| Acharnement | 30 Jan 2009 4:46 p.m. PST |
I wanted a bell curve of results similar you would get on 2d6- but on a single die so I could roll more than one at a time. So, I worked out how to re-number the faces on a d12 to get something close to it. Cut up labels and glue them on the faces of a regular d12 unless you happen to have a blank d12 lying around. (You can also just use a regular d12 and refer to the table below.) Write the New face numbers on the die faces. If you covered all the faces then it doesn't matter which faces you write the numbers on. If you deduced that you could leave the 6 and 7 faces uncovered, then give yourself a point for foresight. Here goes: Original face: New face 1: 1 / 2* / 3 (Yes, 3numbers on one face) 2: 3 / 4 (two numbers on one face) 3: 4 / 5 (two numbers on one face) 4: 5 5: 6 6: 6 7: 7 8: 7 9: 8 10: 8 / 9 (two numbers on one face) 11: 9 / 10 (two numbers on one face) 12: 10 / 11* / 12 (3 numbers on one face) Now, what do we do with the faces with more than one number? On the original faces of 2, 3, 10, and 11 there are now 2 numbers: respectively 3 & 4, 4 & 5, 8 & 9, and 9 & 10. It doesn't really matter where you write the numbers on the die face because it is the orientation of the die, not the numbers, which will determine the value of the roll. When you roll the die, and the top of the die face pentagon ends up facing away from you (up), read the higher number. If the top of the pentagon ends up facing toward you, read the lower number. So, for example, if you get the 4,5 face, and the top of the pentagon faces away from you (up), the result is 5. In the case of the original 1 and 12 faces, which now have 3 numbers on them is gets a little fiddly. If the top of the pentagon lands facing away from you (up), the result is the middle, asterisked result (the 2 or the 11 depending on the face). If the top of the pentagon lands facing toward you and to the left (down, left), the result is the lowest or highest number (the 1 or 12). If the top of the pentagon lands facing toward you and to the right (down, right), the result is 3 or 10, again depending on the face. Since the direction the top of the pentagon is facing may be subject to some heated disagreement, I recommend using the edge of a piece of paper or a line drawn on the table with the die roller and the die rolling place on opposite sides of it. This helps to determine the pentagon top direction. Pre-emptive comments notes: 1. The 6 and the 7 have the same chance of occurring so the bell curve is flat on top but until they make an odd-numbered-sided polyhedral, it is good enough for me. 2. I could get a similar result by rolling several same-colored pairs of d6s but that would mean no 1's and that you would not be squandering your time by reading this post. If this idea/madness can be improved upon/further madified, I would be most interested to hear about it. Your comments and derision are welcome. Thanks! |
| Farstar | 30 Jan 2009 5:13 p.m. PST |
so
1 – 0.33 of a face 2 – 0.33 of a face 3 – 0.83 of a face 4 – 1.0 face 5 – 1.5 faces 6 – 2 faces 7 – 2 faces 8 – 1.5 faces 9 – 1.0 face 10 – 0.83 of a face 11 – 0.33 of a face 12 – 0.33 of a face Seems a bit stilted. |
| Mark Plant | 30 Jan 2009 5:31 p.m. PST |
You have actually flattened the bell curve quite a lot compared to 2d6. You'd get better results using a d20. I recommend using matching pairs of coloured dice and still rolling 2d6. It's a lot less effort. |
| emckinney | 30 Jan 2009 7:00 p.m. PST |
Uh, you do realize that you have 20 results, so you could just renumber a d20? |
| I Jim I | 30 Jan 2009 8:14 p.m. PST |
I still don't know why you want to do this, but to have the exact same distribution as 2D6, the following is the number of sides on a d12 for each number result: 2 – 1/3 of a face 3 – 2/3 of a face 4 – 1 face 5 – 1 & 1/3 faces 6 – 1 & 2/3 faces 7 – 2 faces 8 – 1 & 2/3 faces 9 – 1 & 1/3 faces 10 – 1 face 11 – 2/3 of a face 12 – 1/3 of a face or in your notation: Original face: New face 1: 7 2: 2 / 3 / 3 3: 5 / 6 / 6 4: 4 5: 5 6: 6 7: 7 8: 8 9: 9 10: 10 11: 9 / 8 / 8 12: 12 / 11 / 11 |
| KatieL | 31 Jan 2009 3:49 a.m. PST |
It's funny I should read this. TOH is sat next to me letrasetting numbers onto blank Chessex D20s in order to put a bell curve on them rather than a flat 1-20 distribution. He's plotting on using them in the D20 system RPGs. |
| pphalen | 31 Jan 2009 6:05 a.m. PST |
Use paired colors of d6s, much simpler. Or blow the whole idea out even further and create a binomial distribution table and use percentage dice to determine the entire results from multiple 2d6 rolls
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| pphalen | 31 Jan 2009 6:06 a.m. PST |
And just to be pedantic, it is a "Gaussian Ditribution" and not really, since 2d6 is a step function
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| Acharnement | 31 Jan 2009 6:58 a.m. PST |
Thanks very much for all your answers. I will try the d20 method as recommended. Thanks to James Viel for working out the faces' numbering. I am looking forward to the looks of disgust when I pull out the new die and mention that it is a Gaussian distribution! Lots of fun. |
| Nappy29388 | 31 Jan 2009 8:24 a.m. PST |
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| bobstro | 31 Jan 2009 10:39 a.m. PST |
Well, it makes sense to me. You want to roll a bunch of 2d6 at once. The modified D20 sounds ideal. - Bob |
| Kevin Cook | 01 Feb 2009 6:56 a.m. PST |
Is this perhaps what you are looking for? picture |
| bobstro | 01 Feb 2009 8:53 a.m. PST |
Interesting, Kevin. Gotta wonder about tossing a handful of those though. :) I think I'd just use a computer at some point. - Bob |
| Kevin Cook | 01 Feb 2009 10:49 a.m. PST |
> I think I'd just use a computer at some point. Ditto I was just trying to come up with a reason for inventing another 'sided' die
ie the D36 in this case I think a die to do the curve for 3d6 would be 216 sided wouldn't it? (6^3) |
| Chris PzTp | 02 Feb 2009 11:36 a.m. PST |
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| Kevin Cook | 02 Feb 2009 2:46 p.m. PST |
> Does anyone make a D36? The one in the link above is the only one I have ever seen
and I made that one |
| Farstar | 03 Feb 2009 1:49 p.m. PST |
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