PDA

View Full Version : 2d10 vs. d100 for %

Ent
2009-05-23, 10:58 AM
Where can I find discussions on the difference between rolling 2d10 and d100 for %?

I have searched the forums but think I'm just not using the right keywords.

Flickerdart
2009-05-23, 11:00 AM
Where can I find discussions on the difference between rolling 2d10 and d100 for %?

I have searched the forums but think I'm just not using the right keywords.
You mean, between rolling 1d10 for tens and 1d10 for ones, and rolling a d100 for the whole thing? No difference. In both cases, there is only one combination to produce each number.

Egiam
2009-05-23, 11:00 AM
Maybe because nobody cared to post of discussion like that.

D100 good. D% bad.

Any opposition?

Mando Knight
2009-05-23, 11:01 AM
...I don't think I've seen any. The zoccihedron is a rather finicky object, and doesn't roll quite as smoothly, and d10s have more uses than just rolling d%, so I believe rolling 2d10 for a d% is more common than rolling an actual d100.

Blackjackg
2009-05-23, 11:02 AM
I'm not sure I understand the question. Taking two D10s, designating which is the 10s column and which is the 1s, and synthesizing the result is the standard way of rolling d100. Some clever dice makers have even made d10s where every side is a multiple of ten, thus removing the whole "designating" step.

There are d100s out there, but I don't tend to use them because they're bulky and sometimes difficult to read.

kamikasei
2009-05-23, 11:04 AM
There should be no difference - it's not like 2d6 vs 1d12. That said, there will probably be a difference because actual d100s are finicky to construct.

Zeta Kai
2009-05-23, 11:05 AM
To my knowledge, the odds of hitting a particular number is effectively identical. There might be a very slight difference in the inherent randomness, but I'm sure no human would notice without some serious math under their belts. Meatdolls like us aren't very good at detecting true randomness, even though we ARE good at pattern detection.

As far as practicality is concerned, 2d10 is easier to get your hands on than the golfball of doom. A lot of shops don't carry many d% dice.

arguskos
2009-05-23, 11:12 AM
...I don't think I've seen any. The zoccihedron is a rather finicky object, and doesn't roll quite as smoothly, and d10s have more uses than just rolling d%, so I believe rolling 2d10 for a d% is more common than rolling an actual d100.

There should be no difference - it's not like 2d6 vs 1d12. That said, there will probably be a difference because actual d100s are finicky to construct.
I don't have anything real to add, just that I haven't seen finicky used in a long time, and all of a sudden, it shows up twice in 5 posts. Pattern detection indeed Zeta! :smallbiggrin:

Oh, and screw the zoccihedron. Hate rolling those things, since they take forever to stop and are soooo easy to nudge. :smallannoyed: 2d10 foreva'!!

2009-05-23, 11:23 AM
Well, assuming that you are discussing the difference between 1d10*10 + 1d10 and 1d100:

The d100 (aka Zocchihedron (tm)) is quite biased. The first version of the d100 had all the smallest and largest numbers clumped around the poles of the shape; these were less likely to come up than the numbers in the middle. More recent versions have the numbers scattered around so the bias isn't so noticeable, but it's still present.

The bias is small enough that you'd probably never notice it in normal play (especialy with how uncommon the d% roll is). But, if you use the random treasure tables in the DMG or MIC a lot, you might see some repetitiveness. I find it ironic that GameScience - which prides itself on the fairness and quality of its dice - can't get their flagship product to roll fairly.

On top of this, it's clunky, huge, and doesn't roll well. Just use the d10s.

JeminiZero
2009-05-23, 11:27 AM
Actually, it should be [(1d10-1)*10] + 1d10. That gives you the normal range from 1 to 100.

Zeta Kai
2009-05-23, 11:37 AM
Actually, it should be [(1d10-1)*10] + 1d10. That gives you the normal range from 1 to 100.

I always just say "this one is your tens value, & that one is your ones value". Now a program would probably require the formula you gave, but even the nerdiest human would find "my" way easier to work with.

