Rolling an 11+ on 3d6

Question

I roll three d6, discard the lowest, and need an 11 or 12. I may also reroll one of the dice.

I calculated the probability of rolling an 11+ on 2d6 as .056, and with a reroll as .102.

Where I get hung up is being able to discard the third die.

Example: I roll 3d6 and get a 4, 1, 6. I keep the 4 and 6 and reroll the 1 in hopes of getting a 5 or 6 on it.

Answer 1

Exactly one of the following mutually exclusive outcomes will occur:

• You roll three sixes (occurs with probability $\dfrac{1}{6^3})$

• You roll exactly two sixes (occurs with probability $\dfrac{3\cdot 5}{6^3}$)

• You roll exactly one six and two fives (occurs with probability $\dfrac{3}{6^3}$)

• You roll exactly one six and one five (occurs with probability $\dfrac{6\cdot 4}{6^3}$)

• You roll exactly one six and no fives, followed by rerolling one of the not-fives into a five or six (occurs with probability $\dfrac{3\cdot 4\cdot 4}{6^3}\cdot\dfrac{2}{6}$)

• You roll exactly no sixes and at three fives, followed by rerolling one of the dice into a six (occurs with probability $\dfrac{1}{6^3}\cdot\dfrac{1}{6}$)

etc...

Adding these together gives a final total probability.