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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.

Elric Greywolf
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    3d6? Is that D&D jargon? – Angina Seng Jun 05 '18 at 21:19
  • I suggest just breaking it into cases. If the first roll (of three) yields a maximum $≤4$ then there is no chance. If you already have a pair adding to $11,12$ then you are done. If you have a $5$ alone then the answer is $\frac 16$. If you have a $6$ alone then the answer is $\frac 13$. – lulu Jun 05 '18 at 21:21
  • @LordSharktheUnknown As an aside, yes "roll $xdy$" refers to rolling $x$ fair dice, each of which have $y$ sides with sides labeled $1,2,3,\dots,y$, so $3d6$ refers to rolling three standard six-sided dice. Usually it further implies that you add the results, but the OP apparently is skipping that part of the meaning here with the "discard the lowest" clause. (As an additional aside, common dice shapes in production include d4, d6, d8, d10, d12, d20, and some less common include d3, d30, d100) – JMoravitz Jun 05 '18 at 21:57
  • The question is unclear. In the first line, you say without qualification that you discard the lowest die. In the third line, you say that you're "able to discard the third die" -- that's different from always discarding it. And then in the example you don't mention the discarding aspect at all. – joriki Jun 07 '18 at 12:53
  • @joriki it seemed clear to me. Reworded, roll four dice and look at the sum of the largest two dice. (Since the option of rerolling the lowest can never hurt and can only help the sum of largest two and since the number appearing on smallest doesn't appear in sum we may assume wlog s reroll always occurs). The reroll and discard does appear in his example. Having rolled a 4,1,6 the current sum of largest two is 10. Rerolling the smallest will either keep sum of largest two as 10 (if reroll became 4 or less) or will make sum as 11 (if 5) or 12 (if 6) – JMoravitz Jun 08 '18 at 15:37
  • @JMoravitz: Unfortunately that hasn't done much to clear it up for me :-) Do you really mean "four dice"? Because the OP rather unambiguously says "three d6" twice. Also, it's not any clearer to me from your description than it was from the OP whether reroll and discard are the same action or different actions, It sounds like they're the same action (at least I can't discern two different actions of that kind in your description), but the OP says "discard the lowest ... I may also reroll one of the dice". That doesn't seem compatible with the reroll being the same thing as the discard. – joriki Jun 08 '18 at 17:13
  • Yes, I intentionally said four dice, because the end result of the probability distribution will be the same. I will attempt to reword the original intention then again. You begin by rolling three dice. We are attempting to maximize the sum of the largest two results. Once having rolled the three dice, we have the option to reroll the smallest (and indeed no reason not to). After the reroll of the previous smallest has occurred, we look at the new collection of three dice results we record the sum of the largest two results (i.e. ignore the smallest, said as "discard" the smallest) – JMoravitz Jun 08 '18 at 17:23

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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.

JMoravitz
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