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
 A: Consider we roll the two dice and, conditional on the sum of the face values, roll again (if the sum of face values is 0, 1, or 2) or stop (if the sum of the face values is 3 or 4). The sum of the two dice on the first roll will be 0 (zero) with probability $\dfrac{4}{36}$ as indicated. In this case, roll again and the sum of the two dice on the second roll will be:
outcome - probability
0 - $\dfrac{4}{36}$
1 - $\dfrac{8}{36}$
2 - $\dfrac{12}{36}$
3 - $\dfrac{8}{36}$
4 - $\dfrac{4}{36}$
or 144 possible ways (0-0, 16 ways, with probability $\dfrac{4}{36}$.$\dfrac{4}{36}$ for this (0-0) outcome; 0-1, 32 ways, with probability $\dfrac{4}{36}$.$\dfrac{8}{36}$ for this outcome; 0-2, 48 ways, with probability $\dfrac{4}{36}$.$\dfrac{12}{36}$ for this outcome; 0-3, 32 ways, with probability $\dfrac{4}{36}$.$\dfrac{8}{36}$ for this outcome; and 0-4, 16 ways, with probability $\dfrac{4}{36}$.$\dfrac{4}{36}$ for this outcome). Next, the sum of the two dice on the first roll will be 1 (one) with probability $\dfrac{8}{36}$ as indicated. In this case, roll again and the sum of the two dice on the second roll will be:
outcome - probability
0 - $\dfrac{4}{36}$
1 - $\dfrac{8}{36}$
2 - $\dfrac{12}{36}$
3 - $\dfrac{8}{36}$
4 - $\dfrac{4}{36}$
or 288 possible ways (determined similar to the above). Next, the sum of the two dice on the first roll will be 2 (two) with probability $\dfrac{12}{36}$ as indicated. In this case, roll again and the sum of the two dice on the second roll will be:
outcome - probability
0 - $\dfrac{4}{36}$
1 - $\dfrac{8}{36}$
2 - $\dfrac{12}{36}$
3 - $\dfrac{8}{36}$
4 - $\dfrac{4}{36}$
or 432 possible ways. Next, the sum of the two dice on the first roll will be 3 (three) with probability $\dfrac{8}{36}$ as indicated. In this case, stop. Next, the sum of the two dice on the first roll will be 4 (four) with probability $\dfrac{4}{36}$ as indicated. In this case, also stop. Drawing a probability tree using the above information might prove beneficial. Hope this reply helps.
A: 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 $\frac{48}{876}$.
