to understand this, create a table and count. But the key is, the table has to have all the options equally likely. Just creating a table of $1,2,3$ is not going to work because $1$ has a different probability than $2$, or $2$ from $3$. But each side of the die is equally likely, so your table needs entries for each of the sides:
$$\begin{array}{c|cccccc} &1&1&1&2&2&3\\\hline 1&\color{red}{1,1}&\color{red}{1,1}&\color{red}{1,1}&\color{blue}{1,2}&\color{blue}{1,2}&1,3\\ 1&\color{red}{1,1}&\color{red}{1,1}&\color{red}{1,1}&\color{blue}{1,2}&\color{blue}{1,2}&1,3\\ 1&\color{red}{1,1}&\color{red}{1,1}&\color{red}{1,1}&\color{blue}{1,2}&\color{blue}{1,2}&1,3\\ 2&\color{blue}{2,1}&\color{blue}{2,1}&\color{blue}{2,1}&2,2&2,2&2,3\\ 2&\color{blue}{2,1}&\color{blue}{2,1}&\color{blue}{2,1}&2,2&2,2&2,3\\ 3&3,1&3,1&3,1&3,2&3,2&3,3 \end{array}$$
There are $36$ possibilities total, and each are equally likely to appear. Note that $9$ entries are $1,1$, so the probability of two $1$s is $\frac 9{36} = \frac 14 = 25\%$
There are six entries of $1,2$ and six entries of $2,1$, for a total of 12 entries that have a $1$ and a $2$. Thus the probability is $\frac{12}{36} = \frac 13 = 33.\overline 3\%$.
You don't explain anything about "circles" and I am not going to search through posts in hopes of figuring out what you cannot be bothered to post clearly.