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(1)=0

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.

Ah:

`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

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.

## Comments

zkzkznakorI'm pretty sure that if he were rolling a d8 for this, the end of the cloud would be a Poisson process. With 2d4, it's a mutant Poisson process. I haven't thought about this in too many years to remember its name.

rifmeisterCertainly rerolling every time like this skews toward lower values.

alieraknakortirianmalI think that not only did this show a distinct lack of understanding of probability, but also of the rules of the _game_.

I would have complained to a senior judge purely on that basis.

gorgoAnd changing the probability distribution in order to keep the information hidden is definitely a no-no.

abceRound 1: Do nothing.

Round 2: Roll 2d4. on a 2, terminate effect.

Round 3: Roll a d30. On a 1-4, terminate effect.

Round 4: Roll d20. On a 1-3, terminate effect. On a 14+, reroll.

Round 5: Roll d10. On a 1-4, terminate effect.

Round 6: Roll d8. On a 1-4, terminate effect.

Round 7: Roll d12. On a 1-8, terminate effect.

Round 8: Roll dn. On a 1-n, terminate effect.

See, isn't that simpler?

kvarkoRound 2: 1

Round 3: 2

Round 4: 3

Round 5: 4

Round 6: 3

Round 7: 2

Round 8: 1

So at round N, what is probability of stopping there? It's the number of chances for the current round, divided by the the sum of the chances from that round on-ward (not including any previous rounds):

Round 1: 0

Round 2: 1/16

Round 3: 2/15

Round 4: 3/13

Round 5: 4/10

Round 6: 3/6

Round 7: 2/3

Round 8: 1/1

crsLess completely hilarious terms :)

kvarkostilltook me several attempts (and many minutes) to figure out how to come to that conclusion. (Sorry if this spoils the exercise for the next reader.)I hadn't decided whether round 8 was purely being funny or was a reminder to hide knowledge from the players -- in which case, you could also roll on round 1.

abceI probably could have gone with : Round 1: Flip a coin. If it comes up heads, roll 2d4 and follow the normal rules. If it comes up tails, roll 3d4, but randomly throw one of the dice at the other GM, and follow the normal rules. If it lands on its edge, and refuses to fall over, even after you hit it, then proceed with these rules.

kvarko