1d3???? [Archive] - Wizards Community - Pages Merged by Solauren

1d3???? [Archive] - Wizards Community

Post/Author/DateTime Post
Steel Cobra
06-10-05, 12:43 AM ok where can I find a 3 sided and a 5 sided dice?
Yules
06-10-05, 12:46 AM 1d3

throw a d6
1-2 = 1
3-4 = 2
5-6 = 3

1d5

throw a d10
1-2 = 1
3-4 = 2
5-6 = 3
7-8 = 4
9-0 = 5
Roreck
06-10-05, 12:48 AM what is d5 used for? I never use that. d3 yes d5 ?
Bunny, Conqueror Supreme
06-10-05, 01:03 AM d3 and d5 are larger dice (d6 and d10) with the results halved
Lord Gwydion
06-10-05, 02:48 AM what is d5 used for? I never use that. d3 yes d5 ?

We used to use them in Star Frontiers (a system that used percentages for most things, so only d10's were needed--knives and other weak weapons did d5 damage).

In D&D? I don't know if there's ever been a need for a d5, other than maybe rolling for something on a chart with only 5 choices.
Deflare
06-10-05, 03:42 AM d3: Roll a d4, reroll 4's.
d5: Roll a d6, reroll 6's.

3- and 5-sided dice aren't possible, for obvious geometric reasons.
The nameles one
06-10-05, 04:11 AM Linky for D5 (http://www.advancinghordes.com/product_info.php/products_id/390)

Linky for D3 (http://www.advancinghordes.com/index.php/cPath/2_61_82)
ThoughtfulOwl
06-10-05, 08:46 AM 3- and 5-sided dice aren't possible, for obvious geometric reasons.

3- and 5-sided platonic solids aren't possible; solids with curved surfaces, rounded borders, multiple shapes for sides and the like can yeld the probability distribution for a d3 or d5, plus the simple device of taking a d6 or d10 and numbering them 1,1,2,2,3,3,...
See the links above, for example.
Ramsey
06-10-05, 09:33 AM I actually HAVE a D3. :D

Well, it's actually a wooden 6 sided dice, but with no numbers above 3 on it. I have no idea where I got it, I think I found it with an old board game or something...
J Milburn
06-10-05, 02:00 PM A friend of mine never thought of half a d6, instead prefering to take a d6 as going through a d3 twice, counting 1 and 4 as 1s, 2 and 5 as 2s and 3 and 6 as 3.

That caused some arguments about what they rolled...
Sulaco
06-10-05, 03:25 PM ok where can I find a 3 sided and a 5 sided dice?

http://files.turbosquid.com/Preview/Content_on_1_29_2001_14_43_20/die_thumbnail.jpg2ED24C2A-EA1D-4DDB-ABD9B64A8CEE1B8F.jpgLarge.jpg / 2

http://www.answers.com/main/content/wp/en/thumb/9/9c/180px-10-sided_dice_250.jpg / 2
ThoughtfulOwl
06-11-05, 12:39 PM @ Sulaco: :ayyyy!:
UZI
06-11-05, 12:48 PM Just make sure when you divide by 2 you round up. Conventionally in D&D you round down. Alternatively you can roll (1d6+1)/2 or (1d10+1)/2 and continue to round down.
ILL34GL3
06-11-05, 01:02 PM A d3 is a d6 with the result divided by 2. Round up.

A d5 is a d10 with the result divided by 2. Round up.

Figured I'd go ahead and post that for you since it's only been posted 4 other times in this thread.
Fargren
06-11-05, 01:33 PM Just use a d20 and ignore any result higher than the one that you want. You can play anythign with just a d20 and a lot of pacience. Even a d100, but that is acting plain stupid.
Also any scientific calculator in fix mode can be used to get a random number betwen 1 and the max result you want, and can be fixed to any nmber of decimals(generally you'll want none).
DanjaNinja
06-11-05, 04:15 PM d3: Roll a d4, reroll 4's.
d5: Roll a d6, reroll 6's.
Huh... My innate mathematical sense is screaming at this. Does anyone know whether this works out nicely (i.e. all the numbers of the desired range come up with equal probability)?
ThoughtfulOwl
06-11-05, 04:20 PM Huh... My innate mathematical sense is screaming at this. Does anyone know whether this works out nicely (i.e. all the numbers of the desired range come up with equal probability)?

Yes, it works out; however, divide by half and round up is better, as you don't need to reroll.
braindead345
06-11-05, 04:32 PM Huh... My innate mathematical sense is screaming at this. Does anyone know whether this works out nicely (i.e. all the numbers of the desired range come up with equal probability)?

