4d6, reroll 1's, keep highest 3

I was thinking about how to make this easier and faster (once I had to reroll an incredible amount of dice and of course some came up as 1 about 3 or 4 times), so here's the result:

If you roll 1d6 and reroll 1's until you get something else, you'll have a result of 2, 3, 4, 5, or 6, with equal chances - which is the same as rolling 1d5+1, without any rerolls.

If you roll this 4 times and keep the highest 3, then you have 1d5+1, 1d5+1, 1d5+1, 1d5+1, keep highest 3. The +1 is fixed, so this is the same as rolling 4d5, keep 3 highest, plus the three +1's with the 3 highest rolls - which is the same as

4d5+3, keep 3 highest d5's. Without any rerolls.

Any thoughts or feelings about this?

Honest question: what are you trying to make "easier and faster" with this method?

The maths checks out this grants the same distribution as the parameters you set. Though I still don't why people uses this scheme it's so OP compared to 'normal' stat generation...

If you're interested in how this changes the probability distribution, check out AnyDice

Some Example Input: I'm not sure what you are trying to fix here though, but play around with anydice, and you'll get a better feel for what it does for the rolls.

CancerMan:
I'm trying to avoid lots of rerolls, getting the same result in one roll, without any rerolls.

Azira:
I didn't find how to include rerolls in anydice, but generally I like the site.

Oh, I see what you are doing now. Yes, 1d6, reroll ones indefinetely would be the same as 1d5+1. I read the post a bit quickly earlier and got it in my head that you would be rerolling only once, which would skew the numbers ever so slightly.

If you want to model it in anydice you can define a function. Note that it will stop exploring it after 10 tries, so it basically ignores what happens after 10 ones:
(mean: 4)

Rerolling ones just once would be slightly different however:(mean: 3.92)

In conculsion:
If you want to model 1d6, reroll all ones indefinetely with a dice roller, 1d5+1 is a good alterative, yes. :)

One problem is the feasibility of this method without digital dice -- 1d5's aren't very common. If you're forced to use a digital die roller, then, well, you could use (or program) a roller that performs the basic operation you described without rerolls.

I suppose you could use d10s, and do: floor(1d10/2)+1

And I now have idea to improve the GP dice roller with :)

I spot a dangerful timesink in your immediate future :p

A more powerful die roller would be awesome, but things like that can get complex fast ;)

That's exactly what I was aiming for.
True, but I was looking for a PbP solution, since I only roleplay this way.
Yay.