Tsotha-lanti
2009-05-23, 11:38 AM
If you mean using two d10s to roll a number from 1-100 vs. using a d100 (I've only ever seen one, and it was huge), there's no difference mathematically.

If you mean the odds of results, the spread for d100 is 1% for each result, whereas for 2d10 it's...

2: 1%
3: 2%
4: 3%
5: 4%
6: 5%
7: 6%
8: 7%
9: 8%
10: 9%
11: 10%
12: 9%
13: 8%
14: 7%
15: 6%
16: 5%
17: 4%
18: 3%
19: 2%
20: 1%

So that's a 55% chance to get 11 or below, 55% to get 11 or above, 1% chance to get 20, 10% chance to get 17 or above... and so on.

As usual for multiple-die resolution systems, using 2d10 gives you a situation where higher skills are more resilient to negative modifiers and benefit less from positive modifiers (going from 18+ to succeed to 19+ to succeed is only a 3% change in the odds; going from 11+ to 12+ is a 10% difference; same thing but reversed if you're rolling under the number). This is found preferrable by some people - like me - who think it does a nice job of representing the fact that even regular joes should have an easy time at easy tasks (positive modifiers to attempts), but experts are more reliable when performing difficult tasks.

I always just say "this one is your tens value, & that one is your ones value". Now a program would probably require the formula you gave, but even the nerdiest human would find "my" way easier to work with.

Yeah, but that's not what adanadhel9 said; adanadhel9's formula (1d10*10 + 1d10) gives a range of 11 to 110. JeminiZero was correcting that.

Chronos
2009-05-23, 11:43 AM
Meatdolls like us aren't very good at detecting true randomness, even though we ARE good at pattern detection.In fact, it's precisely because we're so good at finding patterns that we're so bad at recognizing true randomness, since we'll even find patterns that aren't actually there.

I find it ironic that GameScience - which prides itself on the fairness and quality of its dice - can't get their flagship product to roll fairly.Eh, don't judge them too harshly. All of the standard dice are inherently fair, but there's really no way to make a d100 inherently fair: Even if you get it so it rolls fair in your test laboratory, it won't be fair rolled onto some other surface, or rolled from a different height, or at a different speed, or whatever.

Obviously, the proper solution is to make all those d100 tables into d120 (http://www.dicecollector.com/PAPER_D120_MY_DESIGN_DISDYAKIS_TRIACONTAHEDRON.jpg ) tables instead.

Siosilvar
2009-05-23, 11:48 AM
Actually, it should be [(1d10-1)*10] + 1d10. That gives you the normal range from 1 to 100.

I don't know what d10s you use, but mine are labeled 0 to 9. You just roll one for your ones column, and one for your tens column. 00 means 100.

If you want to be really... finicky, the formula could be 1d10*10 + 1d10 + 1. 00=1 and 99=100.

Artanis
2009-05-23, 11:52 AM
I don't know what d10s you use, but mine are labeled 0 to 9. You just roll one for your ones column, and one for your tens column. 00 means 100.

If you want to be really... finicky, the formula could be 1d10*10 + 1d10 + 1. 00=1 and 99=100.
Do you also use d6s that are labeled 0 to 5? Because 1 to 10 is a normal d10.

And your formula will give a range of 12 to 111 with the 99 cut out of it...with normal d10s, at any rate.

Oracle_Hunter
2009-05-23, 11:56 AM
Do you also use d6s that are labeled 0 to 5? Because 1 to 10 is a normal d10.

:confused:

The dice I got in the Basic D&D Box were labeled 0-9; you just read the "0" as a "10" if you were rolling 1d10.

Artanis
2009-05-23, 12:01 PM
But look at his formula: he's treating it as though it really is 0-9

FMArthur
2009-05-23, 12:03 PM
I haven't even been able to find a d10 with an actual 10 on it, and I've searched. :smallconfused:

Tsotha-lanti
2009-05-23, 12:05 PM
Do you also use d6s that are labeled 0 to 5? Because 1 to 10 is a normal d10.

