I asked this question on Reddit a few minutes ago, but maybe one of the three people that reads this blog is good at math/dice stuff:

I want to use two rolls to randomize what is basically a d100 roll. First I roll a d10 to figure out what kind of dice to use for the second roll: 1-2 use a d4 for the second roll, 3-4 use a d6, 5-6 use a d8, 7-8 use a d10, 9-0 use a d12.

For the second roll, instead of just using 2 d10s to make a d100 roll, I’m going to use 2 of the dice generated by the previous roll, which will give some weird results. Like if my first roll was a [2], then I’m using 2 d4s like percentile dice, which gives a pretty limited range: 11-14, 21-24, 31-34, and 41-44. Reading a percentage is the same as if I was using d10s: first die is the tens column, second die is the ones column. So if I rolled 2 d8s and got [6] and [2] that’s a 62. d12 results are treated slightly differently: in the tens column, an [11] is 110 and a [12] is 120; in the one’s column, an [11] adds 11 (which increases the tens column by 1), and a [12] adds 12, so a roll of [12] and [11] would be 131.

So the actual range is 01-132 (right?), but with very weird distributions. Getting 01-09 is pretty hard, actually getting anything that ends in a ‘9’ is hard, but maybe 31 or 33 or 42 is the most common result? I’m hoping that’s where you guys come in – can someone figure out the probability for each result between 1-132? Or is that like some insanely difficult math?

This is for the OD&D supplement CARCOSA, which uses funky dice mechanics, but I’m interested in taking it a little further for some random tables that need some wildly variable results. *edit: The range doesn’t go all the way to 132… there are only a few results you can get above 100. I think…*

My end goal is to create a mutation chart with some wildly variable results, and I want the unique CARCOSA dice mechanics to be a part of it. I don’t know enough math to even know if this is a hugely difficult question.

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scottsz

said:Your first roll is d10, broken up into 5 segments, so right out of the gate you have five sections of 20% each. The dice differ on the second roll only, so each segment of the second roll will show something like normal distribution (http://en.wikipedia.org/wiki/Dice#Probability)

…but… the overall probability for each of the secondary roll distributions (2d4, 2d6, 2d8, 2d10, 2d12) will still represent 20% of the *overall* distribution, so…

The probability of rolling 2 on the secondary 2d6 roll is 1/36 (2.7777778%) x 20% (.2), or 0.5555557% probability overall.

crusssdaddy

said:Okay, I think I’m getting a better handle on it. Thank you.

Daniel

said:You have 720 possible rolls under this system.

The 72 results of 1-10, 103-110, 113-120, 123-132 can turn up in two different ways, so each has a 1 in 360 chance (0.28%) or 1 in 10 (10%) overall.

The 120 results of 19-20, 29-30, 39-40, 49-50, 59-60, 69-70, 79-80, 89-90, 93-102, 111-112, 121-122 can turn up in four different ways, so a 1 in 180 chance (0.56%), 1 in 6 (16.67%) overall.

The 156 results of 17-18, 27-28, 37-38, 47-48, 57-58, 67-68, 73-78, 83-88, 91-92 can turn up in six different ways, so a 1 in 120 chance (0.83%), 13 in 60 (21.67%) overall.

The 160 results of 15-16, 25-26, 35-36, 45-46, 53-56, 63-66, 71-72, 81-82 can turn up in eight different ways, so a 1 in 90 chance (1.11%), 2 in 9 (22.22%) overall.

The 140 results of 11-14, 23-24, 33-34, 43-44, 51-52, 61-62 can turn up in 10 different ways, so a 1 in 72 chance (1.39%), 7 in 36 (19.44%) overall.

The 72 results of 21-22, 31-32, 41-42 can turn up in 12 different ways, so a 1 in 60 chance (1.67), 1 in 10 (10%) overall.

You can see a distribution plot here ( http://gaming.danieldockery.com/carcosadice.png ).

crusssdaddy

said:Holy cow, that’s amazing. How come I haven’t seen your blog before today? Your CARCOSA ritual name generator is fun as hell — ‘The Obstruction of the Litanies’ and ‘The Pendant Fiend from Beyond the Colorless Consuming’ have write-ups in their future.

