Something odd about the Bonesabres

DeathKlokk said:

The odds of passing/failing are exactly the same on the second roll because the first is totally discounted. Your logic is faulty.

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Not sure I agree, but will address soon.

SHarrington said:

I'm actually quite qualified to rebute your assumption.
I'm a math tutor at the college I attend, with a special interest in statistics.
(It's why my friends say I play mathammer, not warhammer!)
Since you said statistics, I will elaborate in said language.
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What your describing is true, but only if your accepting the first rolls have value.
But since the first roll has no elements in any set related to the outcome, and only the second roll has elements in either the PASS set, or the FAIL set, the first roll is discounted completely.

The probabilities related to what you roll in the first rolling, are not related to what you roll in the second set.

I see what your trying to say, that you should have a minimum number of rolls and that if you roll more dice, the mean should hit 3.5, but in fact, it allready does, its just that its not easily recognizable when your rolling less then 30 or 40 dice.

Your assumption that by rolling the first set it will improve the mean of the second set is flawed in this way. The mean already exists, rolling more dice will not cause it to change, only become more visable to us humans.

Fin

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I did quite well in statistics. Not tutor level and over six years ago, but quite well none the less. So, in a limited sample, say 4 of which over a large enough series of trials, only 66% or 2.5ish actually hit and about 2 wound. For each dice, if the invulnerable is a 4+, you have a 50-50 chance of passing. However, what are the odds that two consecutive rolls will both be higher than 4? I am not asking what the odds of a 4+ are because that is 50-50. I am asking what the odds of two consecutive 4+ are. So, the first roll does have a value.

http://arnoldkling.com/apstats/coins.html

What I am talking about is graphed as Pascal's Triangle in this link and fairly well explained.

In the above, I am assuming that the eldar get a 4+ invulnerable save. If they don't. Modify the above to whatever their actual save is.

forkbanger said:

You would reroll every dice, and the results of the second roll are completely independent of the first. There is no point at all in rerolling everything, it makes absolutely no difference.

I mean, you could-
Roll 10 4++ saves.
5/10 pass. 5/10 fail.
Swarmlord forces the 5 passes to reroll.
Fortune forces the 5 failures to reroll.
Reroll all 10 4++ saves.
5/10 pass. 5/10 fail.

Or simply-
Acknowledge everything must be rerolled. Don't bother, as it makes no difference at all.
Roll 10 4++ saves.

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A swarmlord only gets 4 attacks, of which only 2 will wound and those saves should be re-rolled. Unless, as someone mentioned, there is an FAQ stating otherwise. To do otherwise is a house rule.

This argument is as old as the dinosaurs, go google it:

Fortune vs Null Zone
Fateweaver vs Null Zone
Fortune vs Bonesabres
Fateweaver vs Bonesabres
Fortune vs Executioner's Axe
Fateweaver vs Executioner's Axe

In the end they all boil down to:

--->Basically If you have even the slightest ounce of common sense then the two just cancel each other out.

--->But if you're one of those people who likes to play the rules as written,... and I don't just mean a regular RAW player, I mean you're someone who takes the rules beyond common sense or reason,... like you do completely pointless actions just because the rules say you should, even when logic dictates that that action contributes absolutely nothing to the game... For the sake of the hobby I hope The Warhammer 40k 6th Edition Rulebook has a special sidenote at the end of the book:

Warhammer 40 said:

After completing a game of Warhammer 40k go climb the tallest building in view, and hurl yourself off of it.

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Wolf_Lord_Skoll said:

I clicked on your link. And this jumped out at me:

"Many people do not understand the concept of independence. Sportscasters are notorious for this.

For example, suppose that a sequence of seven coin flips came up with five heads and two tails. What is the probability of getting tails on the next coin flip? A sportscaster would say that "the law of averages" says that we are more likely to get tails.

But there is no law of averages! The chance of getting tails on the next coin flip is 1/2."

The first roll has no effect whatsoever on the second. They are completely independant of each other as the first roll is discounted entirely and the second roll is made using the exact same chance of success for each dice.

