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I'm not sure why you say that the probability of an "explodes" event begins to decrease with high volume of shots. Unless you are taking into account the conditional probability of the events

"Explodes and not wrecked"

and

"Wrecked but not explodes".

My model does not take that into account. I was only interested in the probability that the vehicle would be destroyed and did not care about the mode of destruction. So the probability that a vehicle would be destroyed is

P(Explode and not Wrecked) + P(Wrecked and not explode) + P(Explode and Wrecked)

=

1 - P(Not Wrecked and Not Explodes).

=

P(Wrecked or Explodes)

And this is what my model computes.

That graph is way off...

But since you've asked, let me be a little bit more explicit about how the calculation is made for the 6th edition sheet.

% chance to explode column (6th Ed worksheet)


% Chance to Wreck (6th Ed Worksheet)





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elrabin wrote:This is a great tool, but I do think it contains an error. The interaction between wrecked and explodes is a little more complicated on the 6th table than you have modeled. You count explode results in your wrecked probability. And the probability of an Explodes should begin to decrease with the number of shots, especially considering that a penetrating hit will cause a Wrecked result if the vehicle only has one hull point remaining. Overall, it does make a big difference.

The net result of 6th is that it's easier to wreck a vehicle with concentrated fire. Missile Launchers are much weaker overall. And of course, with glances no longer rolling on the damage table, higher AV vehicles will more often be able to move and shoot than they could in 5th.

I can say for sure that your graph is incorrect because the probability of a vehicle being destroyed in 6th Edition (blue line = either wrecked or explodes or both) should be at least the probability of a 5th Edition destruction (red line). Also, I think that maybe I didn't make it clear what the "explodes" line and the "wrecked" line really meant. They are just that, the probability that with N shots you get and Explodes! result or respectively a "wrecked" result. The are NOT, and should not be, the conditional probabilities you have described

"Explodes result but no wrecked result" and respectively " Wrecked result and no explodes result".

Because if you compute it that way, then the sum of these probabilities DOES NOT include the possibility that with N shots, you could get BOTH an explodes AND a wrecked result. The probability of BOTH events happening will grow rapidly with the number N of shots. And that is why your green line begins to decrease.

Think about it.

Your results mirror my experience of the game. Vehicles are more ressiliant early in the game when only a couple of shots are directed at them, but become much less survivable as the game goes on and the number of shots increases.

From a tactical standpoint that means that you should never expect a transport to live past turn 3, I just assume all mine have a self destruct button. And unless you are a fly or AV14 odds are that all vehicles will be dead by the end of the game. From a list building standpoint that means that you should avoid using AV11-12 vehicles as your primarly firepower unless they either make a big splash turn 1-2 that justifies their point, or some sort of increased survivability (fyers).

IMO it's going to take a while, but people should soon realize razor spam is dead.

Grugknuckle wrote:But since you've asked, let me be a little bit more explicit about how the calculation is made for the 6th edition sheet.

Thanks for the clarification.

Grugknuckle wrote:I'm not sure why you say that the probability of an "explodes" event begins to decrease with high volume of shots. Unless you are taking into account the conditional probability of the events ... My model does not take that into account.

But you do! When you combine P(Explode) and P(Wreck), you are looking at a conditional probability. If you look below, you will notice that it doesn't matter which term you apply the conditional. If you look at (1-P(wreck))*P(Explode), you will see that this decreases with the number of shots. (Likewise, the same is true if you look at the conditional wreck and not explode, but it cannot be true that more shots make a wreck less likely!).

P(Destroy) = P(Explode) + (1-P(Explode))*P(Wreck))
P(Destroy) = P(Explode) + P(Wreck) - P(Explode)*P(Wreck)
P(Destroy) = P(Explode) * (1 - P(Wreck)) + P(Wreck)


The real issue is that you are currently computing P(Explodes | not wrecked) and P(Wrecked or explodes). When you combine these into P(Destroy) (adding in the conditional), you arrive at:

P(Destroy) = P(Explodes | not wrecked) * P(not wrecked) + P(Wrecked or explodes)
P(Destroy) = P(Explodes) + P(Wrecked or explodes)

[Note: P(Not wrecked | explodes) = 1, so you can apply Bayes' theorem here]

In order to fix this, you need to calculate P(Wrecked) instead of P(Wrecked or explodes). This is a quick fix -- change the formula in your wrecked binomial distribution to:

A$22*(I$9-1)/6*(D$22 + D$23*SUM(G$21:G$24))
P(Hit) and (Fail cover save) and (Glance or Pen but not explode)

grugknuckle wrote:I can say for sure that your graph is incorrect because the probability of a vehicle being destroyed in 6th Edition (blue line = either wrecked or explodes or both) should be at least the probability of a 5th Edition destruction (red line).