Statisticly speaking, no. It is my understanding that this method is bias. The divide by 2 method is much less so.
warpdragon
06-11-05, 05:23 PM The chances are equal.
25% 1
25% 2
25% 3
25% reroll
Raymond_Luxury_Yacht
06-11-05, 06:14 PM The chances are equal.
25% 1
25% 2
25% 3
25% reroll
It seems to me as if it would work out similar to the Monty Haul problem, that is, result in an uneven distribution. Of course, since I still can't figure out how the hell the Monty Haul problem works out the way it does, my opinion is more or less worthless.
Solaris
06-11-05, 07:18 PM DanjaNinja, braindead345, Raymond_Luxury_Yacht:

It works because successive rolls are independent. Rolling a 4 is equivalent to not having rolled at all.

If you flip a coin 99 times, and you get 99 heads in a row, what is the probability of getting heads on your next flip? It's still 1/2 - the coin doesn't care how many times you've flipped it already. It's the same with this. If you roll a 4 and then re-roll, all outcomes of that next roll are still equally likely; there's no bias. If it were any different, dice wouldn't be useful.

Here's a proof that it works.

Let x be 1, 2, or 3. The probability of getting x on the first roll is clearly 1/4. To this, we must add the probability of getting x in two rolls: 1/4 for rolling a 4, multiplied by 1/4 for then rolling x, is 1/4^2. Likewise, we must add the probability of getting x in three rolls, 1/4^3; four rolls, 1/4^4; five rolls, 1/4^5; and so on, forever (we could roll arbitrarily many 4s before finally rolling x).

The total probability of rolling x is therefore given by the infinite sum 1/4 + 1/4^2 + 1/4^3 + .... This is an instance of a "geometric series", which has the general form a + ar + ar^2 + .... In this case, a = 1/4 and r = 1/4. When the absolute value of r is less than 1, as it is here, a geometric series has a finite sum which is equal to a / (1 - r). Substituting 1/4 for a and r gives a / (1 - r) = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3.

Each outcome therefore occurs with probability 1/3, as expected.

Raymond_Luxury_Yacht:

Here's a really good explanation of the Monty Hall problem
http://www.comedia.com/hot/monty.html

(To hell with the wizards.com "you are leaving" nanny.)
Ehlijen Iswa Tschingjo
06-11-05, 07:31 PM We frequently use d5s for figuring out wether someone hit the monster with 4 mirror images up.
neilthrun
06-11-05, 07:55 PM Has anyone seen those crystal dice? They had pointy ends and as many sides as you wanted on them. I saw a d73, it was pretty wacky. And big.
Count Arioch the 28th
06-11-05, 08:49 PM Has anyone seen those crystal dice? They had pointy ends and as many sides as you wanted on them. I saw a d73, it was pretty wacky. And big.

I have a set. The D20 is obnoxious, it tends to keep rolling forever.

