Here’s the 2nd place winner’s die.

~~There’s one problem. As submitted, the die shouldn’t have won. That’s my oversight. As proposed, the die has 21 sides which obviously doesn’t fit on a d20.~~

~~That’s embarrassing. My apologies to everyone, I should have checked that.~~

Edit: Nope, nope, I'm a dope. Uriah's idea works fine, there was just something wrong with my brain when I made this. I checked several times and I still didn't get it right. I remade the die as per his specs now and he should get it by Saturday.

But I was able to make the whole thing work regardless. When looking at the probabilities the 4’s and the 10’s don’t quite fill the probabilities of two whole sides. The 2 and the 12 (almost) fit with those probabilities to be a little over the 10% chance that the 4’s and a 2 would fill out and the 10’s and the 12 would fill out.

So this is what I did, I put a red pip next to one 4 and one 10 (the ones that aren’t doubles).

If a number with a red pip is rolled, re-roll and if the result is a 5 or under the roll is the 2 (if a 4 was originally rolled) or a 12 (if the 10 was originally rolled). If the second roll is higher than 5, it remains the original number (the 4 or the 10).

It’s functional, and usually not obtrusive.

One down, one to go. My hand is getting steadier but I need more practice, which is why I did the second place first.

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I went back to count and I think it did have only 20 sides. The 2/12 were supposed to be on the same side. If there was a slash between them, the number closest to you would have been the one you got. Sorry if it wasn’t explained correctly.

You’re right, I redact my error. Maybe my brain was having a bad day?