 # 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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Good, then my confusion is cleared up. So whats being argued now anyway? Whether probability of re-rolls changes anything? Personally if you get a re-roll for failed results I'd think yes.
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.
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, 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.

The point is that people claim that you should dispense with the first roll and only roll one dice when whatever you would do would result in a re-roll. I am not a hundred percent positive on this, but I think that you should math wise take the roll anyway. I think the second roll might change the distributive curve because of the small sample size. On average, 2 or 3 re-rolls per combat round. The other point is that you can house rule out the extra roll, but that there is nothing in the rule book to support this action. That by the rule book, you should make the extra roll.
Good, then my confusion is cleared up. So whats being argued now anyway? Whether probability of re-rolls changes anything? Personally if you get a re-roll for failed results I'd think yes.
Basically the argument is to do with a fortune (reroll failed saves) and null zone (reroll passed saves) situation. So basically when you reroll all your passed Invulnerable saves, and all your failed invulnerable saves (and are forced to do so).

So you roll your Seer Counsel's Invulnerable saves, then reroll all the dice. It's pointless and common sense says we simply don't bother with the second roll.
Sorry for the double post, I'm annoyed.

It doesn't prevent a six from happening again. It changes the probability of the six occurring again. Honestly, 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 page or two ago.
Your not honestly telling us that if we roll 20 dice all together, 10 red and 10 black, that the red and black probability density functions will be different?

If you're not then your whole argument is wrong.

If you are then you're coo-coo-crazy.
Sorry for the double post, I'm annoyed.

Your not honestly telling us that if we roll 20 dice all together, 10 red and 10 black, that the red and black probability density functions will be different?

If you're not then your whole argument is wrong.

If you are then you're coo-coo-crazy.
:headbutt: No because what I am talking about is probability over multiple rolls of one dice. What you seem to say is one roll of many dice where the probability of each dice is independent. If so, we are talking about different things.
Yeah I'm sorry but if I roll sixty dice and only accept the 3's then through probability I should have ten of those. Re-rolling all sixty dice would not increase or decrease the number I get.

If all the rolls are forced to be re-rolled, then its like the first roll never happened (probability wise anyway.)

If I roll one die and need to get a six, my chances are one in six. Regardless of the result for that roll, if I roll again my chances are one in six for the re-roll. The probability does not change, my odds of getting that result does not change.

If I had six dice and wanted a six, got one six in the group, and re-rolled the other five, then the probability of getting a second six from those five increases but thats only because the group size has decreased.
Yeah I'm sorry but if I roll sixty dice and only accept the 3's then through probability I should have ten of those. Re-rolling all sixty dice would not increase or decrease the number I get.

If all the rolls are forced to be re-rolled, then its like the first roll never happened (probability wise anyway.)

If I roll one die and need to get a six, my chances are one in six. Regardless of the result for that roll, if I roll again my chances are one in six for the re-roll. The probability does not change, my odds of getting that result does not change.

If I had six dice and wanted a six, got one six in the group, and re-rolled the other five, then the probability of getting a second six from those five increases but thats only because the group size has decreased.
Yes, I agree, but that is a significant sample size. When you throw large numbers of dice, you will get a regular distribution. When you throw 2, you won't. So, I think it is worth having to do so.

Nothing in the rules support one roll. Given the numbers even houseruling it seems silly.
I think I may be getting what your going on about now lokis222. In the case of two dice, you roll them and then re-roll them, are you looking at the overall probability of four rolls as opposed to two?
No because what I am talking about is probability over multiple rolls of one dice. What you seem to say is one roll of many dice where the probability of each dice is independent. If so, we are talking about different things.
If you roll a die, pick it up (no matter what you rolled) and reroll it, the two results are mutually exclusive. No matter what you rolled on the first roll you still rerolled it. I don't think I can be any clearer than that. lets look at the maths:

Prob of rolling a 5+ on a d6 = 2/6
Prob of rolling a 5+ on a d6 with a reroll if you fail = 2/6 + 4/6*2/6
Prob of rolling a 5+ on a d6 with a reroll if you pass = 2/6 - 4/6*2/6
Prob of rolling a 5+ on a d6 with a reroll if you pass and fail =
2/6 + 4/6*2/6 - 4/6*2/6 = 2/6

ChaosRedCorsairLord's Maths said:
So now lets use the basic independence equation, anyone who's done even basic HS mathematics will have seen it.

Two events are independent if the occurrence of one of the events gives us no information about whether or not the other event will occur; that is, the events have no influence on each other.

If the following equation is 'true' then the two events are independent:

P(AnB) = P(A)*P(B), where:

Ok so lets use this equation on a single die situation. You need to roll a 5+, but because of fortune & null zone both passes & fails get rerolled, so:

-P(A) is the probability of the first roll being a 5+ = 2/6 (Pretty fucking obvious.)
-P(B) is the probability of the reroll being a 5+ = 1*2/6 (The '1' is because in order for the reroll to happen you must have passed or failed on the first roll and because a pass or fail are the only two possible outcomes the probability of either occurring is '1'.)
-P(AnB) is the probability of the first roll and the subsequent reroll being a 5+ = So how do we work the probability of this occuring?

