# atomic highway dice probability

Atomic Highway is a game system that uses pools of 6-sided dice to determine player successes. I'm trying to write a probability calculator for this system but have hit a snag. The rules are as follows:

1. You roll 1-5 6-sided dice, counting dice that result in 6 as successes.
2. Reroll the dice that rolled as a 6, keep rerolling 6es until you have no more 6es. Add the new 6es to the successes.
3. You get 0-5 bonus points you can divide as you want among the dice that didn't result in 6es on the first roll. If you manage to bump up one or more dice results to 6 with this, add those to successes.
4. If your total successes is equal to or higher than a treshold, the player succeeds.

I originally found a reddit thread describing a process to calculate the odds of success (thread here) and managed to program this process, but unfortunately the odds they achieve with their method are way off.
Specifically, they claim when rolling 3 dice with 2 bonus points and needing at least 2 successes, you have a 71% chance of succeeding. The rulebook however provides a table with the success chances and cites a 41% chance of success in this case. I simulated 1 million of these dice rolls and averaged the results, coming out at about 37-38% success rate.
Clearly the reddit method is wrong, and I've unfortunately not found alternatives. Can anyone here find where the reddit method goes wrong and explain how you should calculate this for a more accurate result? Thank you in advance!

• The description on the reddit thread differs from what you wrote: 1) If you reroll an in itial 6 and get a 6 again, this is in the reddit thread not described as counting success (maybe an omission) 2) The reddit thread allows adding bonus points to non-6 values that are rerolled initial sixes (the $5$ in the example), which you explicitly forbid. Please check that the rules you describe are really what you want. – Ingix Jun 19 '19 at 14:21
• The rerolls being counted is indeed an omission within the thread, and the game rules explicitly state you cannot add bonus points to dice that were the result of a reroll. The calculation in the thread does count extra 6es as successes, but does indeed add the bonus points to all dice. In my implementation I took care to avoid that and only use the original "failed" dice, but still got a result of 69% chance. – Yannick Vanhoutte Jun 19 '19 at 14:36
• FWIW, my quick-and-dirty simulation matches your results. – amd Jun 19 '19 at 19:22

The basic approach in that thread of working through a probability tree is sound, but the computed values are off. The computation goes south quite early in fact. Following the “rolled only one 6 initially” branch, the probability increment for getting another six on the reroll is only $$\frac{75}{216}\cdot\frac16 = \frac{75}{1296}$$. This isn’t off by very much, but the probability for the last branch is way off: this is the case in which there were no sixes rolled, there is exactly one five and at least one four. The correct value is $$\frac{125}{216}\cdot\frac{48}{125}\cdot\left(1-\frac9{16}\right) = \frac7{72}$$, which is less than half the value given in the post and changes the final value by quite a bit.
All told, the correct probability of at least two successes with three dice and three bonus points is $$\frac{121}{216} \approx 56\%$$, much lower than the claimed $$71\%$$. This number matches my quick-and-dirty simulation.
That’s the easy part. For each of the ways in which not enough sixes were rolled, you’ll have a success deficit $$s$$ and a number $$b$$ of bonus points and remaining dice $$d$$ to make up this deficit. If $$d\lt s$$ or $$b\lt s$$ then you’re done—there’s no way to make up the deficit. Otherwise, you’ll need to go through a similar case analysis to the one in that thread. A systematic way to approach this might be to go through the partitions of the $$b$$ bonus points available into $$s$$ summands and work out the winning (or perhaps easier in many cases, losing) combinations of rolls for each.
• @YannickVanhoutte Working the same sort of tree analysis for getting at least two successes on 3d6+2, I calculate a probability of $163/432 \approx 37.7\%$, matching our simulations. – amd Jun 20 '19 at 7:58