And your formula will give a range of 12 to 111 with the 99 cut out of it...with normal d10s, at any rate.

Actually, yeah, I have a dozen d10s and all of them go 0 to 9. Always figured that was the usual way. 0 is read as 10 by default, which is why I figured d10*10 is 10-100 (rather than 0-90).

Incidentally, my old RuneQuest books tell you that d20 is rolled with one d10 and one anything that has an even number of results (like a coin flip); on an even, you add 10 to the d10 result. On an odd, you keep it as is.

Flickerdart
2009-05-23, 12:05 PM
"double zeroes are 100" wouldn't exist if the dice had tens on them instead of zeroes.

Artanis
2009-05-23, 12:07 PM
I guess I just haven't gotten out much, because all I've ever seen were 1-10, with a special rule used for d100 rolls because they're much more rare than d10.

I stand by my statement though, as explained in my reply to Oracle_Hunter.

Siosilvar
2009-05-23, 12:07 PM
Do you also use d6s that are labeled 0 to 5? Because 1 to 10 is a normal d10.

And your formula will give a range of 12 to 111 with the 99 cut out of it...with normal d10s, at any rate.

I've also got a d10 or two that's an icosahedron labeled 0-9 twice.

Mando Knight
2009-05-23, 12:07 PM
I'm not sure I understand the question. Taking two D10s, designating which is the 10s column and which is the 1s, and synthesizing the result is the standard way of rolling d100. Some clever dice makers have even made d10s where every side is a multiple of ten, thus removing the whole "designating" step.

I've even got a matched set of d10s, with one carrying the extra 0 on each spot. It's the only d10 in my collection marked with a "10" anywhere on the die.

charl
2009-05-23, 01:27 PM
I've also got a d10 or two that's an icosahedron labeled 0-9 twice.

I've got one of those. I got it from this very old board game box that was too old to be used anymore (all the paper stuff was falling apart).

I've never seen any d10s actually marked with 1-10. It's always 0-9 (and half of them come with values multiplied by 10, with the zero becoming 00. I thought that was the way d10s were sold: in pairs with one of them being x10). I do have a couple of old Vampire the Masquerade dice that have ankh symbols instead of 0/10 though, if that counts (and a friend of mine had similar dice for old Mage that had some kind of made-up magical glyph instead of the ankhs).

Yuki Akuma
2009-05-23, 01:33 PM
I have three d10s - one labelled 1-10, one labelled 0-9 and one labelled 00-90.

They all come from WotC D&D products. Not even WotC is consistent!

Mando Knight
2009-05-23, 01:42 PM
There's also Munchkin Dice, which have a Munchkin instead of a 0 or 10...

2009-05-23, 01:48 PM
You mean, between rolling 1d10 for tens and 1d10 for ones, and rolling a d100 for the whole thing? No difference. In both cases, there is only one combination to produce each number.

Except this is not in the least bit true.

1d100 offers equal chance to roll each number because of the linear probability.

2d10 for d% is about the same for 2d6 for craps with a bell-curve for probability.

Anytime you roll more than one die you will no have an equal chance to roll the numbers, even when each die represents part of the resulting numbers.

Zukhramm
2009-05-23, 02:05 PM
Except this is not in the least bit true.

1d100 offers equal chance to roll each number because of the linear probability.

2d10 for d% is about the same for 2d6 for craps with a bell-curve for probability.

Anytime you roll more than one die you will no have an equal chance to roll the numbers, even when each die represents part of the resulting numbers.

While 2d10 would indeed give the bell curve, rolling 1d10 for the tens and 1d10 for the ones is not the same thing as rolling 2d10 and will have the same probability for every number between 1 and 100.

Tsotha-lanti
2009-05-23, 02:08 PM
Except this is not in the least bit true.

1d100 offers equal chance to roll each number because of the linear probability.

2d10 for d% is about the same for 2d6 for craps with a bell-curve for probability.

Anytime you roll more than one die you will no have an equal chance to roll the numbers, even when each die represents part of the resulting numbers.

