Dice rolling

Are you asking the "median" of the dice being identical to the median of the previous roll?

>>8419515
different, removing the smallest number will create a higher average than re-rolling a die

>>8419519
Close enough to make a difference. Like of done enough times, will the 4 dice method give me better ( or worse, or the same) numbers than reroling the lowest

Its 50/50. Either the medium will be the same, or it won't.

>>8419521
Sorry I should specify, rerole and take the higher number

I might be a retard... I still don't understand the question

>>8419527
> Like of done enough times

Enough times would be on the order of thousands, and i doubt you have.

Statistically speaking removing the lowest value will lead to a higher overall average than re-rolling a dice. Think about it this way, by removing a dice you are necessarily getting rid of the lowest roll. If however you re-roll a 3 because it is the lowest, then you could end up with an even worse average than you started with. While, removing a piece guarantees a higher average.

I did a simplistic distribution and they were drastically different.

>>8419541
We want a set of three dice, and we are trying to find how to create the highest average within any given set of three dice. There are two ways OP discusses approaching it.

1) Roll four dice and remove the lowest dice, then calculate the average of the remaining three highest dice.

2) Roll three dice only, then re-roll the lowest die. Then you can take whatever number was highest between the first and second roll of that third dice. Then take the average.

Which method produces a higher average set of three numbers.

>>8419541
I have 4 dice, if I role them and drop the lowest, will the results be different ( better or worse) or roughly the same compared to rolling 3 and reroling the lowest and taking the lager of the 2. This came about for rolling stats for dnd, we are trying to figure out which one gives the higher numbers on average.

>>8419544
I messed up, you rerole but take the higher

>>8419515
Yes. You can see this even with two dice. With two dice you have

P(Max=1)=1/36
P(Max=2)=3/36
P(Max=3)=5/36
P(Max=4)=7/36
P(Max=5)=9/36
P(Max=6)=11/36

So if you take away the smaller roll, then on average you get about 4.5, compared to 3.5 with just one die.

>>8419549
The probabilities/averages are obviously identical, I don't think this requires further explanation

>>8419564
But how does this compare to re-rolling the smaller of two dice. That is the question. It isnt clear when using a single dice to relate.

I did simplified distributions again considering: >>8419562


I considered two dice that are each three sided.

In one case i rolled 3 dice and discarded the lowest value and got an average of 2.3

I followed the same procedure but re-rolled the lower of the two answers only considering the higher of the re-rolled dice. And the average was also 2.3.

Mind you both results were exactly 2.3 with no repeating decimals. SO, both these methods are better than simply rolling dice freely with no extra step. But they are exactly the same in terms of their relative advantage over the regular average. Im sure there's some stats theorem out there proving this more concretely.

>>8419549
Assume wlog that of the four dice, three are colored red and the last is blue

In 1) you simply roll all 4 and drop the lowest, color plays no part
In 2) you drop the blue die if it is the lowest among all 4 die, otherwise you drop the lowest red die

In both cases you end up dropping the lowest among all four dice, so both produce identical (longrun average) results.

>>8419574
Oh sorry I misread the question.

When you roll the 4 dice, you have

P(Min=6)=1/6^4
P(Min=5)=(2^4-1)/6^4
P(Min=4)=(3^4-2^4)/6^4
P(Min=3)=(4^4-3^4)/6^4
P(Min=2)=(5^4-4^4)/6^4
P(Min=1)=(6^4-5^4)/6^4

So (Average of 3 highest)=4*3.5-(Average of Minimum)=14-1.75=12.25

For 3 dice, we want to compute (Average of 2 highest)+3.5

In this case,

P(Min=6)=1/6^3
P(Min=5)=(2^3-1)/6^3
etc.

so like before (Average of 2 highest)=3*3.5-(Average of Minimum)=10.5-2.04=8.46
and (Average of 2 highest)+(Average of rerolled)=8.46+3.5=11.96

(I encourage you to check the computations of the averages, since I may well have made an arithmetic mistake!)

First off, op is looking at 2 common ways to roll for stats in a d20 based game. D&D being the most popular.

If you roll 3 die, on average, you get 3.5x3 = 10.5
Rerolling the lowest (3.5) would still result in a 3.5 on average.

If you roll 4 and drop the lowest, on average you expect 3.5x4 - 3.5 = 10.5

Which is the same.

I prefer doing 3 and re rolling but it doesn't matter.

Point buy is going to give higher stats on average, but then you can't live the 18 18 18 dream.

Better yet, our dm would have us roll 5 sets of stats using 3 die, no re rolling and using the best set.

Each set consists of 6 rolls of 3 die. Sometimes your class will want rounded stats, and sometimes you get something silly like 15,17,17,17,12,8.

If you aren't starting below level 6 or are playing 3.5 and older, it doesn't matter because you can just wish your stats up to 18s.

>>8419592
Boy did i over complicate that..

>>8419605
Something is wrong there.

This:>>8419592


Roll three dice, set aside the lowest, then roll a fourth separate die. You then get rid of the lowest of a total of four rolls, which is the same as rolling 4 initially and dropping the lowest.

>>8419617
Wrong. It is not the same as rolling 4 and dropping the the lowest, because if you include the 4th die, then the roll you set aside originally (that was the lowest of the 3) may not be the lowest of all 4 rolls. Consider what happens if the 4th roll is lower than than the lowest of the original 3 rolls.

>>8419643
Gotcha, which is why it was originally a more complicated problem but OP specifically states that the individual takes the larger of the first lowest roll and re-roll.

>>8419643
>Sorry I should specify, rerole and take the higher number
>>8419647

>>8419647
Oh I see now, he changed the question in a comment. Then what you said makes sense.

>tfw solved the original problem in >>8419604
but then OP changed it

>>8419564
>>8419604
Can you explain what you're doing here?
I'm not familiar with this notation.

>>8419679
Sure. Imagine rolling 2 dice and displaying all possible outcomes on a 6x6 grid. To get a minimum of 6, both have to be 6, and this corresponds to the square in the upper right corner.

The outcomes with a minimum of 5 form an "L" shape around this square.


...(6,5) (6,6)

...(5,5) (5,6)

So the number of outcomes with a minimum of 5 is 2^2-1. The others are computed in the same way

'Average of 3 highest' is a little confusing the way I wrote it, since I really mean the average of the sum of the 3 highest.

Like
>>8419564
said, the word "dice" is the plural form of die.