Possible simple way to roll for random players

[Shadow Monkey]

Not sure if anyone has thought of this before or not. I know there are d16's out there, but for those who don't want to bother with special ordering an obscure type of dice, I've thought of another way.

How about a d44? Pretty much the same as a d66 roll in Blood Bowl, but using d4's. There are 16 possible outcomes. Just roll 1d4 times 1d4 (roll the first d4, and mutiply it by a second d4 roll. If you roll a 3 on the first, and a 2 on the second, the outcome is 6 (3 x 2 = 6)). I know d4's are a lot easier to come by than a d16, it's very simple, and very random. It also beats trying to gather up those number counters to do a random draw all the time too...especially if you lose some of them.

Oh, and in case you need another type of odd roll like that, the same proccess can be used for just about any size of die. Simply multiply one die by another for the number of possible results. A d33 offers 9 possible results, d66 offers 36 possible results, d88 offers 64 possible results, and so on.

You could also use different sizes of dice in the same way. A d3 and a d10 could give you a d30 roll. A d6 and a d12 could give you a d72 roll. A d4 and a d6 could give you a d24 roll. You're only limited by the dice you have.

[sean newboy]

Isnt that a bit simple?

[Shadow Monkey]

You have a point there. Perhaps building some sort of robot with random number generators that follows you around would work better. You might want to give it a laser cannon to blast those who oppose your dreams of world domination as well.

[plasmoid]

Hi,
D16 - we do the classic hi/lo roll:

roll d6 + d8.
If d6 = 1-3, then use d8 roll.
If d6 = 4-6, add 8 to d8 roll.

Gives a number from 1-16 which isn't slanted.
Martin

[Old Man Draco]

I've got a d16 now, but before that i used a D20 (from AD&D games)

Your theory about using 1 d4 * 1 d4 has a flaw.

1*4 = 4
2*2 = 4

this can be thrown more often than

1*1 = 1

So I'm sorry to say it is not a good system.

[Duke Jan]

Just roll 1d4 times 1d4 (roll the first d4, and mutiply it by a second d4 roll. If you roll a 3 on the first, and a 2 on the second, the outcome is 6 (3 x 2 = 6)).

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So if the first d4 roll is 1
the outcome is 1 2 3 or 4
if the first die comes up 2
the outcome is 2 4 6 or 8
if you roll a 3 first
you get 3 6 9 or 12
if you roll a 4 on the first one
it's 4 8 12 or 16

So your chances of rolling are

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1 - 1/16   9  - 1/16
2 - 2/16   10 - 0
3 - 2/16   11 - 0
4 - 3/16   12 - 2/16
5 - 0      13 - 0
6 - 2/16   14 - 0
7 - 0      15 - 0
8 - 2/16   16 - 1/16

So it's not that easy you could add 4 times the result of the first to the second dice roll and then subtract 4.

[Gorbad]

I'd say, why bother bringing along extra d4's, when you already have a d8 and a d6?

Use the system as outlined by plasmoid, I use it all the time as well, and it's dead easy. 4+ on the d6 means d8+8, else it's d8. A perfect d16.

[sean newboy]

No the perfect d16 is a d16, which i have.

[Gorbad]

rrrrrrighttt....

A mathematically and statistically ideal d16 then (provided your d8 and d6 are mathematically and statistically ideal that is)

[Vero]

With (d4-1)*4+d4 it can be done.

But the more obvius problem is to use d6 to roll for anything between 1-16 because the number of players in a team varies.

[Shadow Monkey]

ahhh...yeah, but you could still use the dice types to have 16 variations the same as with a d66 offering 36 possibilities. I see how simple multiplication might screw some results. It's not as simple, but it could be done as...

11 = 1
12 = 2
13 = 3
14 = 4
21 = 5
22 = 6
23 = 7
24 = 8
31 = 9
32 = 10
33 = 11
34 = 12
41 = 13
42 = 14
43 = 15
44 = 16

So...roll the first d4 -1 x 4, and then add the second number to the total. So rolling a 3 on the first roll and a 2 on the second, you get 10 (3-1x4=8+2=10). Not as simple as standing systems, but as I said before, this type of system can be used to offer a lot more possible die rolls.

The universal formula is... die #1 - 1 x die size of second die + outcome of second die. So a d72 roll could be made as 1d12 - 1 x 6 + d6 = d72. Not too complex. I'm mainly looking for new ways to open up random number generating through dice to greater versatility. So I admit this wasn't really just an idea for Blood Bowl, but rather gaming in general. I just figured I'd try to find an existing game to work it out for actual use.

Thanks for spotting the probability flaws. I now have a more firm grasp on working out my probabilities.

P.S. I've editted this post, because I wrote the formula wrong. I was a little stoned at the time, and calculated it wrong. Anyway, I fixed it to work properly now. Remember kids, winners don't use drugs