Reroll 2 dice sum probability My statistics are very far in my memory and I am not a math guru so I do not understand half of fuzy symbols used in most post that could have the answer I am looking for. So I would ask for a very simple and easy to understand answer pretty please :)
I have 2 dice, numbered {0,0,1,1,2,2}.
It gives me 36 possible results with the following distribution:
0 : 4 
1 : 8 
2 : 12
3 : 8 
4 : 4 
Now, if I want to reroll once when I do not get at least a sum of 3, what would be:
1) The amount of possible results?
2) The result distribution?
Thanks

If you want to reroll just one time when you get a sum lesser to 3 then you must multiply the probability for each roll, this let to you directly the probability distribution of these results.
I will show you: you have the distribution of 0*4, 1*8, 2*12, 3*8, 4*4 over 36 different possible results. The probability for a sum will the times of this sum can happen divided all possible cases.
For 0 and 4 the probability will be $\frac{4}{36}$, for 1 and 3 the probability to happen on a throw will be $\frac{8}{36}$, and the case for 2 the probability will be $\frac{12}{36}$.
You will reroll if you take a 0, 1 or 2, i.e., anything lesser to 3 as you said. For each result (0, 1 or 2) you roll the dice again taking again the same probability. By example you roll and the sum is 2 so you must reroll again and you can have a 0, 1, 2, 3 or 4. You must multiply both probability to know the final probability for this sum of 2 rolls: probability of roll 1* probability of roll 2.
If you take a 2 in the first roll you reroll again and you get a 3: the total sum of the 2 rerolls is 2+3 and it probability will be $\frac{12}{36}\cdot\frac{8}{36}$. When you have 0 and you reroll again the sums will be just the same that a normal reroll because if you sum zero you sum nothing. For 1 you will have the same original sums +1, and for 2 the original sums +2 (the original sums are 0, 1, 2, 3 and 4).
Doing all the work by hand is really painful and boring, I recommend to you try to understand this formula based on a generating function for your game
$$f(x)=(4x^0+8x^1+12x^2)(4x^0+8x^1+12x^2+8x^3+4x^4)+8x^3+4x^4=\\=16x^0+64 x^1+160 x^2+232 x^3+228 x^4+128 x^5+48 x^6$$
This represent the distribution of the $ax^s$ elements, where $a$ is the frequency of the sum $s$. If you sum all frequencies $a$ of all $s$ you have the total number of possible outcomes for your game
$\sum a_i=876$
The probability for a sum of 6 will be then $\frac{48}{876}$.