Daniel

said:Possibly because I only get around to updating when the stars are right. I’ll look forward to those write-ups—I’ve saved a number of those names for future use myself but so far haven’t had a chance to do more than contemplate the possibilities! I was intrigued by “The Primordial Chiming of the Golden Corpses” and “The Dirge of the Encrusted Shamblers of the Lightless Chasm”. One that seemed somewhat generic but still managed to inspire some ideas was “Priest of the Nine-Thousand Nameless Ones”, and I couldn’t help but wonder what might come “From the Deranged Grimoire of the Sunless Charnel-Pit”. That last one gave me the idea to possibly work up a script to generate random grimoires in the same fashion, but fleshing out the idea remains on the to-do list.

James Hutchings

said:Here’s how I’d work it out.

Each possible result of 2d4 has a 1/5 x 1/16 = 1/80 chance of happening (1/5 chance of getting a 1 or 2 on the initial d10 roll, then 16 possible combinations of 2 results of a d4)

Each possible result of 2d6 has a 1/5 x 1/36 = 1/180 chance of happening.

Each possible result of 2d8 has a 1/5 x 1/64 = 1/320 chance of happening.

Each possible result of 2d10 has a 1/5 x 1/100 = 1/500 chance of happening.

Each possible result of 2d12 has a 1/5 x 1/144 = 1/720 chance of happening.

However there are many possibilities that can result from more than one shape of dice.

00 to 10 – you can only get these results from a d10 (because no other dice has a zero). So these results have a 1/500 chance of happening.

20,30,40,50,60,70,80 and 90 – You can only get these from a d10, again because of the zero. So these results have a 1/500 chance of happening.

21,22,31,32,41,42 – you can get these results from any of the five dice, and there are two ways to get it from a d12 (for example twenty-one can be 2, 1, or 1,11). So the chance of these results is 1/80+ 1/180 + 1/320 + 1/500 + 2/720.

11-14,23-24,33-34 and 43-44 – you can get these results from any of the five dice. So the chance of these results is 1/80+ 1/180 + 1/320 + 1/500 + 1/720.

51,52,61,62 – you can get these results from any dice except a d4, and there are two ways to get them on a d12 (for example 62 can be 6,2 or 5,12). So the chance is 1/180 + 1/320 + 1/500 + 2/720.

15-16,25-26,35-36,45-46,53-56,63-66 – you can get these results from any dice except a d4. So the chance is 1/180 + 1/320 + 1/500 + 1/720.

71,72,81,82 – you can get these results from any dice except a d4 or d6, and there are two ways to get them on a d12. So the chance is 1/320 + 1/500 + 2/720.

17-18,27-28,37-38,47-48,57-58,67-68,73-78,83-88 – you can get these results from any dice except a d4 or d6. So the chance is 1/320 + 1/500 + 1/720.

91,92 – you can get these results from a d10 or a d12, and there are two ways to get them on a d12. So the chance is 1/500 + 2/720.

19,29,39,49,59,69,79,89,99 – you can get these results from a d10 or d12. So the chance is 1/500 + 1/720.

100 – 1/500 if you count ’00’ as 100, no chance otherwise.

101,102,111,112,121,122,131,132 – you can only get these results on a d12. So 1/720.

103-110, 113-120, 123-130 – there’s no way of getting these results.

I think that’s right.

Daniel

said:For those you list as not possible, I think you’ve overlooked the possibility of the d12. For instance, you note the 100 as only possible with a 00 on the percentile roll, but you can also land 100 via rolling a 9 in the tens column and a 10 in the ones column with 2d12. Since you can get the 00 once each for a roll of 7 or 8 on the first roll, and can get the d12’s 9 and 10 roll once each for the 9 and 0 on the initial roll, that gives 100 four chances to turn up out of 720 total, which is a 1 in 180 chance. The 103-110 you can get with d12 with a 10 in the tens column and 3 to 10 in the ones, etc.

