Did you know that the maximum number of sides for a mathematically fair die is 120?
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Did you know that the maximum number of sides for a mathematically fair die is 120?
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Or maybe he does, and just gives me grief about it not being fair.
Edit: I have blasted a few rolls in my Discobot topic, and it seems he isn’t capped out at 120 after all. I have been working on a faulty assumption.
Okay - that makes things simpler. ![]()
Hi! To find out what I can do, say @discobot display help.
I guess we won’t know for sure until he comes up with a number higher than 120 (haven’t seen the rolls in discobot topic - has he??) but since he’s not actually rolling a dice and should just be working on a random number generator algorithm there’s no reason he can’t go much higher than 120 ![]()
ETA I think this should be the fairest way of random choice each time
it works! ![]()
I did think it was an odd limitation to impose on him.
As yet, I haven’t actually seen him go higher than 120 in my test rolls (I’m hoping the ‘yay, it works’ means that you’ve succeeded?).
If it works I’ll do the 1d140 (141, 142 etc).
Bit bummed I capped off the Role Play game at 120 now though…
I think there is plenty of variety with the number of roll play options listed! ![]()
Oh, of course. 120 is loads. Who would possibly need/want more than that?
*quietly shelves plans for a 250-slot mega-grid…*
Knock yourself out!
Small update. I don’t think he does go higher than 120…
@discobot roll 20d500
Did you know that the maximum number of sides for a mathematically fair die is 120?
55, 82, 9, 86, 80, 82, 27, 31, 45, 3, 25, 69, 49, 57, 108, 114, 46, 46, 3, 87
Oh well that’s disappointing and very stupid since no real dice is being rolled!!! ![]()
Oh well. As you were. ![]()
When Brenna edits a post she normally leaves an edit message to tell you why. It also leaves a note in your edit notification too, so if you missed it the first time you can have a look back and see. ![]()
It’s also in the edit history of the post as well. They’re only visible to you and the site staff, but you can click on the pencil icon in the top right of an edited post to see all the revisions ![]()
It was over a week ago, so I don’t know why except it said it was offensive to others (it was something I posted in the fantasy thread). I guess I need to read the words you can’t say on the forum.
I didn’t see it until today. I’ve been away with my husband tending to my mother-in-law’s funeral arrangements, etc. She died the Saturday night before Easter. Just now getting back to somewhat normal.
I’m sorry to hear that. That must have been a tough Easter for you all.
I wouldn’t worry too much about the odd edit here and there, especially if you’re a newer member. If it was on overly descriptive fantasy they can often get a nip and a tweak here and there if they’ve erred to the salacious side. ![]()
@Ian_Chimp @Peitho could you have him role a die to see which group I. E. 1-9, 10-19 etc… And then roll another die to choose number in that range.
Thinking 10 would be 0, 10, 20 etc…
Ideally it’d be good for it to be done in one Discobot-ing, so the dice would have to have the same number of sides. I think that way would need a 14 sided and a 10 sided? (it’s early, so I may have misunderstood
)
I think using two dice like I have been makes it easier to roll a mid-range number, as there are multiple combinations to make each. Whereas there are fewer combinations to make numbers at the lower/higher ends (eg 1 + 1 vs 49+51/48+52/47+53 etc etc). So the probability of rolling 100 is much greater than rolling a 2.
I’m thinking 1 die is the only way to go. I’m thinking alternate weeks will dictate the start point. So even rounds are 1-120, and odd 20-140 (the top number will increase by 1 each week, but this method should cover us until there’s 240 posts
).
Ooh, comes with complications when it isn’t a total of multiple of 10 (…though 141 could be a 3 sided and a 47 sided, 142 could be a 2 sided and a 71 sided?) but I like it because it does result in a completely fair roll combining the two numbers this way with the first dictating the section of numbers to use and the second the specific number within it. Means, despite rolling the 2 dice every outcome is weighted equally. Impossible for prime numbers but all others would work.
Do you think you could demonstrate it? I mean, obviously I totally understand it 100%. But just so everyone else can follow along too…
Does have to be two rolls for discobot but the outcome is a fair chance for every possible number, for example:
For 141 possible answers (using as example of slightly trickier numbers)
Roll one would be 1d3 to let you know whether to use numbers 1-47, 48-94 or 95-141
Roll 2 would be 1d47 which tells you which number within the section to use.
(For 140, would be 1d14 then 1d10, etc.)
ETA 149 would be the first prime but you maybe just roll it as 150 and hope 150 doesn’t come up ![]()
Yayyy I’ve learnt something @Ian_Chimp