Arkham Horror and Bathtubs of D6

Matt Miller 

In Arkham, extra dice is always better. Every die in Arkham adds +1/3. But Arkham does a very nice job of limiting the dice you can add through the use of 'hands'. You only get two, and you you an only use two hands worth of spells/weapons* in a single attack.  In Arkham, things which do add piles of dice (withering, etc) also require a roll to activate, and include an additional cost. 

I recall some game that played with d10s. It made doing the mental math easy...but I think it was based on the ShadowRun style 'bathtubs of D6', where only a 1 counts as a success...which makes it worse than ShadowRun. ShadowRun, a 6d6 makes a success very likely. With a d10 system, you need 10d10. Making it a 1 or a 2 on a D10 doesn't alter it that much: 1/5 chance, vs a 1/6 on a d6--not worth the extra dice required. 

**Recalls to me, before the 'Christmas Tree' twinking out, there was a rule that players could only have one magic item.

Arkham Horror Math

Playing a great deal of Arkham Horror, which has a 1/3 success rate for dice; just came across the math for predicting count of successes, given X dice, an a X success rate. Begs to be used for an RPG.

Arkham Horror Mechanics

Matt Miller 

I've been playing a great deal of Arkham Horror, where a 5 or 6 on a d6 count as a success. Rolling 8 or 9 dice, it takes a bit to pick out successes. But estimating how likely something is to succeed is easy; every dice you roll gives you 1/3 of a success. Which makes it tempting to adapt as a base mechanic for an RPG. But I'd like more than two ways (1) Extra dice, 2) Extra success) to affect a players roll.

Matt Miller 

Arkham has a mechanic which asks you to roll a success. Mentally, it's hard to imagine how that affects the outcome, as it is deeply unintuitive. In effect, it squares your chances. If you roll 1 dice, your chance is 1 in 9 (1/3*1/3). But if you roll 2 dice...I become confused. The probabilities get wonky. Chance of getting 1 success, on X dice, squared, is the match. X*(1/3)^2...I think. If I have 5 dice, the chances of getting 1 success, twice, is not the same as the chance of getting 2 successes on 10 dice; the first is harder.