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

d3? [Archive] - Wizards Community

Post/Author/DateTime Post
Darkxarth
06-27-06, 01:30 PM Okay, I've got a d4, a whole lot of d6s, 3 d8s, 2 d10s, 1 d12, 1 d20, and a d%. But no d3? Now, my question is; Is the d3 an actual die, or just a hypothetical probability generator. Is my best bet the Automatic Dice Roller (from this website) or should I just roll a d4 and reroll on 4s?

One of my players asked what it was, and I just couldn't answer them...
cwslyclgh
06-27-06, 01:52 PM roll D6 and divide by 2 (rounding up)

rolling a d4 rerolling 4's gives you a different probablility curve.

thre are actualy dice alled D3's that you can buy, but they are realy just six-sided with the numbers 1,2, and 3 appearing twice each.
Ashlan
06-27-06, 01:53 PM Roll a D6. 1 and 2 = 1, 3 and 4 = 2, 5 and 6 = 3 (ie divide by 2 and round up).
Darkxarth
06-27-06, 02:00 PM Thanks, I knew that rolling the d4 wasn't the same, but I couldn't think of anything else. d6 / 2... gotcha!
Amaril
06-27-06, 02:07 PM Or just roll a d6 and subtract 3 from anything that's above a 3.
4=1
5=2
6=3
Solaris
06-27-06, 03:43 PM rolling a d4 rerolling 4's gives you a different probablility curve.No it doesn't.

I knew that rolling the d4 wasn't the sameYes it is.

How could that possibly be the case?

If I roll a d4, then all four outcomes are equally likely. In the event that I roll a 1, 2, or 3, then I'm done. Otherwise, I re-roll, and I'm back where I started: rolling a d4 with all four outcomes being equally likely.

The d4 doesn't suddenly become loaded on the second roll. It doesn't know whether or not I've rolled it before. If I roll a 4, then for all intents and purposes it's as if I haven't rolled it before; the next roll will have the same probabilities as the first.

...

here's a proof that the odds are identical.

We want to find the probability of obtaining any given result by rolling a d4 and re-rolling 4s.

Obviously we can never obtain a final roll of 4. For any other result, 1, 2, or 3, the probability of getting it on the first roll is 1/4. To this, we must add the probability of getting it in two rolls: 1/4 for rolling a 4, multiplied by 1/4 for then rolling the result, is 1/4^2. Likewise, we must add the probability of getting it 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 the particular result).

The total probability of rolling it 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 of the outcomes 1, 2, and 3 therefore occurs with probability 1/3, as expected.
Coridan
06-27-06, 04:08 PM Actually the D4's shape gives it a bit less "randomness" with each roll, although marginal ;)

Why does this same topic keep popping up once a month? Is it really that hard to figure out on one's own?
Darkxarth
06-27-06, 04:14 PM Why does this same topic keep popping up once a month? Is it really that hard to figure out on one's own?
I just wanted to know if I had missed something, one of my players asked and I wanted to make sure I knew the whole truth before telling her...
Solaris
06-27-06, 04:21 PM Actually the D4's shape gives it a bit less "randomness" with each roll, although marginal ;)

That depends entirely on how it's thrown, though. Bounce it around a little bit in a cup, and it'll be fair (neglecting manufacturing defects, etc).
xandat
06-27-06, 04:22 PM There are D3s, there simply a 3 sided cyl.
hmmmm
06-27-06, 04:25 PM There was a thread (Probally a couple pages back by now) with pictures of a d3 in it. If you can't buy one you can always use the ways described above.
sizbut
06-27-06, 04:29 PM pssst... D3 do exist, I can think of at least 4 designs, one of which is truely 3 sided (think American football ball stitched together from 3 curved panels). If this lovely forum had a search engine you could find some threads where people have included pictures.
hmmmm
06-27-06, 04:45 PM When you get some one with no time and internet you don't need a search engine. Here is the link. (http://boards1.wizards.com/showthread.php?t=648289)Of one that I know of.

http://shop.rpg.net/images/uploaded/d3.jpg
speeblefreak
06-27-06, 11:30 PM D3s exist. I own one. I has 3 ovaloid surfaces connected by curves.
GoldLeader13
06-28-06, 12:15 AM There are also d3s that are simply d6s with 1, 2, and 3 each printed twice on it, instead of going from 1-6. No different, really, than a d6 divided by 2, but more aesthetically pleasing.
pres_man
06-28-06, 12:32 AM Don't forget poor old mister d12.
1-4 = 1
5-8 = 2
9-12 = 3
ArwenTwoPointO
06-28-06, 12:35 AM when is it really necessary to roll a d3 anyway...hardly ever, which is why they aren't packed in with the set of 'common' dice
stealth_monkey
06-28-06, 02:13 AM when is it really necessary to roll a d3 anyway...hardly ever, which is why they aren't packed in with the set of 'common' dice

I'd say they aren't packed into the regular sets because they're one of those specialised dice that really serve no purpose. A d6 has the exact same probabilities if you half it

For those interested, this is a d3 http://d89156.i50.quadrahosting.com.au/images/chxxt0304.jpg