Yep! I can say for sure it is wrong as well! That's what I get for hastily making changes. Adjusting the P(Wrecked) equation should result in a much more subtle change.

grugknuckle wrote:Because if you compute it that way, then the sum of these probabilities DOES NOT include the possibility that with N shots, you could get BOTH an explodes AND a wrecked result. The probability of BOTH events happening will grow rapidly with the number N of shots. And that is why your green line begins to decrease.

If you got explodes AND wrecked results, then the vehicle would be wrecked. Penetrating hits only roll on the damage table after removing hull points. Since you can only explode from a roll on the damage table, you cannot simultaneously wreck and explode.


Though, please feel free to check my math.

When it comes to computing P(explodes), consider this. I don't care if I get 1 explodes result or 10 explodes results. I want to know what the probability is of getting AT LEAST one explodes result. That is a cumulative distribution and therefore an increasing function. There is no way the probability of getting at least one explode result could be lower with more shots. You see? That's why I question your graph.

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elrabin wrote:
If you got explodes AND wrecked results, then the vehicle would be wrecked. Penetrating hits only roll on the damage table after removing hull points. Since you can only explode from a roll on the damage table, you cannot simultaneously wreck and explode.

Though, please feel free to check my math.

What you said here is not true. When we play the actual game, we evaluate the damage sequentially and we stop evaluating as soon as we get a destroyed result. For example, If I shoot 10 lascannons at your rhino and get, say 3 penetrating hits and 2 glances, then the vehicle is going to be destroyed because I've removed 5 hull points. The rhino is at least wrecked. But... you still have to roll on the damage table and see if you get an explodes result too! In game terms, it matters because a wreck is LOS blocking terrain and an explodes result can kill passengers and bystanders. If the event is explodes and wrecked, then the vehicle explodes.

But in any case, I don't think this affects the math and I think I've found the mistake. Or rather that there is no mistake.

Let's make some definitions so we can shorten the notation. (And now I really wish Dakka would support LaTeX - wouldn't that be awesome?)

Definition:
Fix the number of hull points, weapon strength, AP etc. And most importantly, fix N to be the number of shots.

E = { Events of N trials with at least one explodes result }
W= { Events of N trials with at least H pens or glances }
D = E union W = { Events of N trials with a destroyed result }

Then the probability of an event occurring in the subset A of all possible events is P( A ). Now, you have pointed out that (E intersect W) is a non-empty set. Therefore,

P( D ) = P( E union W ) = P( E ) + P(W) - P(E intersect W) = P( E ) + P (W) - P(E)*P(W) = P( E ) + (1 - P(E) )*P(W)

This is how the probability to destroy column is computed. For example, if you look at the row for 5 shots, under the column "% Chance to Destroy" you'll see the computation,

=B34 + (1-B34)*D34

Where the B column is P(E) and the D column is P(W). I'm pretty sure my model is correct.

Grugknuckle wrote:I'm going to take some time to look this over carefully, but here are my first impressions (which are sometimes wrong so take with salt).

Disregard what I was saying about your P(Wrecked) table It just dawned on me that you are assuming all of these shots are resolved at the same time. Your P(Destroy) and P(Wrecked) are correct (except for double immobilized results ), though I do still think that your P(Explode) table should be adjusted.

Grugknuckle wrote:My spreadsheet never computes

P( Explodes | wrecked)

I only compute P(Explodes) and P(wrecked). Then add them to get P(Destroy). That last step might be questionable.

Since you can't explode a vehicle if it wrecks first, and your P(Explodes) doesn't take into account wrecking, then you are calculating P(Explodes | not wrecked). When you calculate P(Destroy), you are implicitly changing this to P(Explodes).

Grugknuckle wrote:Now, when it comes to computing P(explodes), consider this. I don't care if I get 1 explodes result or 10 explodes results. I want to know what the probability is of getting AT LEAST one explodes result. That is a cumulative distribution and therefore an increasing function. There is no way the probability of getting at least one explode result could be lower with more shots. You see? That's why I question your graph.

That's true -- however you will only roll on the damage chart if you don't wreck the vehicle first! As more shots increase the likelihood of a wreck, the probability that you will even roll on the damage table will decrease. Rolling twenty shots at once, it's very unlikely to get at most two pens (or a glance and a pen), with at least one pen being an Explode result. Like I said earlier, you are already calculating this probability -- it's just hidden in your P(Destroy) equation.


Edit: I am very bad about reading! Missed the part in the rulebook that you can explode a vehicle even if you wreck it.

