Random Math Question

[Guest Agaib]

Ok, so here is the situation, once upon a time in a D&D game from long ago I played rouge. This rouge was really quite silly in battle, because for his damage, he often have to roll sixty six sided dice to calculate damage. Per round. Unfortunately both Me and the dungeon master decided that this was slowing the game down way too much (You try adding up sixty values within a short amount of time). Anyway what We decided is this.

We wanted to know what the expected value of the dice was. Fortunately when rolling one six sided dice, a little bit of mathematical knowledge will tell you that the expected value of a single die roll will be 3.5.

Of course you can’t roll three and a half on a six sided die, but this is an approximation anyway. Because I roll so many dice, the average value per die is going to be very close to three and a half.

HOWEVER...

There was a big catch. A horrible catch. One that made Me want to shove a hand up My metaphorical butt.

Not only did My rouge roll sixty six sided dice every turn. He also did something very obscene because of a special ability. Every time he rolled a one with one of the six sided he got to re-roll it, ONCE.

So to simplify it all my question is this.

What is the expected value if you roll a single six sided die, and then in the case it rolls a one you may re-roll that die a single time? Once I know the answer to that question I can multiply by sixy to find the expected value of the entire thing.

Think long and hard My friends.

On a side note I've actually already figured it out, I just want to test you guys.

EDIT: I would like to restate that it is not impossible to roll a one. What the thing did was it let me re-roll it once. Just once. If it rolled a one twice in a row then I kept the one.

[Guest Agaib]

Whaaat? No takers? I've only had one person even try.

Whimps

[StoryJunkie]

Do you mean the probability of getting a one is a one in six chance, so if done 60 times, the probablility of one coming up is 10 times, so you have 10 free rolls, of which 1.4 is another one, so you actually have 71.4 rolls, the average of which is 3.5, nes pas? This gives 249.9 damage points. Give or take 10, I would randomly surmise. The probability of you rolling 60 ones is very high.

FYI, I have no idea about D&D

[Guest Agaib]

I seem to have come up with a different answer. I suppose I'll explain it.

Problem

Expected Damage per roll.

Complications

Sixy Six sided dice are rolled each time

The ones that come from it are seperated and rerolled. Only once though.

Solution.

To start. Pretend I'm just figuring out the average damage per die just normally. Each six sided die will give me 3.5. Now I multiply this by sixy to find the average damage done if I have not re-rolled the ones yet. So before re-rolling the ones in the set I have a total of two-hundred and ten points of damage. Now I think about what exactly I'm doing when I re-roll ones. I'm taking the ones away and making it go again. I can justify on average that from sixty dice I will get about ten ones. So I subtract ten from my damage because each of those dice are contributing one point to it giving Me 200 points of damage. Then I re-roll the ones, each die giving Me yet again, an average of three point five. So I multiply three point five times ten (the amount of dice being rolled during this step) and add it to the two-hundred I have from the other fifty dice.

Answer: 235 points of damage.

[StoryJunkie]

Ah, so what you are saying, is that when you roll 1, that doesn't count, so you go again and if it rolls even another one, you take that one rather than roll again. Well, that skews things, because obviously, the chances of counting 1 in a roll are actually one in 71.4. Your average roll would not include 1. So your average roll, per se is not 3.5, but 3.3333333etc. (the sum of the sides minus 1 divide by 6)