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275647 Posts in 27717 Topics by 4285 Members Latest Member: - Jason DAngelo Most online today: 70 - most online ever: 565 (October 17, 2020, 02:08:06 PM)
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Author Topic: Dice distribution calculator  (Read 12578 times)
Jasper Flick
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« Reply #30 on: May 19, 2009, 02:11:31 PM »

You're in luck! This feature is in the pipeline. It will replace the old way of doing it. I might implement it tomorrow, otherwise later this week.
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Jasper Flick
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« Reply #31 on: May 23, 2009, 05:45:54 AM »

The calculator now has a more powerful set notation!
You can do stuff like 1d{3=5} which is 1d{3,3,3,3,3} and 1d{2..5} which is 1d{2,3,4,5}. You can combine it too, like 1d{1..3=2,4,6=3}. See the instructions for details.
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Akos Szederjei
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« Reply #32 on: May 24, 2009, 11:57:25 AM »

Thanks for the great tool. It is really helpful!
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Devenger
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« Reply #33 on: June 14, 2009, 11:29:58 AM »

This has been really handy for visualising the likelihood of very unusual things happening (for example, I now realise I have a mechanic for something that will only happen on 0.02% of rolls... so I'll probably not include that). So, thank you very much!

For something else I've been experimenting with: with the current notation, can you ask it the probability that, when rolling XdY dice, Z dice show the same number, and that number is above V? For example, could you somehow ask 'what is the chance that, when rolling 5d6, you get a pair of dice showing the same number, a number above 3?'

I'll probably try coding something in Python if that isn't possible - it is a pretty obscure thing. But I've been amazed at some of the things you can do...

Thanks again.
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Jasper Flick
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« Reply #34 on: June 15, 2009, 11:05:51 AM »

Great it's of use!

A 0.02% is a bit small for a special mechanic indeed! But if it's a fun mechanic, perhaps you can increase its chances of triggering instead of removing it?

The calculator can't tell you want you want to know, because it doesn't compare individual dice (except for finding highest/lowest X). Well, perhaps you can get close, but that would be so contrived it'll be practically useless. There's just too much to track: how many pairs occur (up to two), and which numbers they are.

The calculator is really designed to deal with sums of dice, with a rewrite-gimmick thrown in to get something like counting successes. It's fun people are pushing it to its limits, but now those limits are painfully obvious to me. It's getting time to create a version 2!
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dindenver
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« Reply #35 on: June 15, 2009, 11:49:46 AM »

Jasper,
  Two questions:
1) How do you model Exalted dice (Where 7, 8 and 9 are one success, 0 is two successes and it is a botch if you roll all ones)? I think I can do it without the botch, but I have no clue how to isolate the both
2) Can you get it to work in such a way that shows the die faces instead of summing it? Like what are the odds of getting 6, 6 5 on 3d6 and what if it was 4d6 keep 3? Obviously it would need to sort them or every instance would be one chance.
  If you don't want to code your web page to do this, can you at least tell me how to calculate it, my system uses that and I have no idea what my die mods do. I have a pretty good handle on it, but it is all anecdotal.

  Either way, its a cool program, good job man!
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Dave M
Author of Legends of Lanasia RPG (Still in beta)
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Jasper Flick
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« Reply #36 on: June 15, 2009, 01:03:23 PM »

Hi Dave!

1) Ignoring the botches, default target-7 Exalted dice are 1d{0,0,0,0,0,0,1,1,1,2} or 1d{0=6,1=3,2}. You can roll multiple dice without problem.
Botches are trouble though. You need to know if, in case of no successes, you rolled any 1s, and optionally the number of 1s. You can't just weigh the 1s as -1 because that would disturb the success count, so they need to be tracked separately. This is something the calculator can't do. You can sneak additional information in the result, but the input string gets unwieldy and won't scale. Regarless, here it is:

1d(5d{100,0=5,1=3,2})c{0=0,1=1,2=2,3=3,4=4,5=5,6=6,7=7,8=8,9=9,10=10,100=-1,101=1,102=2,103=3,104=4,105=5,106=6,107=7,108=8,200=-2,201=1,202=2,203=3,204=4,205=5,206=6,300=-3,301=1,302=2,303=3,304=4,400=-4,401=1,402=2,500=-5}

This works for up to five dice. Decrease the red number to see the results for fewer dice.

2) Unfortunately, the data of indivual dice gets lots, so it can't do that. Closest is the cartesian product, but that one preserves order so it's not what you want.

For the specific case of getting {6,6,5}: there are 3 ways to order that sequence ({6,6,5}, {6,5,6} and {5,6,6}) and there are 6*6*6 sequences in total, so it's 3/216, a 1.39% chance. Note that this is the same for all sequences of the form {X,X,Y} on 3d6.

Also note that you can only get a total of 17 by rolling {6,6,5}, so you can look that up in the distribution graph of 3d6. Or you can input 1d(3d6)=17 in the calculator to get only the value you want (and 0 for the rest). That way you can get the answer for 4d6h3 as well: 1d(4d6h3)=17. It has a 4.17% chance of occuring. Unfortunately, that trick only works for a few totals.

I'm definitely going to make a new calculator that should make this stuff easy, but that's gonna take a while to complete.
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dindenver
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« Reply #37 on: June 15, 2009, 01:31:57 PM »

Jasper,
  Thanks for the prompt response. I just used 6, 6, 5 as an example, how do I calculate them for other possibilities, like 6, 5, 4? I think I can do it with straight 3 dice, but then it gets all fuzzy when I throw 4 and keep 3. I tried mapping all the die faces in excel (over 1200 rows, ug), but I still get thwarted by my inability to take 3...
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Dave M
Author of Legends of Lanasia RPG (Still in beta)
My blog
Free Demo
Jasper Flick
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Posts: 161


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« Reply #38 on: June 16, 2009, 01:28:10 AM »

What you're specifically looking for here are orderer sequences. In the case of roll 4 keep 3 highest, you're looking at a sequence with a wildcard. So getting {6,5,4} out of 4d6h3 requires a sequence of {6,5,4,X}, where X is at most 4. You can calculate this by hand by considering two cases:

1) The case {6,5,4,4}. This is a sequence of the form {A,B,C,C}. There are 12 ways to order this sequence: 4 ways to place the A, times 3 ways to place the B, with the holles filled with the Cs.

2) The cases {6,5,4,3}, {6,5,4,2}, and {6,5,4,1}. These are three sequences of the form {A,B,C,D}. There are 24 ways to order this sequence: 4!.

So {6,5,4} on 4d6h3 can show up in 12 + 3 * 24 = 84 different ways, out of 6^4 = 1296 possible outcomes. That's a 6.48% chance.

Now I'll do the same for the other two ways to get 15:

Case 6+6+3 is {A,A,B,B} (6 ways) + 2 * {A,B,C,C} (12 ways) = 30 out of 1296 is a 2.31% chance.

Case 5+5+5 is {A,A,A,A} (1 way) + 4 * {A,B,B,B} (4 ways) = 17 out of 1296 is a 1.31% chance.

Taking them together, there are 131 ways out of 1296 to get 15 on 4d6h3, a 10.11% chance. Fortunately, the calculator agrees.
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