That's nonsense. 2d10 produces a bell curve of results if you add them up. If you read the first as 10s and the second as 1s (and 00 as 100), you have 100 different results (the numbers 1 through 100), all equally probable.

The 2d10 odds I listed above are for adding them up. Your statement makes no sense, especially seeing as how using two d10s to roll 1-100 is the standard approach. If it didn't produce 100 equally likely and different results, it wouldn't be the standard.

The different results possible, and their odds, when rolling d100/d% with two d10s are...

01: 1%
02: 1%
03: 1%
04: 1%
...
97: 1%
98: 1%
99: 1%
00: 1%

2009-05-23, 02:08 PM
shadzar, 2d10 gives a bell curve distribution from 2 to 20 in the same way that 2d6 gives a bell curve from 2 to 12. But each sequence of two numbers has the same probability as any other sequence. And when you roll a tens column and then a ones column with 2d10, the result is defined by the sequence. You aren't adding.

For example, the odds of getting a 79 are (the odds of rolling a 7 on the first die)*(the odds of rolling a 9 on the second die) = (1/10)*(1/10) = 1/100.

Same deal with any number from 00 to 99, assuming that 10s are counted as 0s.

So, it's a bell curve distribution of totals from 2 to 20, but a liner distribution of sequences from 00 to 99. The latter is what we're concerned with here. :smallsmile:

artaxerxes
2009-05-23, 02:15 PM
FINICKETY people.

Oh. Well it is here in Scotland.

Also Decade (00 10 20...) + 1d10 for d%

Tsotha-lanti
2009-05-23, 02:22 PM
FINICKETY people.

What dictionary is that word in? The actual word is indeed "finicky."

2009-05-23, 03:57 PM
While 2d10 would indeed give the bell curve, rolling 1d10 for the tens and 1d10 for the ones is not the same thing as rolling 2d10 and will have the same probability for every number between 1 and 100.

However with more than one die the linear progression is not there because of the dependence on each of the other die.

You don't have 1:100 chance of getting any number because when one die has stopped it already reduces your range of possibilities by 90%.

For example instead of rolling 2d10 you could roll 1d10 twice for the same effect and yield the same results, but the dependence each dice rolls chances the probability outcome each time a new dice ir rolled.

Say you roll a 2 on the tens place dice. You can only get a result on the second rolling of the d10 from 20-29.

While you still have a 1:10 chance on each dice of getting each number, the probability of rolling a specific number is not the same as a linear probability because of that second die.

Dogmantra
2009-05-23, 04:01 PM
While you still have a 1:10 chance on each dice of getting each number
Exactly. It's an AND probability (i.e. you need to get an X on the first die, AND a Y on the second), so you multiply the two. 0.1x0.1 = 0.01 = 1% chance

Tiki Snakes
2009-05-23, 04:07 PM
My usual dice sets go 1 to http://komiformguiden.se/wp-content/blogs.dir/1/files/avatars/3/elder_sign_small-avatar1.jpg, personally.

Apparently if I ever roll them all at once and they all come up tops, bad things happen. Good to know, really. :smalleek:

Talic
2009-05-23, 04:43 PM
However with more than one die the linear progression is not there because of the dependence on each of the other die.

You don't have 1:100 chance of getting any number because when one die has stopped it already reduces your range of possibilities by 90%.

For example instead of rolling 2d10 you could roll 1d10 twice for the same effect and yield the same results, but the dependence each dice rolls chances the probability outcome each time a new dice ir rolled.

Say you roll a 2 on the tens place dice. You can only get a result on the second rolling of the d10 from 20-29.

While you still have a 1:10 chance on each dice of getting each number, the probability of rolling a specific number is not the same as a linear probability because of that second die.
Shad... You're wrong. And right. All at once.

Yes, you use 2 dice. Yes, each die has an impact on the final number. If the "10's" place die rolls a 6, you can't get a 32. However, we are looking at the odds prior to either die being rolled.

Here's your challenge. When you use 2 seperate d10's, and designate 1 die as the 10's place, and 1 die as the 1's place...