Post/Author/DateTime	Post
Darkxarth 06-27-06, 01:30 PM	Okay, I've got a d4, a whole lot of d6s, 3 d8s, 2 d10s, 1 d12, 1 d20, and a d%. But no d3? Now, my question is; Is the d3 an actual die, or just a hypothetical probability generator. Is my best bet the Automatic Dice Roller (from this website) or should I just roll a d4 and reroll on 4s? One of my players asked what it was, and I just couldn't answer them...
cwslyclgh 06-27-06, 01:52 PM	roll D6 and divide by 2 (rounding up) rolling a d4 rerolling 4's gives you a different probablility curve. thre are actualy dice alled D3's that you can buy, but they are realy just six-sided with the numbers 1,2, and 3 appearing twice each.
Ashlan 06-27-06, 01:53 PM	Roll a D6. 1 and 2 = 1, 3 and 4 = 2, 5 and 6 = 3 (ie divide by 2 and round up).
Darkxarth 06-27-06, 02:00 PM	Thanks, I knew that rolling the d4 wasn't the same, but I couldn't think of anything else. d6 / 2... gotcha!
Amaril 06-27-06, 02:07 PM	Or just roll a d6 and subtract 3 from anything that's above a 3. 4=1 5=2 6=3
Solaris 06-27-06, 03:43 PM	rolling a d4 rerolling 4's gives you a different probablility curve.No it doesn't. I knew that rolling the d4 wasn't the sameYes it is. How could that possibly be the case? If I roll a d4, then all four outcomes are equally likely. In the event that I roll a 1, 2, or 3, then I'm done. Otherwise, I re-roll, and I'm back where I started: rolling a d4 with all four outcomes being equally likely. The d4 doesn't suddenly become loaded on the second roll. It doesn't know whether or not I've rolled it before. If I roll a 4, then for all intents and purposes it's as if I haven't rolled it before; the next roll will have the same probabilities as the first. ... here's a proof that the odds are identical. We want to find the probability of obtaining any given result by rolling a d4 and re-rolling 4s. Obviously we can never obtain a final roll of 4. For any other result, 1, 2, or 3, the probability of getting it on the first roll is 1/4. To this, we must add the probability of getting it in two rolls: 1/4 for rolling a 4, multiplied by 1/4 for then rolling the result, is 1/4^2. Likewise, we must add the probability of getting it 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 the particular result). The total probability of rolling it 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 of the outcomes 1, 2, and 3 therefore occurs with probability 1/3, as expected.
Coridan 06-27-06, 04:08 PM	Actually the D4's shape gives it a bit less "randomness" with each roll, although marginal ;) Why does this same topic keep popping up once a month? Is it really that hard to figure out on one's own?
Darkxarth 06-27-06, 04:14 PM	Why does this same topic keep popping up once a month? Is it really that hard to figure out on one's own? I just wanted to know if I had missed something, one of my players asked and I wanted to make sure I knew the whole truth before telling her...
Solaris 06-27-06, 04:21 PM	Actually the D4's shape gives it a bit less "randomness" with each roll, although marginal ;) That depends entirely on how it's thrown, though. Bounce it around a little bit in a cup, and it'll be fair (neglecting manufacturing defects, etc).
xandat 06-27-06, 04:22 PM	There are D3s, there simply a 3 sided cyl.
hmmmm 06-27-06, 04:25 PM	There was a thread (Probally a couple pages back by now) with pictures of a d3 in it. If you can't buy one you can always use the ways described above.
sizbut 06-27-06, 04:29 PM	pssst... D3 do exist, I can think of at least 4 designs, one of which is truely 3 sided (think American football ball stitched together from 3 curved panels). If this lovely forum had a search engine you could find some threads where people have included pictures.
hmmmm 06-27-06, 04:45 PM	When you get some one with no time and internet you don't need a search engine. Here is the link. (http://boards1.wizards.com/showthread.php?t=648289)Of one that I know of. http://shop.rpg.net/images/uploaded/d3.jpg
speeblefreak 06-27-06, 11:30 PM	D3s exist. I own one. I has 3 ovaloid surfaces connected by curves.
GoldLeader13 06-28-06, 12:15 AM	There are also d3s that are simply d6s with 1, 2, and 3 each printed twice on it, instead of going from 1-6. No different, really, than a d6 divided by 2, but more aesthetically pleasing.
pres_man 06-28-06, 12:32 AM	Don't forget poor old mister d12. 1-4 = 1 5-8 = 2 9-12 = 3
ArwenTwoPointO 06-28-06, 12:35 AM	when is it really necessary to roll a d3 anyway...hardly ever, which is why they aren't packed in with the set of 'common' dice
stealth_monkey 06-28-06, 02:13 AM	when is it really necessary to roll a d3 anyway...hardly ever, which is why they aren't packed in with the set of 'common' dice I'd say they aren't packed into the regular sets because they're one of those specialised dice that really serve no purpose. A d6 has the exact same probabilities if you half it For those interested, this is a d3 http://d89156.i50.quadrahosting.com.au/images/chxxt0304.jpg