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Subject: Killing the Broodlord rss

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soeren kepler
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Just to make sure i got this right.

The marines need to put two hits on the broodlord in one action to kill it. Ie - the flamer can´t kill it, but a 2d6 roll of 6,6 - 5,6 - 6,5 - 5,5 sustained firing stormbolter would kill it (equals 1/8 chance to kill the broodlord every sustained firing shooting action w. stormbolter).
First round of firing a stormbolter would need a 6,6 = 1/32 chance to rip the bug. (I´ll scream LUCKER when it happens....)

Or - can the broodlord be wounded? I argued no ingame but still found the question interesting.I play with friends, who play 40k, and they are accustommed to big bugs having wounds.

Example: marine with stormbolter shoots at broodlord.
First action point - 1,6 = 1 hit
Second action point (sustained fire) - 5,2 = 1 hit
= dead broodlord?!

Could someone do the math on the assault cannon? 3d6 confuses me heh
Cheers,
Soeren
 
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FEB
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We recently had the same question in a game and agreed, that the broodlord can take 2 hits in total, so wounding him in one round and finishing him in a subsequent round is fine.
 
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Calvin Wong
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That's not how I see it working in the spirit of the rules.

It says specifically that you require two hits on the same ranged weapon to kill the broodlord, which is why flamers are ineffective.

Disclaimer: I completely respect your right to house-rule.
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Paul
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Containerguy wrote:
We recently had the same question in a game and agreed, that the broodlord can take 2 hits in total, so wounding him in one round and finishing him in a subsequent round is fine.


While you are fully entitled to play the game anyway you want, I agree with the above poster: I dont think that that is the intention of the rule.

No figure in Space Hulk has ever had multiple "wounds", there is no bookeeping in that regard. Its a bit like the jam mechanism, if you roll a 6 this round and and a 6 in the next while shooting a Storm Bolter, it doesnt cause two "half jams". Not an ideal example, but whatever.

Similarly, the Broodlord is either 100% alive or 100% dead: no half measures. I dont have the rules handy but I am pretty sure that that is explicitly stated in the Broodlord rules.

The Broodlord is nasty.

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Michael H.
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I think you made the broodlord way too easy. It must be a hard kill and imho hitting with two separate shots isn't really a challenge, even if they had to be subsequent.
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Stewart Thain
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soerenkepler wrote:
Just to make sure i got this right.

The marines need to put two hits on the broodlord in one action to kill it. Ie - the flamer can´t kill it, but a 2d6 roll of 6,6 - 5,6 - 6,5 - 5,5 sustained firing stormbolter would kill it (equals 1/8 chance to kill the broodlord every sustained firing shooting action w. stormbolter).
First round of firing a stormbolter would need a 6,6 = 1/32 chance to rip the bug. (I´ll scream LUCKER when it happens....)

Or - can the broodlord be wounded? I argued no ingame but still found the question interesting.I play with friends, who play 40k, and they are accustommed to big bugs having wounds.

Example: marine with stormbolter shoots at broodlord.
First action point - 1,6 = 1 hit
Second action point (sustained fire) - 5,2 = 1 hit
= dead broodlord?!

Could someone do the math on the assault cannon? 3d6 confuses me heh
Cheers,
Soeren


I believe that your probability calculation is in error. Throwing a 6,6 is a 1/36 chance, not 1/32. Similarly, 5,5 5,6 6,5 and 6,6 is a 1/9 chance not 1/8.

It is difficult to kill the broodlord without the assault cannon.

I've done the assault cannon probability calculations very quickly, so they may well be in error, but I make the kill probability of a normal shot as 7/27 and for sustained fire as 1/2.
 
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Simon Lundström
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Edit: This chain of thought was wrong.
The probability that two of three dice are either 5 or 6 is 1/5.4 (5/27)
The probability that two of three dice are either 4, 5, or 6 is roughly 1/2.67 (3/8)

The probability that a broodlord will survive 2 rounds of subsequent fire from the assault cannon (counting the sustained fire bonus) is roughly 51%


I thought like this:
The number of outcomes are 6x6x6=216. If you write them all down (111, 112, 113 etc) in 6 columns depending on the first digit, you'll get 4 outcomes for the columns where the first digit is 1, 2, 3 or 4. For the 5 and 6 column, you get 12 20 outcomes, total 56.
Think the same way for the second case: It's 9 outcomes for the first 3 columns, and 27 outcomes for the last 3, total 108.

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Andreas Bøttger
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Odense SV
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Pure fluff, but the way I see it, the Broodlord has a much thicker carapace, and/or is much faster than a vanilla GS, thus requiring an extremely well-aimed shot to bring him down - which is why you either have to be really, really lucky with a storm bolter, or douse him in hot lead with the assault cannon.
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Thomas Koziatek
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Can also consider it an enhanced regeneration rate. 1 wound heals between rounds, thus you require 2 wounds in 1 round to kill it. Otherwise it just heals...
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Neil Christiansen
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Except that it is not just regenerating between rounds, but within. You can hit it four times from different shots and not kill it.