Well it's easy. All we need to do is draw up the following table:

O 1 2 3 4 5 6
1 N N N N N N
2 N N N N N N
3 N N N N N N
4 N N N N N N
5 N N N N Y Y
6 N N N N Y Y

The Orange represents all the single 5+s rolled. But we want it when both rolls are a 5+, which is represented in the small green area. There are 4 green entries, out of a possible 36. So the probability of both occuring is 4/36 or 1/9.

So we now have:
P(AnB) = P(A)*P(B)
And:
P(A) = 2/6
P(B) = 1*2/6 = 2/6
P(AnB) = 1/9
So we substitute the values into the equation and get:
1/9 =?= 2/6*2/6
which is the same as:
1/9 =?= (2*2)/(6*6)
1/9 =?= 4/36
1/9 =?= 1/9
1/9 = 1/9
So because this equation holds true the two events are independent!

I think I may be getting what your going on about now lokis222. In the case of two dice, you roll them and then re-roll them, are you looking at the overall probability of four rolls as opposed to two?
The law of large numbers does not affect this situation.
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Loki, your assumptions are overlooking one thing. You can pass the Inv save even if the first roll is a fail.

Chance of two successes in a row is indeed 25%
Chances of two failures in a row is also 25%

You now have to take into account the fact that pass/fail is a fail and fail/pass is a pass

ie:
Pass, followed by a fail- 25%- Fail
Fail, followed by a Success- 25%- Success

So your total for failure and success are still 50/50.

Now let's look at a scenerio. You roll 4 dice. 2 fail, 2 pass.

Boneswords says you re-roll passes. So that makes the two passes a pass and a fail.

However, Fortune let's you re-roll those fails. SO they become one pass and one fail.

Your total number of passes is 2 and total number of failures is also 2. Which is the same as the first roll.
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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.

Aramoro
Some posters had been arguing that, because you are re-rolling based on two rules requiring dice to be re-rolled it is conditional probability. My point was that whilst it is a conditional probability, it is a unique case in which the conditional and discrete probabilities for the second roll are identical.

So (unlikely as it might seem :grin I was agreeing with you that you could ignore previous rolls.
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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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.

Aramoro
Wow, I didn't expect 8 pages of rule debate...kind of forgot I asked this question :biggrin:

As I said in the initial post, I'm certain that you'd only re-roll once, and being a fair player that's what I'd ask. Should you make your save twice in a row, then I'd just live with it. I've come across the scenario where the Swarmlord and Fateweaver fought each other in cc, and we ruled it that both re-roll rules are negated.

I do believe that there is no difference in probability when figuring out the likelihood of passing or failing a save when asked to do it multiple times. Since you are rolling each dice separately I don't see how the probability of getting a certain roll would be different than the 1st. I'm no expert on Stat, but it doesn't seem that complicated to me
I think I may be getting what your going on about now lokis222. In the case of two dice, you roll them and then re-roll them, are you looking at the overall probability of four rolls as opposed to two?
Yeah. In the case where you roll regardless, with a large enough sample, the first and second roll would cancel out. However, in small numbers, probability isn't going to actually work in such a visible fashion. There isn't a large enough sample size. So, the math won't be quite as tight. If you throw thirty dice, you will come up with a pretty even spread... law of averages in regards to the independence of each throw. When you have two dice, you probably won't.

Since you have only one re-roll and the rules are diametrically opposed, I figure just take them. It is by the rules and it is fluffy. Especially in the case where you roll two saves or fail two saves. And, like I said, if you want to, house rules it out.
Yeah. In the case where you roll regardless, with a large enough sample, the first and second roll would cancel out. However, in small numbers, probability isn't going to actually work in such a visible fashion. There isn't a large enough sample size. So, the math won't be quite as tight. If you throw thirty dice, you will come up with a pretty even spread... law of averages in regards to the independence of each throw. When you have two dice, you probably won't.

Since you have only one re-roll and the rules are diametrically opposed, I figure just take them. It is by the rules and it is fluffy. Especially in the case where you roll two saves or fail two saves. And, like I said, if you want to, house rules it out.
The sample sizes are exactly the same:

-In one situation you roll X amount of dice and take the results.

-In the other situation you also roll X amount of dice. You then ignore whatever is rolled on them pick all the dice up again and reroll. You then take the results of the reroll. The results from the first roll have been ignored.

The law of large numbers is not applicable because in the end you've sill only taken X amount of results from each situation.

The law of large numbers comes into mathhammer in situations like this:

-Roll 30 dice and take all the 6+ results = approximately 5
-Roll 6 dice and take all the 2+ results = approximately 5

So which is better? The first situation is more reliable, because you are rolling more dice the standard deviation of the results will much smaller. The second situation is more prone to lucky/unlucky results.
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