Since there were so few (computationally speaking) possible results (2*4^2 + 2*6^2 + 2*8^2 + 2*10^2 + 2*12^2 = 720), I didn’t work out the problem statistically but simply computed every possibility with a quick enumerating script and tabulated. The numbers indicated in my first post show how many times those results actually turn up in the list of all possibilities for this dice configuration. HTH

James Hutchings

said:You’re right. I didn’t add in the possibility of getting a 10 on the d12. Adding that in:

Each possible result of 2d6 has a 1/5 x 1/36 = 1/180 chance of happening.

Each possible result of 2d8 has a 1/5 x 1/64 = 1/320 chance of happening.

Each possible result of 2d10 has a 1/5 x 1/100 = 1/500 chance of happening.

Each possible result of 2d12 has a 1/5 x 1/144 = 1/720 chance of happening.

However there are many possibilities that can result from more than one shape of dice.

0 – The only way to get zero is by rolling a double zero on 2d10, and then only if you count double zero as zero rather than as 100. So the chance is 1/500 or none.

1 to 10 – you can only get these results from a d10 (because no other dice has a zero). So these results have a 1/500 chance of happening.

20,30,40,50,60,70,80 and 90 – You can only get these from a d10 or a d12 (the latter with a 10 on the second roll). So these results have a 1/500 + 1/720 chance of happening.

21,22,31,32,41,42 – you can get these results from any of the five dice, and there are two ways to get it from a d12 (for example twenty-one can be 2, 1, or 1,11). So the chance of these results is 1/80+ 1/180 + 1/320 + 1/500 + 2/720.

11-14,23-24,33-34 and 43-44 – you can get these results from any of the five dice. So the chance of these results is 1/80+ 1/180 + 1/320 + 1/500 + 1/720.

51,52,61,62 – you can get these results from any dice except a d4, and there are two ways to get them on a d12 (for example 62 can be 6,2 or 5,12). So the chance is 1/180 + 1/320 + 1/500 + 2/720.

15-16,25-26,35-36,45-46,53-56,63-66 – you can get these results from any dice except a d4. So the chance is 1/180 + 1/320 + 1/500 + 1/720.

71,72,81,82 – you can get these results from any dice except a d4 or d6, and there are two ways to get them on a d12. So the chance is 1/320 + 1/500 + 2/720.

17-18,27-28,37-38,47-48,57-58,67-68,73-78,83-88 – you can get these results from any dice except a d4 or d6. So the chance is 1/320 + 1/500 + 1/720.

91,92 – you can get these results from a d10 or a d12, and there are two ways to get them on a d12. So the chance is 1/500 + 2/720.

19,29,39,49,59,69,79,89,99 – you can get these results from a d10 or d12. So the chance is 1/500 + 1/720.

100 – You can get this on d12s (9, 10). You can also get it on d10s if you count 0, 0 as 100. So 1/500 + 1/720 if double zero is counted as 100, or 1/720 otherwise.

101,102, 110, 111,112,120, 121,122,130, 131,132 – you can only get these results on a d12. So 1/720.

103-109, 113-119, 123-129 – there’s no way of getting these results.

James Hutchings

said:I wish you could edit these. I didn’t count getting a 10 then something else on the d12.

Each possible result of 2d6 has a 1/5 x 1/36 = 1/180 chance of happening.

Each possible result of 2d8 has a 1/5 x 1/64 = 1/320 chance of happening.

Each possible result of 2d10 has a 1/5 x 1/100 = 1/500 chance of happening.

Each possible result of 2d12 has a 1/5 x 1/144 = 1/720 chance of happening.

However there are many possibilities that can result from more than one shape of dice.

0 – The only way to get zero is by rolling a double zero on 2d10, and then only if you count double zero as zero rather than as 100. So the chance is 1/500 or none.

1 to 10 – you can only get these results from a d10 (because no other dice has a zero). So these results have a 1/500 chance of happening.

