4 ms·

The 3-sided coin cannot work, the probability to fall over depends on the surface conditions. For 1d3 roll 1d6 and divide by 2.

jerf 4y ago
Given the nature of the question, an answer that includes parameterization on the relevant "surface conditions" would be perfectly acceptable and well within recreational mathematics bounds.
robocat 4y ago
Or buy 1d3 dice: https://duckduckgo.com/?q=1d3+dice https://duckduckgo.com/?q=1d3+dice Edit: apparently for some 1d3 designs it is difficult to roll them fairly: as per https://news.ycombinator.com/item?id=33778669 https://news.ycombinator.com/item?id=33778669
- hinkley 4y ago
  One of the hits for that search: https://boardgamegeek.com/thread/1673702/roll-1d3/page/2 https://boardgamegeek.com/thread/1673702/roll-1d3/page/2 Is just a six sided die with three edges rounded off.
mc32 4y ago
Also it depends more on the rotational axis; if instead of flipping it you basically spin the coin, the likelihood of landing on the edge side is enhanced --if it's landing on a flat surface.
TylerE 4y ago
You can use any fair dN as a fair dN-X. Just roll, and if you roll bigger than N-X, just discard and roll again (Don't do modulo division unless it's an exact multiple)
- nemo1618 4y ago
  You can also use any fair dN as a fair dN+X, by rolling ceil(log_N(N+X)) times and interpreting the result as a base-N number. (This is commonly done in D&D: instead of rolling a d100, you roll two d10s.) For example, say you want to roll a d20, but all you have is a coin. Flip the coin five times, producing a 5-bit integer (again discarding if the sum is not an exact multiple of N+X). A weirder example would be using a d6 to simulate a d100: ceil(log6(100)) = 3, so roll 3 times, then compute 36*r0 + 6*r1 + r2. Edit: Even more generally, you can use any set of fair dice. Imagine you have a d2, d4, and d6, and you want to simulate a d40. Roll the d6 and multiply by 4*2; roll the d4 and multiply by 2; then roll (flip) the d2. The resulting sum is a fair roll in the range (0, 48]. This is a significant improvement over rolling a d6 three times: you can roll all of the dice together, and you're much less likely to need a reroll.
  - TylerE 4y ago
    Interesting. I'm going to have to build a table to convince myself that the 2/4/6 example works. I'm not saying your wrong - just that combinatorial math (or really math beyond programmer stuff) isn't something I'm an expert in. The bit I'm having trouble with...aren't you missing the low end of the range? Like say you roll a 1 (the lowest possible roll), you'd get, if I understand right, you'd get 14 + 12 + 1*1 = 7. Maybe this works if you subtract one from each roll so they can roll 0?
    - aidenn0 4y ago
      yes, you have to subtract one from each roll
jbverschoor 4y ago
The best part about d&d, is that the dice really don’t matter too much if you have a fun dm

jerf 4y ago

Given the nature of the question, an answer that includes parameterization on the relevant "surface conditions" would be perfectly acceptable and well within recreational mathematics bounds.

robocat 4y ago

Or buy 1d3 dice: https://duckduckgo.com/?q=1d3+dice https://duckduckgo.com/?q=1d3+dice Edit: apparently for some 1d3 designs it is difficult to roll them fairly: as per https://news.ycombinator.com/item?id=33778669 https://news.ycombinator.com/item?id=33778669

hinkley 4y ago
One of the hits for that search: https://boardgamegeek.com/thread/1673702/roll-1d3/page/2 https://boardgamegeek.com/thread/1673702/roll-1d3/page/2 Is just a six sided die with three edges rounded off.

mc32 4y ago

Also it depends more on the rotational axis; if instead of flipping it you basically spin the coin, the likelihood of landing on the edge side is enhanced --if it's landing on a flat surface.

TylerE 4y ago

You can use any fair dN as a fair dN-X. Just roll, and if you roll bigger than N-X, just discard and roll again (Don't do modulo division unless it's an exact multiple)

nemo1618 4y ago
You can also use any fair dN as a fair dN+X, by rolling ceil(log_N(N+X)) times and interpreting the result as a base-N number. (This is commonly done in D&D: instead of rolling a d100, you roll two d10s.) For example, say you want to roll a d20, but all you have is a coin. Flip the coin five times, producing a 5-bit integer (again discarding if the sum is not an exact multiple of N+X). A weirder example would be using a d6 to simulate a d100: ceil(log6(100)) = 3, so roll 3 times, then compute 36*r0 + 6*r1 + r2. Edit: Even more generally, you can use any set of fair dice. Imagine you have a d2, d4, and d6, and you want to simulate a d40. Roll the d6 and multiply by 4*2; roll the d4 and multiply by 2; then roll (flip) the d2. The resulting sum is a fair roll in the range (0, 48]. This is a significant improvement over rolling a d6 three times: you can roll all of the dice together, and you're much less likely to need a reroll.
- TylerE 4y ago
  Interesting. I'm going to have to build a table to convince myself that the 2/4/6 example works. I'm not saying your wrong - just that combinatorial math (or really math beyond programmer stuff) isn't something I'm an expert in. The bit I'm having trouble with...aren't you missing the low end of the range? Like say you roll a 1 (the lowest possible roll), you'd get, if I understand right, you'd get 14 + 12 + 1*1 = 7. Maybe this works if you subtract one from each roll so they can roll 0?
  - aidenn0 4y ago
    yes, you have to subtract one from each roll

jbverschoor 4y ago

The best part about d&d, is that the dice really don’t matter too much if you have a fun dm