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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.


( 13 comments — Leave a comment )
Feb. 19th, 2007 09:14 am (UTC)
too many numbers. you should graph the two distributions against each other.
Feb. 19th, 2007 02:53 pm (UTC)
Sure it's weighting towards lower values: imagine if the first roll is an 8. If you're following the rules, it lasts for 8 rounds. Following this GM's approach, you've got to roll 7 more 8s to get it to last that long!

I'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.
Feb. 19th, 2007 03:23 pm (UTC)
I don't think it can be a Poisson process. In a Poisson process, the arrivals are memoryless --- you always expect to wait the same amount of time for the next arrival, regardless of how long you've waited so far. This is not memoryless (it's bounded by 8 rounds in any case, and that's still true with a d8).

Certainly rerolling every time like this skews toward lower values.
Feb. 19th, 2007 03:41 pm (UTC)
No, in this GM's approach, to make the effect last 8 rounds, you need only roll anything above a 1, followed by anything above a 2, followed by anything above a 3, and so on. If it were about rolling 8 consecutive 8s, p(8) would be *much* lower.
Feb. 19th, 2007 04:19 pm (UTC)
Yes, that's so. I was trying to say something about the relative value of an 8 rolled early. But I think I was pretty confused when I wrote Poisson.
Feb. 19th, 2007 06:19 pm (UTC)
You really have to ask yourself ... why didn't he just roll 2d4 out of sight if he didn't want folks to know what the end round was?

I 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.
Feb. 20th, 2007 12:39 am (UTC)
The GM's approach has the advantage that it's double-blind: neither the GM nor the player know how long the effect is going to last. Just rolling out of sight means that the players don't know how long it will last, but that the GM does. Now, if the GM isn't capable of playing NPCs as if he/she didn't know information that isn't available to the NPCs, he/she probably shouldn't be GMing a competitive event.

And changing the probability distribution in order to keep the information hidden is definitely a no-no.
Feb. 20th, 2007 04:56 am (UTC)
If he'd wanted to keep it unknown, he simply could have:
Round 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?
Feb. 21st, 2007 12:15 am (UTC)
Indeed! The distribution of 2d4 is:

Round 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

Feb. 21st, 2007 12:20 am (UTC)
Well, yes, you're saying what Andy said, only in different terms.

Less completely hilarious terms :)
Feb. 21st, 2007 12:53 am (UTC)
Well, yes, that was my intention. After reading the proposed die-rolling process, it still took 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.
Feb. 21st, 2007 01:49 am (UTC)
I realized after I had posted that I meant for round 8 to actually be a d6, which would have resulted in every type of die in my bag being used.

I 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.
Feb. 20th, 2007 11:34 pm (UTC)
You might find this interesting.
( 13 comments — Leave a comment )