We used to include Tyrranid figures that took 4 total hit points in 1st edition. But to make BL have 2 HPs makes him WAY too weak.

We need good scenarios that make clever use of the BL.

Right now, the dominant GS strategy is to wait until a 3 blip is in position to attack the Assault Cannon to declare the BL, as if you can kill the AC you are in good shape.

I can see some great games where there is no AC, 1 BL that starts on the board (call him Grendel) and a 1 blip per turn reinforcement draw.

I have seen some argue that if BL is in play the SM player needs a Librarian. I have to disagree as the Librarians are now so tough as to be almost broken. The problem is being able to seal a square off 10 times. Can just keep doing that to BL and poor guy can't do anything.

 
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Stewart Thain
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Zimeon wrote:
The probability that two of three dice are either 5 or 6 is 1/5.4 (5/27)
The probability that two of three dice are either 4, 5, or 6 is roughly 1/2.67 (3/8)

The probability that a broodlord will survive 2 rounds of subsequent fire from the assault cannon (counting the sustained fire bonus) is roughly 51%

I thought like this:
The number of outcomes are 6x6x6=216. If you write them all down (111, 112, 113 etc) in 6 columns depending on the first digit, you'll get 4 outcomes for the columns where the first digit is 1, 2, 3 or 4. For the 5 and 6 column, you get 12 outcomes. 4x4+12x2=40.
Think the same way for the second case: It's 9 outcomes for the first 3 columns, and 18 outcomes for the last 3. 9x3+18x3=81.
Revert the numbers for survival: The Broodlord has 176/216 to survive the first and then 135/216 to survive the second. Combined it's 110/216, which is about 51%


Edited for clarity.


Simon,

You are calculating probabilites for the wrong question -
It should be at least two out of three dice are hits. Results when all three dice are 5 or 6 (or 4, 5 or 6 for sustained fire) also kill the Broodlord.

The probability of a 1st shot killing the Broodlord is about 26%
The probability of a 2nd (sustained) shot killing the Broodlord is 50%
The probability of a Broodlord surviving both shots is about 37%

Stewart
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Kris Vezner
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The rule is definitely two hits from the same fire action. No "wounds" like in 40K.
 
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Wes Clyburn
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Right. No such thing as "wounding", Your action either kills it, or it doesn't. And the rule for how a Broodlord is killed is pretty clear:
"Shooting attacks will only kill the Broodlord if two or more of the dice roll high enough to kill the target".

 
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soeren kepler
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You guys are the best! Thx! :-) I like this forum
sincerely,
Soeren
 
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Gordon Adams
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I tried for the fifth time to kill that ^****^* Broodlord last night and failed again He is lurking and laughing at my miserable attempts, but I will kill him, I WILL angrylaugh

 
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FEB
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Uglytruth wrote:
Right. No such thing as "wounding", Your action either kills it, or it doesn't. And the rule for how a Broodlord is killed is pretty clear:
"Shooting attacks will only kill the Broodlord if two or more of the dice roll high enough to kill the target".



Alright - then we house-ruled wrong. I´ll correct that the next time. whistle
 
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Simon Lundström
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Now who are these five?
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macbeth77 wrote:
Simon,

You are calculating probabilites for the wrong question -
It should be at least two out of three dice are hits. Results when all three dice are 5 or 6 (or 4, 5 or 6 for sustained fire) also kill the Broodlord.

I did count them like that but botched a place, you're right. I counted only a select few of the outcomes with more than 2 dice showing the right numbers. I recounted, and came to your conclusion. Thanks for the pointer.

Clarification: Out of 216 possible outcomes for 3d6, 56 outcomes (not 40 like I said before) result in two or more dice being 5 or 6, which is roughly 26%.

Out of 216 possible outcomes for 3d6, 108 outcomes (not 81 like I said before) result in two or more dice being 4, 5 or 6, which is exactly 50%.

Two-shot survival for the broodlord is thus 50% of 74%, which is 37%.
 
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Mike Miller
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The sure way to kill the Broodlord is to have the Librarian hit him with the Force Axe while spending 11 points of Psi energy.
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Neil Christiansen
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Not sure this would actually work. Does it say ranged weapon?
 
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Rami Finkelshtein
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In close combat you don't need to actually roll 2 hits you just need to beat the boordlords dice (like normal combat) the only problem is that the broodlord adds both max and min in close combat so very often his score is over 6 which means an average marine cannot beat him. Since the librarian can roll as high as he has psi points (using force axe) in the worst case scenario (librarian rolls a 1 and broodlord rolls 3 sixes) adding 12 psi points would give you a total of 13 beating the broodlords 12.
 
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