Simplified Rolls

[Lancelot]

If I am wrong just tell me I am wrong... and I wont bother with it any more

You're wrong.

If you want to simplify the roll, make it 1d10+2, as suggested by keterys. That has an identical distribution to 1d12 (reroll all 1's and 2's). You have an equal chance of getting any number between 3 and 12.

Rolling 3d4 produces a bell curve. This might be easier to see if you think about rolling 2d6...

• There is only one possible way of getting a 12. You have to roll a 6 on both dice.
• There are six possible ways of getting a 7. You could roll a 1 on the 1st die and a 6 on the 2nd die. You could roll a 2 on the 1st die and 5 on the 2nd die. And so on...
• Hence, you have a much higher chance of rolling a 7 when you're throwing two dice, as opposed to rolling a 2 or a 12.

This situation gets more exaggerated the more dice you throw. When you're throwing 3d4, you need all three dice to come up with a 4 on the same throw to get a 12. That's a 1-in-64 chance (1/4 x 1/4 x 1/4).

However, if you roll 1d12 (re-roll all 1's and 2's) or 1d10+2, you get an equal chance (1-in-10, or 10%) of getting any single number.

Note that there's nothing inherently wrong with rolling 3d4 to generate a number between 3 and 12 as long as you're okay with getting an "average" result more often than not, and almost never getting either a very high or a very low number. So, even though 1d10+2 is a more mathematically accurate roll to use, you should use whatever works best for your game and your preferences (as long as the DM is okay with it).

 

[Oldtimer]

The maths you are talking about is going over my head...

Think like this:
Imagine three different d4's - one blue, one green, one red. Each can show four different numbers, which means that the total number of unique results is 64 (4*4*4). How many of those 64 gives the sum of 3? Only one. How many gives the sum of 4? That's three (1 1 2, 1 2 1, and 2 1 1). How many gives a sum of 5?

If you write out all the results (hint: use Excel for this) and the sums they give, you'll find that the sums 7 and 8 have the highest frequency and are therefore more likely to occur.

 

[mudlock]

I love how rpgs, by way of dice, are such a wonderful gateway to the world of statistics. (Damn you Amber!)

To gain your next experience level, answer the following question:

What--precisely--is the average result of a "vorpal" (i.e., whenever you roll the highest number, re-roll and add) d10? Show your work.

 

[Locutus Zero]

Another example, "roll" a d2. Heads count as 1, tails as 0. Roll it 100 times. For you to get a 100, you'll have to get 100 heads. The odds of that are really small (1/2^100).

Roll a d100. The odds of getting 100 are 1/100.

 

[FireLance]

What--precisely--is the average result of a "vorpal" (i.e., whenever you roll the highest number, re-roll and add) d10? Show your work.

I'm a little rusty, but I think it should be:
[SBLOCK]Let the Average be A.

When you roll from 1-9 the result is as shown on the die, i.e. 1-9.

When you roll a 10, the result is 10 + another roll on the die, i.e. 10 + A.

Hence, A = Sum of (Probability of each result x the result) = 0.1 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + A)
A = 0.1 x (55 + A)
10A = 55 + A
9A = 55
A = 55/9 = 6.1[/SBLOCK]

 

[Nichwee]

Not quite right there, as you ignore the possiblity of cascading "max rolls".

Vorpal 1d10:
Average = 0.1*(1+2+3+...+9) + 0.01*(11+12+13+..+19) + 0.001*(21+22+..+29) + .....
Average = 0.1*45 + 0.01*(10*9 + 45) + 0.001*(2*10*9 + 45) + ....
This is the Sum[10^(-n)*(90*[n-1]+45)]
Where n=1 to infinity. As the initial multiplication is 10^(-n), which exponentially tends to zero as n tends to infinity (versus 90*[n-1] which only multiplicatively tends to infinity) this summation has a limit it tends towards that would be the average roll on a Vorpal 1d10.

This is very close to what FireLance got but is fractionally higher as s/he only included the 0.01 stage of the equation not all the following parts - but as they get reduced in significance by a factor of 10 each time they don't make much difference.
I think the average of Vorpal 1d10 = 6.111111 recurring judging by quick assessment of the maths above. But I admit I am not riguorously applying the summation so I may well have errors in that assessment.

 

[FireLance]

This is very close to what FireLance got but is fractionally higher as s/he only included the 0.01 stage of the equation not all the following parts - but as they get reduced in significance by a factor of 10 each time they don't make much difference.
I think the average of Vorpal 1d10 = 6.111111 recurring judging by quick assessment of the maths above. But I admit I am not riguorously applying the summation so I may well have errors in that assessment.

For the record, I'm a he.

And the algebraic method I used above does take into account subsequent iterations. The actual answer is 55/9 or 6.11 recurring - I just rounded it off to one decimal place in the last step (I guess I should have mentioned that).

 

[Oldtimer]

Not quite right there, as you ignore the possiblity of cascading "max rolls".

No, he has that covered.
He just preferred an algebraic method rather than an arithmetic one.

 

[erf_beto]

I'm a little rusty, but I think it should be:
[SBLOCK]Let the Average be A.

When you roll from 1-9 the result is as shown on the die, i.e. 1-9.

When you roll a 10, the result is 10 + another roll on the die, i.e. 10 + A.

Hence, A = Sum of (Probability of each result x the result) = 0.1 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + A)
A = 0.1 x (55 + A)
10A = 55 + A
9A = 55
A = 55/9 = 6.1[/SBLOCK]

Why are you doing math? Are you, like, old?

Remember how google (and the Internetz) is ruining everything? There's no avoiding that, embrace the dark side...

For all your math dice troubles, please refer to AnyDice*.
You're welcome.

*Programming knowledge not included. Sorry, but you gotta be a geek these days...

 

[Locutus Zero]

Actually, I don't think that website does what he was doing math for (add d10 for every roll of a 10 on a d10).

Explode almost does it, but only finitely and it says the average is 6.11, not 6.[1 repeating].