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Something odd about the Bonesabres

5077 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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I don't know. Statistically, whichever rule applies, you should re-roll. You pass, go swarmlord. You fail, go eldar whatchamafrigger. The odds of failing two 4+ saves is 25%. The odds of failing a 2+ roll twice are not the same as failing a 4+ twice. It is 10%. Or at least a big bottle of sake is making a strong argument in favor of this.

Think about it this way. Guy has a 6+ cover save and makes it. The odds of guy making it again are pretty slim. So, guy should re-roll. Since you ever only get one re-roll as of the Big Bloody Rule Book, the one that should apply is the one that's rules counts. Both rules should be applied depending on the outcome. That is what they are designed for. They are rules for different areas. If you pass, Swarmlord. If you fail, Eldar. They don't cancel. They can both apply, but only when their rule is invoked. Since you only ever get one re-roll, only one rule can be invoked per incident.

Now to be shouted down.... :p
The odds of passing/failing are exactly the same on the second roll because the first is totally discounted. Your logic is faulty.
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.
What you fail to grasp is that ALL cases of the first roll are re-rolled making it a 100% probability of being re-rolled. So you're multiplying by 1 essentially.
The argument Loki is making is that re-rolling the initial roll (regardless of it's result) somehow alters the probabilities of the second roll. Which is false.

Rolling a 6 on the first roll somehow prevents you from rolling another 6 even though neither roll is in any way dependent on the other. Statistics Fail.
It doesn't prevent a six from happening again. It changes the probability of the six occurring again. Honestly,
NO it does NOT. Your assumption is based upon the second roll being dependent on the value of the first, which it does not. If there were just one rule affecting the initial roll (Fortune, Swarmlord, Astorath, Fateweaver, etc...) then your assertion would be correct. Having two opposite rules working for the initial roll (canceling themselves out) renders the initial result moot. The initial probability is %100 for a re-roll so it has no bearing whatsoever on the probability of the second roll.

unless you understand the math, it probably won't make much sense. I provided a link that provided a pretty decent explanation of the math a bunch of pages ago.
Oh, I understand the math and what it actually is saying in your link. The whole idea of Pascal's Pyramid is based upon each coin flip being dependent on the others. It is showing the probability of getting a certain amount of heads in a row.

The odds of 4 Heads in a row?

1st flip 1/2 X 2nd Flip 1/2 X 3rd Flip 1/2 X 4th Flip 1/2= 1/16= 0.0625

Which is not what is happening with these rules interactions. What is happening with the dice is the probability of any value being rolled (100%) followed by the probability of a certain value being rolled (4+=50% or whatever the save may be).

Odds of getting a 4+ save after rolling any value on the inital roll?

Initial Roll X Re-roll

("1" 1/6+ "2" 1/6+ "3" 1/6+ "4" 1/6+ "5" 1/6+ "6" 1/6) X ("4" 1/6+ "5" 1/6+ "6" 1/6)

(1) X (3/6)= 1/2 = 50%

Exactly the same probability as just one roll.
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