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

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.