20,30,40,50,60,70,80 and 90 – You can only get these from a d10 or a d12 (the latter with a 10 on the second roll). So these results have a 1/500 + 1/720 chance of happening.

21,22,31,32,41,42 – you can get these results from any of the five dice, and there are two ways to get it from a d12 (for example twenty-one can be 2, 1, or 1,11). So the chance of these results is 1/80+ 1/180 + 1/320 + 1/500 + 2/720.

11-14,23-24,33-34 and 43-44 – you can get these results from any of the five dice. So the chance of these results is 1/80+ 1/180 + 1/320 + 1/500 + 1/720.

51,52,61,62 – you can get these results from any dice except a d4, and there are two ways to get them on a d12 (for example 62 can be 6,2 or 5,12). So the chance is 1/180 + 1/320 + 1/500 + 2/720.

15-16,25-26,35-36,45-46,53-56,63-66 – you can get these results from any dice except a d4. So the chance is 1/180 + 1/320 + 1/500 + 1/720.

71,72,81,82 – you can get these results from any dice except a d4 or d6, and there are two ways to get them on a d12. So the chance is 1/320 + 1/500 + 2/720.

17-18,27-28,37-38,47-48,57-58,67-68,73-78,83-88 – you can get these results from any dice except a d4 or d6. So the chance is 1/320 + 1/500 + 1/720.

91,92 – you can get these results from a d10 or a d12, and there are two ways to get them on a d12. So the chance is 1/500 + 2/720.

19,29,39,49,59,69,79,89,99 – you can get these results from a d10 or d12. So the chance is 1/500 + 1/720.

100 – You can get this on d12s (9, 10). You can also get it on d10s if you count 0, 0 as 100. So 1/500 + 1/720 if double zero is counted as 100, or 1/720 otherwise.

103-109 – You can get these on d12s (10 followed by 3-9). So 1/720.

101,102, 110, 111,112 – there are two ways to get these on d12s. For example 101 can be 10, 1 or 9, 11. So 2/720.

120, 121,122,130, 131,132 – you can only get these results on a d12. So 1/720.

113-119, 123-129 – there’s no way of getting these results.

Daniel

said:You need to bear the concept of the 10 in the tens column on a d12 in mind and apply it to also rolling an 11 or 12 in the tens column on your 2d12 rolls. E.g., 11 and 3 is a 113, 11 and 9 is a 119; 12 and 3 is a 123, etc., in the same way that rolling the 10 and 3 through 9 made 103-109.

Also, your numbers don’t always seem to be considering multiple ways of reaching a value, or the fact that each potential 2dN result can occur in two possible places, so there is no roll that only occurs 1 in 720, since any roll that can occur can occur at least twice. E.g., you list 122 as 1/720. While 122 can only be had via 2d12, it can happen in two ways: a roll of 11 and 12, and a roll of 12 and 2. But both of those can happen in two different ways, since one possibility is rolling 2d12 after rolling a 9 on the first d10, and the other is when rolling 2d12 after rolling a 0 on the first d10. If you’re not counting the two possibilities like that, then the total number of possibilities is only 360, not 720. Either way, you get 360 as your maximum denominator after reduction and 1/360 as the smallest chance, or in the case of 122 specifically, 1/180 (4/720).

For illustration, here’s a list of every possible roll sorted to show how many ways each result can turn up: http://gaming.danieldockery.com/carcosaTabulation.txt If you rotate it 90° counterclockwise, you’ll see it matches the image plot shown earlier. In parentheses before each dice pair is the die type rolled for that result, so “(10) 02 09” means rolling a 2 and 9 with 2d10, while “(12) 02 09” is rolling 2 and 9 with 2d12. Each occurs twice, as above, to reflect getting it via either of the first d10 possibilities.

David Macauley

said:My brain hurts.

crusssdaddy

said:You don’t know the half of it. This ritual is actually the Contemplation of the Desiccated Variable – now all I need is the pale green incense that focuses the mind with singular clarity upon a sole chosen task and a Jale leper in order to bind the Lurker Amidst the Integers to my will… then I purchase lotto tickets.