Dice problems [Archive] - Wizards Community

Dice problems [Archive] - Wizards Community

Post/Author/DateTime Post
newbie_DM
04-27-07, 07:26 AM Hi, I got into dnd a month ago and got my friends to join in, I will be the DM and I have already wrestled my way through the players guidebook aswell as the DM's but I have a really big problem. I live in Spain and maybe its because of the translation but I just can't make up what I have to do when it says: throw a 1d3 or : throw a 1d2
I understand the part of just one dice but as far as I know there are no dices of 3 nor 2 sides. The only solution I thought of was to use the d4 (for the d2) and the d6 (for the d3) and assining new values to the sides.
BUT later in the DM's guidebook I read something like "he has to throw a d3 but if he misses by 5 or more something bad will happen" which, of course, cancels out my solution.

If someone could be so helpfull to explain this I will finally be ready to start my campain.

Thank you very much

Stijn
VampireHunterD4
04-27-07, 02:03 PM The best way to do it is either to use a d4(Ignoring the highest roll) for a d3 and dividing what you roll by 2(round down) for a d2.

When it talks about missing something by X, it means your entire roll with modifiers. If it IS talking about it, it means a D20 roll, almost without fail.

For example, take Disable Device. The party rogue is level 2, and have 5 ranks in disable device, a Dex modifier of +3, and a masterwork trapkit for +2, for a total of 10 as his Disable Device Skill check. He finds a tripwire in the dungeon, and attempts to disable it. As a simple trap, you give it a DC of 20. He rolls a D20 and adds his Disable Device skill check to it, or 1D20+10. If the result is 20 or higher, then he disables it. However, as with all traps, if he rolls a 5 or less (getting a total of 15 or less), he has missed the DC by 5 or more. In such a case, he accidentially sets off the trap. If he had missed by only four, however, he would have only failed to disable it- he would not have set it off. Several skill checks work that way- it represents the possible 'bad events' for not being good enough.
Ranger REG
04-27-07, 08:37 PM Hi, I got into dnd a month ago and got my friends to join in, I will be the DM and I have already wrestled my way through the players guidebook aswell as the DM's but I have a really big problem. I live in Spain and maybe its because of the translation but I just can't make up what I have to do when it says: throw a 1d3 or : throw a 1d2
I understand the part of just one dice but as far as I know there are no dices of 3 nor 2 sides. The only solution I thought of was to use the d4 (for the d2) and the d6 (for the d3) and assining new values to the sides.
BUT later in the DM's guidebook I read something like "he has to throw a d3 but if he misses by 5 or more something bad will happen" which, of course, cancels out my solution.

If someone could be so helpfull to explain this I will finally be ready to start my campain.

Thank you very much

Stijn
Rare but you might find a d6 with only three numbers printed twice.

If you can't, just roll a d6 and divide the number by 2, round up to get a result of 1-3.

d2 is very simple: flip a coin. Head is "1", Tail is "2."
VrecknidjX
05-07-07, 03:02 PM There are many solutions, and some of them imply solutions to problems that have not yet come up for you.

The easiest d2 solution is to roll any die: you have a 1 if the result is odd, and 2 if the result is even.

My own dice collection includes several d3 dice that are simply a d6 with the numbers 1, 2, 3 repeated. Each number is opposite itself. If you do not have one of these, perhaps the most common solution is to roll a d6 and use this system: 1 or 2 = 1, 3 or 4 = 2, 5 or 6 = 3. But, you could roll a d12 and use 1 though 4 = 1, 5 through 8 = 2, and 9 through 12 = 3.

If you run into other situations, like needing a result from 1 to 7, the best thing to do is roll the next higher die and re-roll in case of an impossible result.

Sometimes you'll find this terribly inconvenient or impossible.

Suppose, for instance, you need to generate a random number between 1 and 16. Rolling d10+d6 or 2d8 simply won't do because it doesn't generate an equal chance at all results.

So, this is a nice alternative. Roll a single d8. Roll any other die. If the other die is odd, add zero to your first roll. If the other die is even, add 8 to yoru first roll. This gives you an equal chance at all 16 results.

Suppose you need a single result from 1 to 15.

1) Roll a d10, and divide the result by 2 (rounding up) (i.e. 1 or 2 = 1, 3 or 4 = 2, etc.).
2) Roll a d6. If the d6 is 1 or 2, add zero to the first result. If the d6 is 3 or 4, add 5 to the first result. If the d6 is 5 or 6, add 10 to the first result.

This gets you an equal chance at all numbers between 1 and 15.

