Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**Falesh****Guest**

Hi there,

I've been working on a programme that calculates the chance of success of rolling different sets of D6 (dice). I think I have it right but I am not a mathematician, I just enjoy maths, so I would be very grateful if someone can confirm my calculations are correct. The chance of success should be all of the dice getting at least the number required. For example:

Each of there can be re-rolled once if they fail: 2, 3, 4

One of these can be re-rolled if it fails: 2, 3 (e.g. I roll a 2, 2 & re-roll the second 2 as it failed)

One of these can be re-rolled if it fails: 4, 5

None of these can be re-rolled: 2, 3, 4

The result I get is a 5.417905044962658% probability that all of the above will succeed.

The tricky part, to explain as well as code, is when there is an extra re-roll that can be used only once but can be used by any of the "One of these can be re-rolled if it fails:" or "None of these can be re-rolled:" blocks. Also once a dice has been re-rolled it cannot be re-rolled a second time so the extra re-roll is only useful for the "One of these can be re-rolled if it fails:" blocks if two dice fail. The result I get when adding this extra re-roll to the rolls required about is: 11.869997092809927%.

I hope I explained that well enough. You can find the code here: probability.java, look for the calculate() function at the bottom as a starting point.

Many thanks