Post/Author/DateTime	Post
Steel Cobra 06-10-05, 12:43 AM	ok where can I find a 3 sided and a 5 sided dice?
Yules 06-10-05, 12:46 AM	1d3 throw a d6 1-2 = 1 3-4 = 2 5-6 = 3 1d5 throw a d10 1-2 = 1 3-4 = 2 5-6 = 3 7-8 = 4 9-0 = 5
Roreck 06-10-05, 12:48 AM	what is d5 used for? I never use that. d3 yes d5 ?
Bunny, Conqueror Supreme 06-10-05, 01:03 AM	d3 and d5 are larger dice (d6 and d10) with the results halved
Lord Gwydion 06-10-05, 02:48 AM	what is d5 used for? I never use that. d3 yes d5 ? We used to use them in Star Frontiers (a system that used percentages for most things, so only d10's were needed--knives and other weak weapons did d5 damage). In D&D? I don't know if there's ever been a need for a d5, other than maybe rolling for something on a chart with only 5 choices.
Deflare 06-10-05, 03:42 AM	d3: Roll a d4, reroll 4's. d5: Roll a d6, reroll 6's. 3- and 5-sided dice aren't possible, for obvious geometric reasons.
The nameles one 06-10-05, 04:11 AM	Linky for D5 (http://www.advancinghordes.com/product_info.php/products_id/390) Linky for D3 (http://www.advancinghordes.com/index.php/cPath/2_61_82)
ThoughtfulOwl 06-10-05, 08:46 AM	3- and 5-sided dice aren't possible, for obvious geometric reasons. 3- and 5-sided platonic solids aren't possible; solids with curved surfaces, rounded borders, multiple shapes for sides and the like can yeld the probability distribution for a d3 or d5, plus the simple device of taking a d6 or d10 and numbering them 1,1,2,2,3,3,... See the links above, for example.
Ramsey 06-10-05, 09:33 AM	I actually HAVE a D3. :D Well, it's actually a wooden 6 sided dice, but with no numbers above 3 on it. I have no idea where I got it, I think I found it with an old board game or something...
J Milburn 06-10-05, 02:00 PM	A friend of mine never thought of half a d6, instead prefering to take a d6 as going through a d3 twice, counting 1 and 4 as 1s, 2 and 5 as 2s and 3 and 6 as 3. That caused some arguments about what they rolled...
Sulaco 06-10-05, 03:25 PM	ok where can I find a 3 sided and a 5 sided dice? http://files.turbosquid.com/Preview/Content_on_1_29_2001_14_43_20/die_thumbnail.jpg2ED24C2A-EA1D-4DDB-ABD9B64A8CEE1B8F.jpgLarge.jpg / 2 http://www.answers.com/main/content/wp/en/thumb/9/9c/180px-10-sided_dice_250.jpg / 2
ThoughtfulOwl 06-11-05, 12:39 PM	@ Sulaco: :ayyyy!:
UZI 06-11-05, 12:48 PM	Just make sure when you divide by 2 you round up. Conventionally in D&D you round down. Alternatively you can roll (1d6+1)/2 or (1d10+1)/2 and continue to round down.
ILL34GL3 06-11-05, 01:02 PM	A d3 is a d6 with the result divided by 2. Round up. A d5 is a d10 with the result divided by 2. Round up. Figured I'd go ahead and post that for you since it's only been posted 4 other times in this thread.
Fargren 06-11-05, 01:33 PM	Just use a d20 and ignore any result higher than the one that you want. You can play anythign with just a d20 and a lot of pacience. Even a d100, but that is acting plain stupid. Also any scientific calculator in fix mode can be used to get a random number betwen 1 and the max result you want, and can be fixed to any nmber of decimals(generally you'll want none).
DanjaNinja 06-11-05, 04:15 PM	d3: Roll a d4, reroll 4's. d5: Roll a d6, reroll 6's. Huh... My innate mathematical sense is screaming at this. Does anyone know whether this works out nicely (i.e. all the numbers of the desired range come up with equal probability)?
ThoughtfulOwl 06-11-05, 04:20 PM	Huh... My innate mathematical sense is screaming at this. Does anyone know whether this works out nicely (i.e. all the numbers of the desired range come up with equal probability)? Yes, it works out; however, divide by half and round up is better, as you don't need to reroll.
braindead345 06-11-05, 04:32 PM	Huh... My innate mathematical sense is screaming at this. Does anyone know whether this works out nicely (i.e. all the numbers of the desired range come up with equal probability)? Statisticly speaking, no. It is my understanding that this method is bias. The divide by 2 method is much less so.
warpdragon 06-11-05, 05:23 PM	The chances are equal. 25% 1 25% 2 25% 3 25% reroll
Raymond_Luxury_Yacht 06-11-05, 06:14 PM	The chances are equal. 25% 1 25% 2 25% 3 25% reroll It seems to me as if it would work out similar to the Monty Haul problem, that is, result in an uneven distribution. Of course, since I still can't figure out how the hell the Monty Haul problem works out the way it does, my opinion is more or less worthless.
Solaris 06-11-05, 07:18 PM	DanjaNinja, braindead345, Raymond_Luxury_Yacht: It works because successive rolls are independent. Rolling a 4 is equivalent to not having rolled at all. If you flip a coin 99 times, and you get 99 heads in a row, what is the probability of getting heads on your next flip? It's still 1/2 - the coin doesn't care how many times you've flipped it already. It's the same with this. If you roll a 4 and then re-roll, all outcomes of that next roll are still equally likely; there's no bias. If it were any different, dice wouldn't be useful. Here's a proof that it works. Let x be 1, 2, or 3. The probability of getting x on the first roll is clearly 1/4. To this, we must add the probability of getting x in two rolls: 1/4 for rolling a 4, multiplied by 1/4 for then rolling x, is 1/4^2. Likewise, we must add the probability of getting x in three rolls, 1/4^3; four rolls, 1/4^4; five rolls, 1/4^5; and so on, forever (we could roll arbitrarily many 4s before finally rolling x). The total probability of rolling x is therefore given by the infinite sum 1/4 + 1/4^2 + 1/4^3 + .... This is an instance of a "geometric series", which has the general form a + ar + ar^2 + .... In this case, a = 1/4 and r = 1/4. When the absolute value of r is less than 1, as it is here, a geometric series has a finite sum which is equal to a / (1 - r). Substituting 1/4 for a and r gives a / (1 - r) = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3. Each outcome therefore occurs with probability 1/3, as expected. Raymond_Luxury_Yacht: Here's a really good explanation of the Monty Hall problem http://www.comedia.com/hot/monty.html (To hell with the wizards.com "you are leaving" nanny.)
Ehlijen Iswa Tschingjo 06-11-05, 07:31 PM	We frequently use d5s for figuring out wether someone hit the monster with 4 mirror images up.
neilthrun 06-11-05, 07:55 PM	Has anyone seen those crystal dice? They had pointy ends and as many sides as you wanted on them. I saw a d73, it was pretty wacky. And big.
Count Arioch the 28th 06-11-05, 08:49 PM	Has anyone seen those crystal dice? They had pointy ends and as many sides as you wanted on them. I saw a d73, it was pretty wacky. And big. I have a set. The D20 is obnoxious, it tends to keep rolling forever.