Sorry Grugknuckle -- You are right. Disregard everything I've said

For what it's worth, I'll give your spreadsheet the "Elrabin's 'Bad at reading comprehension and math' seal of approval"

Here I am checking your math now.

elrabin wrote:
But you do! When you combine P(Explode) and P(Wreck), you are looking at a conditional probability. If you look below, you will notice that it doesn't matter which term you apply the conditional. If you look at (1-P(wreck))*P(Explode), you will see that this decreases with the number of shots. (Likewise, the same is true if you look at the conditional wreck and not explode, but it cannot be true that more shots make a wreck less likely!).

P(Destroy) = P(Explode) + (1-P(Explode))*P(Wreck))
P(Destroy) = P(Explode) + P(Wreck) - P(Explode)*P(Wreck)
P(Destroy) = P(Explode) * (1 - P(Wreck)) + P(Wreck)

This computation is correct.

The real issue is that you are currently computing P(Explodes | not wrecked) and P(Wrecked or explodes). When you combine these into P(Destroy) (adding in the conditional), you arrive at:

P(Destroy) = P(Explodes | not wrecked) * P(not wrecked) + P(Wrecked or explodes)
P(Destroy) = P(Explodes) + P(Wrecked or explodes)

This is incorrect. The probabilities I am computing are the

P(E ) = P (explodes) = P( N trials produces at least one explodes result and it doesn't matter if you also get a wreck or not )
P(W) = P (wrecked) = P( N trials produces at least H penetrating hits or glancing hits and it doesn't matter what is rolled on the damage table )

Those ARE NOT conditional probabilities.

In order to fix this, you need to calculate P(Wrecked) instead of P(Wrecked or explodes). This is a quick fix -- change the formula in your wrecked binomial distribution to:

No. Because I have made the adjustment in the probability to destroy column.

P(D) = P(E union W) = P(E) + P(W) - P(E intersect W) = P(E) + P(W) - P(E)*P(W) = P(E) + ( 1 - P(E) )*P(W)


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Ah yes... All shots are resolved simultaneously. That's where we weren't communicating. I made that decision in order to simplify computation. Also I made it with the intent of answering the question,

"How many shots of weapon X do I need to have a given chance to destroy vehicle V?"

So I implicitly assumed that all of the shots would be resolved simultaneously. I probably should have told you that. Sorry.

Sorry Grugknuckle -- You are right. Disregard everything I've said.

No worries man! I am actually really happy that someone will actually talk about math with me! Just out of curiosity, are you a mathematician?

I can say for sure that your graph is incorrect because the probability of a vehicle being destroyed in 6th Edition (blue line = either wrecked or explodes or both) should be at least the probability of a 5th Edition destruction (red line).

You are talking about the graph above (ap3 vs rhino) yes?

Am I misremembering? but in 5th you had a 1/3rd chance of being wrecked/exploded on a penetrating hit. In 6th you only have a 1/6th chance. As the chance to penetrate hasn't changed, that I can think of, I do not beleive the above statement is correct, at least for lower number of shots.

5th
Chance to hit = 2/3
Chance to penetrate = 1/2
Chance to kill = 1/3

6th
Chance to hit = 2/3
Chance to penetrate = 1/2
Chance to kill = 1/6

I fire 2 shots then my chance to kill in 5th = 21%
I fire 2 shots then my chance to kill in 6th = 10.2%

clearly the line in 5th edition should be noticeably higher than in 6th edition at least for the lower number of shots, where getting 3 hull hits isn't possible, or very unlikely.

Ah yes... All shots are resolved simultaneously. That's where we weren't communicating.

aha, that explains some stuff that was confusing me as well.

Be careful! You have to remember that in 6th edition you only roll on the damage table for penetrating hits - not for glances. While in 5th edition you rolled for both glancing AND penetrating hits. You might say this doesn't matter because you get a -2 mod for a glance, but you could also get a +1 for AP1 weapons making it possible to get a destroyed result with a glancing hit in 5th. That can't happen in 6th.But in 6th Edition, you can eventually remove enough hull points and wreck the vehicle.

So your computation for a single shot should be

5th ed.
%Chance to destroy = (% chance to hit)*( % chance to Pen)* (% chance to destroy with pen) + (% chance to hit)*( % chance to glance)* (% chance to destroy with glance)

6th Ed.
%Chance to destroy = (% chance to hit)*( % chance to Pen)* (% chance to destroy with pen)

And for multiple shots you have to add into the 6ht ed calculation the possibility that you'll get enough glances and pens to remove all of the hull points.

What actually happens for Krak Missiles against a Rhino is that the 6th edition destroy curve is below the 5th edition destroy curve at first and the two curves cross somewhere around 4 shots and then the 6th edition destroy curve rapidly increased beyond 90% at 9 shots, while the 5ht edition destroy curves doesn't get to 90% until almost 20 shots.