I challenge you to find any number from the range of 1-100 that has any more, or less, than 1 combination.

Because we all know that the total number of chances is obtained by multiplying all the distinct variable ranges together. In this case, we have 2 variables, each with a range of 10. Thus, we have a total of 100 distinct combinations.

Now, the reason for the bellcurve lies with additive results. When you ADD 2d10 together, there is 1 result that yields a 2 (a roll of 1, and another roll of 1), for a 1% chance.

There are 10 chances of rolling an 11 (1 and 10, 2 and 9, 3 and 8, 4 and 7, 5 and 6, 6 and 5, 7 and 4, 8 and 3, 9 and 2, 10 and 1)

This is because while we have 100 distinct possibilities for the roll, we only have 19 total final options, and each gets a different number of combinations in the roll that yields that result, with numbers in the middle range being more likely, as there are more results that yield them.

However, when you are not additive, and when you use the "10's place, 1's place, method", you still have 100 distinct roll possibilities, but it's spread evenly over 100 possible results. A roll of 3 and 7 is different than a roll of 7 and 3, and yields a distinct result.

As always, the odds of any one roll landing are the number of roll results that yield the result, divided by the total number of possible roll results.

So, with 2d10 additive, the odds of rolling an 11 are 10% (10 combinations that yield 11, and 100 total combinations = 10% chance).

With the place system, the odds of rolling an 11 are 1% (1 combination that yields 11, and 100 total combinations = 1% chance).

Tsotha-lanti
2009-05-23, 06:30 PM
However with more than one die the linear progression is not there because of the dependence on each of the other die.

You don't have 1:100 chance of getting any number because when one die has stopped it already reduces your range of possibilities by 90%.

For example instead of rolling 2d10 you could roll 1d10 twice for the same effect and yield the same results, but the dependence each dice rolls chances the probability outcome each time a new dice ir rolled.

Say you roll a 2 on the tens place dice. You can only get a result on the second rolling of the d10 from 20-29.

While you still have a 1:10 chance on each dice of getting each number, the probability of rolling a specific number is not the same as a linear probability because of that second die.

You have to be kidding. d100 is really simple. There's a total of 100 different results, all equally likely when nothing is known. That's it.

You can even make a matrix out of it. It'll have 100 (10x10 fields), each with a different number, since you put them in sequence rather than add them up.

.. 0 1 2 3 4 5 6 7 8 9
0 00 01 02 03 04 05 06 07 08 09
1 10 11 12 13 14 15 16 17 18 19
2 20 21 22 23 24 25 26 27 28 29
...
9 90 91 92 93 94 95 96 97 98 99

That's 100 results, no repeats. That means each result is 1/100, or 1% likely.

Edit: How the heck does rolling d10 twice differ from rolling 2d10? It's the same thing.
If you rolled a 2 for tens (a 1/10 chance), and whatever for ones (a 1/10 chance for any specific number), you got a 1/100 result. That was a 1% chance, same as all the others. That's it.

2009-05-23, 06:30 PM
However with more than one die the linear progression is not there because of the dependence on each of the other die.
You're wrong. The prior probability for each 00 - 99 combination is 1%.

Obviously, after you make a roll and observe its results, the roll you made has 100% probability of being made and the other rolls have 0% probability.

when one die has stopped it already reduces your range of possibilities by 90%.
Yes. But each possible reduction in possibilities is equally likely, so the prior distribution is still linear.

Say you roll a 2 on the tens place dice. You can only get a result on the second rolling of the d10 from 20-29.
Well, yes. And after you roll a 3 for the ones place, you can only have gotten a result of 23. At that point, 23 has 100% probability and other numbers have 0% probability.

However, before you roll either die, each number has the same probability as each other number.

(If you don't believe that's true, what probability do you believe each number has prior to either die being rolled?)

Trodon
2009-05-23, 09:32 PM
http://i664.photobucket.com/albums/vv2/trodon13/100_0336.jpg

here is a D10 with a 10 instead of a 0 (spoilered because of the size)