You'll find your own creative solutions if you have other situations. Hopefully, you don't need, say, d37 anytime soon.

Dave

Post/Author/DateTime	Post
newbie_DM 04-27-07, 07:26 AM	Hi, I got into dnd a month ago and got my friends to join in, I will be the DM and I have already wrestled my way through the players guidebook aswell as the DM's but I have a really big problem. I live in Spain and maybe its because of the translation but I just can't make up what I have to do when it says: throw a 1d3 or : throw a 1d2 I understand the part of just one dice but as far as I know there are no dices of 3 nor 2 sides. The only solution I thought of was to use the d4 (for the d2) and the d6 (for the d3) and assining new values to the sides. BUT later in the DM's guidebook I read something like "he has to throw a d3 but if he misses by 5 or more something bad will happen" which, of course, cancels out my solution. If someone could be so helpfull to explain this I will finally be ready to start my campain. Thank you very much Stijn
VampireHunterD4 04-27-07, 02:03 PM	The best way to do it is either to use a d4(Ignoring the highest roll) for a d3 and dividing what you roll by 2(round down) for a d2. When it talks about missing something by X, it means your entire roll with modifiers. If it IS talking about it, it means a D20 roll, almost without fail. For example, take Disable Device. The party rogue is level 2, and have 5 ranks in disable device, a Dex modifier of +3, and a masterwork trapkit for +2, for a total of 10 as his Disable Device Skill check. He finds a tripwire in the dungeon, and attempts to disable it. As a simple trap, you give it a DC of 20. He rolls a D20 and adds his Disable Device skill check to it, or 1D20+10. If the result is 20 or higher, then he disables it. However, as with all traps, if he rolls a 5 or less (getting a total of 15 or less), he has missed the DC by 5 or more. In such a case, he accidentially sets off the trap. If he had missed by only four, however, he would have only failed to disable it- he would not have set it off. Several skill checks work that way- it represents the possible 'bad events' for not being good enough.
Ranger REG 04-27-07, 08:37 PM	Hi, I got into dnd a month ago and got my friends to join in, I will be the DM and I have already wrestled my way through the players guidebook aswell as the DM's but I have a really big problem. I live in Spain and maybe its because of the translation but I just can't make up what I have to do when it says: throw a 1d3 or : throw a 1d2 I understand the part of just one dice but as far as I know there are no dices of 3 nor 2 sides. The only solution I thought of was to use the d4 (for the d2) and the d6 (for the d3) and assining new values to the sides. BUT later in the DM's guidebook I read something like "he has to throw a d3 but if he misses by 5 or more something bad will happen" which, of course, cancels out my solution. If someone could be so helpfull to explain this I will finally be ready to start my campain. Thank you very much Stijn Rare but you might find a d6 with only three numbers printed twice. If you can't, just roll a d6 and divide the number by 2, round up to get a result of 1-3. d2 is very simple: flip a coin. Head is "1", Tail is "2."
VrecknidjX 05-07-07, 03:02 PM	There are many solutions, and some of them imply solutions to problems that have not yet come up for you. The easiest d2 solution is to roll any die: you have a 1 if the result is odd, and 2 if the result is even. My own dice collection includes several d3 dice that are simply a d6 with the numbers 1, 2, 3 repeated. Each number is opposite itself. If you do not have one of these, perhaps the most common solution is to roll a d6 and use this system: 1 or 2 = 1, 3 or 4 = 2, 5 or 6 = 3. But, you could roll a d12 and use 1 though 4 = 1, 5 through 8 = 2, and 9 through 12 = 3. If you run into other situations, like needing a result from 1 to 7, the best thing to do is roll the next higher die and re-roll in case of an impossible result. Sometimes you'll find this terribly inconvenient or impossible. Suppose, for instance, you need to generate a random number between 1 and 16. Rolling d10+d6 or 2d8 simply won't do because it doesn't generate an equal chance at all results. So, this is a nice alternative. Roll a single d8. Roll any other die. If the other die is odd, add zero to your first roll. If the other die is even, add 8 to yoru first roll. This gives you an equal chance at all 16 results. Suppose you need a single result from 1 to 15. 1) Roll a d10, and divide the result by 2 (rounding up) (i.e. 1 or 2 = 1, 3 or 4 = 2, etc.). 2) Roll a d6. If the d6 is 1 or 2, add zero to the first result. If the d6 is 3 or 4, add 5 to the first result. If the d6 is 5 or 6, add 10 to the first result. This gets you an equal chance at all numbers between 1 and 15. You'll find your own creative solutions if you have other situations. Hopefully, you don't need, say, d37 anytime soon. Dave