Help with probabilities

**ThiagoMartell** · 2012-09-09, 10:07 AM (ISO 8601)

I'm designing my own system and I'm havin trouble calculating probabilities.
I'm using the 'roll X, keep Y' dice roll method, which seems like the most commons. However, instead of adding together, I'm just using counting - kind of like old Storyteller, but instead of a single number, you count duplicates.
So if you roll 2, 2, 3, 4, 5 you have one success (the 2s), if roll 1, 1, 1, 1, 6, 6 you have three degrees of success (the 1s).
How can I calculate the probabilities for this system?

**TuggyNE** · 2012-09-09, 04:51 PM (ISO 8601)

I've rigged up an AnyDice program that should do the job. It could still use a bit more testing to be sure, but 1 and 2 dice are correct, and 3, 4, and 5 seem plausible.

The basic principle derived from that is that any of the six numbers has an equal chance of being most common, so you can decrement the number of dice, and count how many are equal to each chosen initial number in the remaining dice, then take the highest count. (I suspect there's a way to make this recursive, but I didn't bother.)

**Yitzi** · 2012-09-09, 08:02 PM (ISO 8601)

You have to compensate, though, for the chance that there will be a tie as to which is the most common. "three 1s and three 6s" doesn't get to count twice.

**TuggyNE** · 2012-09-09, 08:24 PM (ISO 8601)

The given code takes that into account; it returns the highest count, and nothing more. So if you roll 6d6 and get three 1s and three 6s, it returns 3 from the 6s, which happens to be valid enough.

**ThiagoMartell** · 2012-09-11, 01:27 AM (ISO 8601)

Thanks a lot, that's very helpful.
How would the possibility to roll more dice than you keep change that curve?

**TuggyNE** · 2012-09-11, 02:30 AM (ISO 8601)

Originally Posted by ThiagoMartell

Thanks a lot, that's very helpful.
How would the possibility to roll more dice than you keep change that curve?

Well, specifically? If it's just "keep X dice that you like out of Y", then it would basically mean that you'd take the existing curve for Y dice and cap it at X - 1 successes (which changes very little). If it's something more complicated you'll have to spell it out, sorry.

**ThiagoMartell** · 2012-09-11, 05:09 AM (ISO 8601)

Originally Posted by tuggyne

Well, specifically? If it's just "keep X dice that you like out of Y", then it would basically mean that you'd take the existing curve for Y dice and cap it at X - 1 successes (which changes very little). If it's something more complicated you'll have to spell it out, sorry.

That's pretty much it, yeah. Thanks, that helped a lot.

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