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Yeah. I read that too. You didn't understand what you were reading if you read it from the beginning by the looks of things. Take a closer read of the first part.

The part you were reading deals with consecutive flips. Every time you flip a coin, previous flips do not impact on the probability that flip. So, every time you flip a coin, you have a fifty-fifty chance of get heads or tails. Hence, there is no law of probabilities.

In regards to the discussion, the first part of the link applies. The one detailing the probability of consecutive tosses coming up the same. The first time you flip a coin, it is fifty-fifty. The second time you flip that coin, the odds of it coming up the same side are twenty-five percent. That is because the flips are linked, as it is in this case as well.

They are linked due to whichever power is invoked by the roll. You are always re-rolling and your chances of rolling 4+ is always 50%, but your chances of doing so twice in a row is 25%. Therefore, in a case where you are trying to repeat the first roll, one way or the other, the first roll is not insignificant.

Aramoro said:

I thought you were hilariously bad at maths but it seems I was wrong, you're hilariously bad at understanding.

Say Swarmlord hitting an Un-Fortuned Warlock. He scores 4 hits and 4 wounds. The Warlock needs to make 4 4++ saves, everyone he passes he needs to retake meaning he has a paltry 25% chance of passing each save, sad panda. This is because the probability of success is dependant on the results of previous rolls.

Let say not the Warlock is Fortuned so he rerolls all his fail saves. Now his chance of passing each save is back to 50% as his rerolll has essentially removed the dependence on the previous results. Irrelevant of what the previous result is you re-roll it, pass or fail.

Do you get it now?

Aramoro

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A swarmlord over time will never on average hit that warlord with every attack. You need to read the rule book again to learn about WS and redo you math. You obviously don't know what the odds of four attacks at WS 9 are. You also fail to grasp how wounds work as well. Laughable, isn't it?

The only way it would cancel out is if you could re-roll a re-roll. You can house rule it as you want, but those skills don't cancel out. If they FAQ to say that they do, fine. It is no skin off of my nose, but until they do, streamlining it is not supported by the rules and are mathematically wrong.

If he passes all, his chances are 25% to pass again. If he fails all, his chances to fail all again are 25%. Whichever rule is invoked, it will follow. When you are statistically rolling 2 dice, those count.

forkbanger said:

Really?[/url]

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Yup.

A swarmlord over time will never on average hit that warlord with every attack. You need to read the rule book again to learn about WS and redo you math. You obviously don't know what the odds of four attacks at WS 9 are. You also fail to grasp how wounds work as well. Laughable, isn't it?

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Wait what? Are you reading the words you are writing? I'm going to go ahead and assume you are drunk, in the same way I said 'Lets Assume he hits with all four attacks'

The only way it would cancel out is if you could re-roll a re-roll. You can house rule it as you want, but those skills don't cancel out. If they FAQ to say that they do, fine. It is no skin off of my nose, but until they do, streamlining it is not supported by the rules and are mathematically wrong.

If he passes all, his chances are 25% to pass again. If he fails all, his chances to fail all again are 25%. Whichever rule is invoked, it will follow. When you are statistically rolling 2 dice, those count.

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Again i'm going to assume you're drunk, lets do this in baby steps for you.

Assumptions, The Warlock is fortuned, the Swarmlord hits with all 4 attacks. These are the base assumptions which have no effect on the probability of making an individual save.

The Warlock rolls 4 saves, 4 individual 6 sided dice. Probability states each dice has a 50/50 chance of passing his 4+ save on any one dice.

Thus he should pass 2 saves and fail two saves. He picks up his two successes and the Swarmlord makes him reroll them, should get one pass and one fail.

Fortune makes him pick up his 2 fails and reroll them, should get 1 pass and one fail.

End result, the chance of each save is exactly the same had no re rolls been made.

Lets look at what happens if the Swarmlord gets 1 wound, The Warlock picks up a dice and rolls it, no matter what result is rolled he picks up the dice and rolls it again. The probability is identical.

Aramoro