just a guy made of dots and lines (crs) wrote,
just a guy made of dots and lines

math dumbness at DDXP?

So there was this effect that was supposed to last "2d4 rounds".

The DM decided he was going to figure out how long this lasted by keeping count of the number of rounds so far, and each round rolling a 2d4 to see if the number that came out was the current number of rounds or lower. I assume this was an attempt to make the number of rounds of the effect's duration unknown to everyone at the table, so that it wouldn't unconsciously affect his decisions about the opponents' actions.

Process A: "roll 2d4, that's the number of rounds". This gives the triangular distribution so familiar to players of Settlers of Catan, with the slight adjustment in the size of the triangle.

Process B: "each round, roll 2d4, see if that number is less than or equal to the number of rounds so far."

What kind of distribution does process B give?

P(2)=1/16 (same as process A)
P(3)=(1-P(2))*3/16 = .17578125 (much higher than process A's .125)
P(4)=(1-(P(3)+P(2)))*6/16 = .28564453... (much higher than process A's .1875)

...seeming weighting the value towards lower values. (Maybe I should write a quick program to do the calculations instead of just typing numbers into "dc", and see.)

On one hand, this is kind of a weird, wrong process to go through...

On the other hand, it means players spend less time blinded by the dust in that mod, which is more fun for people. So, like,... maybe it was a conscious decision to weight the values lower. That's not bad. And if the DM doesn't even know when the blindness will lift, maybe that's a good thing for everyone involved.

I'll be mildly embarrassed if my math is wrong, but I have to admit I'm feeling a litle rusty... Maybe I should write that program and make sure P(2...8) add up to 1, as a sanity check.


p(0) = 0.0
p(1) = 0.0
p(2) = 0.0625
p(3) = 0.17578125
p(4) = 0.28564453125
p(5) = 0.29754638671875
p(6) = 0.14505386352539062
p(7) = 0.031381845474243164
p(8) = 0.002092123031616211
total = 1.0

I feel better about posting this now.
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