Search results

iZombie episode "twenty sided, die"

I generally allow it to bring the party size up, especially if it's just a handful of players. I used to dislike it for newish players, but some are quite savvy around the rules and it deals with some of the decisions of "which class" if they can try several at once. There's clearly no right or...

Yes, I received it too.

Thanks, I'll look into that some more.

I wonder if Lulu could do that.... (printme1 appear to be US only??)

I've found that Pandemic works nicely for single and two-player.

I'm in your third category. They are expensive and represent a new investment. (This isn't a dig at WotC's pricing -- most similar hardback volumes are similarly pricey.)

If you end up with a ratio like 7-2 = 3½-1, you'd round it down to 3-1, as 3-1 is better for the defender than 4-1. Hope that helps.

The Laundry Files. Relies on a decent knowledge of Stross's (thoroughly excellent) Laundry books.

If a humanoid is under a charm person spell, what happens if a subsequent charm person is successfully cast on them? How do other people resolve this? Options I can think of are that the later charm replaces the earlier; or that the later takes no effect; or that both are in effect at the same...

Nanvaent I have fond memories of Nanvaent. It looks like it's still going, too.

It looks really interesting. Are there any reviews? (I've started looking through the rules from your link.)

I've found the Labyrinth Lord rules to be fairly accessible to those a bit older than 5 (7 and 10). They had also discovered the old Fighting Fantasy rules: if you have access to those, they're simple enough and fairly good fun. One that I bought but never tried was RpgKids -- but it seems...

I could see a more abstract game using something like this sort of mechanism. So rather than counting individual rounds or arrows, etc., you have a probability of having expended them. Makes more sense for firearms where you reload an entire magazine at once.

Diminishing returns? The more dice you throw, the more likely you'll get a 6 and end up dropping at least one die?

Ah, good, my code matches. Phew! Brain's not failed that badly today, then. :)

How did you calculate it? My brain's a bit tired, but I got as far as: g(x) = expected number of rolls starting with xd6 g(0) = 0 Then for n >= 1, solve for g(n): g(n) = (n! / 6^n) Sum_i=0..n ( 5^i (g(i)+1) / (n-i)! i! ) ---- My explanation: g(0) = 0 because you've got no dice. With n...

So it'd work something like: 10d6 => 1,1,2,3,4,5,5,5,6,6 => 2 6's so down to 8d6 8d6 => 1,3,4,4,5,6,6,6 => 3 6's so down to 5d6 5d6 => 2,2,3,5,6 => 1 6 so down to 4d6 4d6 => 1,3,6,6 => 2d6 ?

Did you manage to find a group?

PM'd you. Phil.