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Something odd about the Bonesabres

5075 Views 76 Replies 19 Participants Last post by  ChaosRedCorsairLord
Thus not familiar, the Tyranid Special character named the Swarmlord has weapons that if you pass an invulnerable save made against inflicted wounds by the swarmlord, you must reroll them.

So I was looking at the bonesabre rule, I noticed that it makes no mention of you having to reroll only once. I just checked the official GW Tyranid FAQ and they make no mention of it either. Now I believe you'd only have to reroll a passed save once since I don't think I've ever come across an instance where you had to keep rerolling until you fail.

However, I'm wondering that since there is no official ruling that states that the bonesabre's wounds only force one reroll, could you argue that your opponent must keep rerolling his Inv until he receives a wound? Or is there a rule in the BGB that I'm overlooking that supersedes this?

Just a thought, what do you all think?
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Re-rolling all the fails and all the saves is exactly the same as not re-rolling anything.

You're confusing the probability of rolling a save on the same dice twice with the general probability of making your save. If you have a 2+ save to make and 6 of them to do, you will fail 1 and pass 5, if you reroll everything you will still fail 1 and pass 5.

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.
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?

Care to share whatever you found, cause I'm not seeing it. Only thing that even comes halfway close is the Runes of Witnessing/Shadow of the Warp answer. Course it deals with additional dice being used in a roll, not in a reroll............ And (IIRC) doesn't even apply anymore since SotW has changed?
Interestingly the Eldar FAQ still stands, it's still Shadow of the Warp and it says they just cancel out so they do. Cancelling is just the easiest way to play it as well.

But only against the Shadow from a Hive Tyrant. :toot:
Indeed against the other ones you roll 9 dice, discard the low third, divide the result by two pick two dice and the Eldar win because that's just how it works.

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?
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.
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.

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Math failure. :laugh:
No, Maths Success!.

I roll a dice and it's a 1, what's my chance of rolling a 1 the next time? 1 in 6 as the second roll is not dependant on the first.

I roll 2 dice and they're both 1, what's my chance or rolling 2 dice and them both being 1? The same, 1 in 36 as the second roll is not dependant on the first.

And so on.

What you're confusing is the probability of rolling 4 1's and then rolling 4 1's again so in effect rolling 8 1's. Rolling 8 1's is very unlikely, but that does not change the probability of rolling 4 dice and them all being 1.

I think it's time you admitted you were wrong.

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He is right; once you say probability of {event B} given {event A} you are in conditional probabilities.

The odds of rolling 4 ones on the re-roll are the same as the odds of rolling of 4 ones the first time; the odds of one following the other are the product, i.e. P1 * P2.

Applied to the original question P1 (the odds of rolling either a success or a failure) is 4/4, i.e. 1, so the product is 1 * P2, so there is no difference in rolling and re-rolling, or just rolling once.
The conditional probability would matter if we care about previous results, but we don't. If the event we were looking for was roll 4 1's and then roll another 4 1's then you can combine the probability of rolling 4 1's together to give you the probability of that event.

But in this case we don't care what the previous result was, only what this result is. Rolling 4 dice will always have the same probability of rolling 4 1's no matter how many times you've done it previously. So if I roll 4 1's and then have to re-roll them all I have not decreased my chances of rolling another 4 1's. The probability of the whole event however is different.

I have to assume Loki is just trolling the forum now, he is so spectacularly wrong that I have to assume he is being deliberatly obtuse to annoy people. Maybe if we ignore him he'll go away.

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