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dF Chips
Another wacky idea. This one stems from the realization that 4dF have a total of 24 faces. Dispersing the faces on 12 chips as shown below gives the same distribution. Grab them all, shake them, and toss them. These chips are the size of nickels, but any would do.

Counting the chips up takes quite a bit longer, so I'd only recommend this if you want to make the "roll" more focal, or it's this, or nothing else.

A variant I was considering is to put these in a hat, let people draw 4, then flip them. That would cut down on the tabulation time I think.

Also, tossing a live scorpion in the hat would add tension. Try a mouse trap instead if you want the lethality of the game to be lower : )
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Jay Cie
So you draw 4 and flip them?
But if you are going to have 12 chips for 4D3 couldn't you just use chips marked the same on both sides or 6 you flip?
That would be one way, yes. Just drawing them wouldn't be enough. Alternately one might make 24 of these, and each had the same symbol on both sides. Then you could simply draw 4.

In either case you'd want to put them back for the next draw.
I don't think the  spread is the same. I've not done the math, but from what I can see, it is possible to throw anything from "-8" to "+8", which should not be the case. I think 6 chips would give you the spread you're looking for, I've run this through excel, and this is what it gives me for 6 of your coins [2 ("+"/"0"), 2 ("+"/"-") and 2("-"/"0")].

-4: 1,56%
-3: 6,25%
-2: 12,5%
-1: 18,75%
0: 21,88%
1: 18,75%
2: 12,5%
3: 6,25%
4: 1,56%

That is, if you throw them all at once. Which is what I understood on the first read, but might not be what you intended.
It's not possible to get +8, or -8. Any time all 8 of anything are face up the other 4 counter half of them to make it either +4, -4, or 0. That's why I show both sides. If the distro is different than what is depicted there will be problems.

It does look like 6 would suffice though. Awesome!
You could get all "+", which leaves the ("-"/"0") chips. If they all land on "0", you have 8 "+" and 4 "0", resulting in +8.
+Paul Kießhauer, thanks for that. It makes this technique better, easier, and more accessible : )

Oh, and yes, I do mean that they should be thrown at once, unless the hat is used.
Dang, you're right! I thought I had that sorted. Well, I am glad I showed you. It has been much improved by your suggestion!
turns out that making the chips as above creates two sets of the "coins" needed. Not a complete waste : )

I will have to do a new pic though for people that don't want two sets.
I am less concerned about stats, and more concerned with methods that can be obtained at little to no cost. If the numbers line up well that's great though : )
Actually the numbers line up very well. Slightly larger deviation (less likely to have 0, more likely to have +/-4).

Edit: derp. Had it the wrong way around. Fixed.
Here's the 4dF vs 12chips vs 6chips in anydice:

6chips is close enough
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