So...Just like GW said, vehicles (at least Rhinos) are actually MORE resilient in the short run but they go down very quickly to massed firepower.

EDIT : I wanted to post a plot, but couldn't figure out how to make Excel spit it out as an image. Try it yourself with the file available for DL.

Grugknuckle wrote:
What actually happens for Krak Missiles against a Rhino is that the 6th edition destroy curve is below the 5th edition destroy curve at first and the two curves cross somewhere around 4 shots and then the 6th edition destroy curve rapidly increased beyond 90% at 9 shots, while the 5ht edition destroy curves doesn't get to 90% until almost 20 shots.

That's all I was saying, though I couldn't be bothered working out where they cross, I was aware of everything else you said.

I was merely responding to the previous point of yours that 6th is at least as high as 5th which is clearly not the case entirely.

Let me point out one more thing that is a weakness in my model.

My model doesn't take into account the possibility of wrecking a vehicle in 5th edition by immobilizing it and destroying all of it's weapons. So the true 5th edition vehicle destroyed curve should be just a little higher and would have that characteristic "S" shape that the 6th Edition curve has.

But now I know how to model that now. You could consider each weapon to be a 5th edition hull point. So for example, A land raider crusader with a multi-melta would have 5 hull points in 5th edition. One for each weapon (Left sponson , Right sponson, TL Assault cannon, Multimelta) and one more for the immobilize result.

Furthermore, in 5th Edition a vehicle would lose a "hull point" anytime it suffered an immobilized or weapon destroyed result. which only happens on 2 out of 6 events in the damage table. If you wanted to model this, you could add a second column to the 5th edition sheet for "hull points' and the computation would be

Chance to lose a 5th Ed HP = (1/3)*(chance to Pen + chance to glance)

then the true chance to destroy a 5th edition vehicle would be computed just like it was for the column on the 6th edition worksheet. Just remember that the number of 5th ed hull points is likely to be different from the 6ht ed version.





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I was merely responding to the previous point of yours that 6th is at least as high as 5th which is clearly not the case entirely.

Ah...I understand now. You're right. It's not the case entirely, but certainly true after some fixed lower bound on the number of shots.

GK, just to let you know, you can never get an exploded and wrecked result in the same shooting step.

If you roll 3 pens or glances in the same shooting step vs a rhino, the vehicle is immediately wrecked.

This is resolved ifferently if you shoot at a squadron.

imweasel wrote:GK, just to let you know, you can never get an exploded and wrecked result in the same shooting step.

If you roll 3 pens or glances in the same shooting step vs a rhino, the vehicle is immediately wrecked.

This is resolved ifferently if you shoot at a squadron.

I thought the same thing, which is why there is now half a page of incorrect math

Take a second look at the paragraph concerning penetrating hits. You roll to see if the vehicle explodes even if the vehicle is first wrecked.

I won't argue rules with you on this because it's actually irrelevant to how the math is done.

If you get 3 pens and 2 glances in one shooting step, the rules might say that the vehicle is immediately wrecked. So fine, you can resolve it in the game as wrecked. Mathematically, my model still has to take into account what happens to the 3 penetrating hit damage dice that won't get rolled. There is a non-zero probability that some of them will roll an explodes! result. Even if the players don't resolve it that way in the game, my mathematical model still has to take those events into account.


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EDIT : Just so we're clear. The only graph which is correct is the one for %chance to destroy (both 5th and 6th). The Explodes column and the Wrecked column do NOT represent the actually probability of getting one of those results or the other BECAUSE in the game, we roll our dice sequentially but in the model all of the damage results are simultaneous. Which result actually happens depends on which one you rolled FIRST. Since the model assumes that the rolls are simultaneous, it cannot distinguish between which event happened first.

However, as long as you don't care how the vehicle is destroyed, then the probability of getting a destroyed result (wrecked, explodes or both) is correctly computed.




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In game terms, a vehicle is either resolved as a wreck or as an explosion. It can never be both. But when doing the statistics, I have to take into account the possibility of both a wrecked result and an explodes result.

elrabin wrote:

imweasel wrote:GK, just to let you know, you can never get an exploded and wrecked result in the same shooting step.

If you roll 3 pens or glances in the same shooting step vs a rhino, the vehicle is immediately wrecked.

This is resolved ifferently if you shoot at a squadron.

I thought the same thing, which is why there is now half a page of incorrect math

Take a second look at the paragraph concerning penetrating hits. You roll to see if the vehicle explodes even if the vehicle is first wrecked.

You are correct!

*edit*
And this is why vehicles are sucking even more!

It just goes to show, even if you offer free ice cream, people will still complain.
.

Soooo damn true

Diezel wrote:

It just goes to show, even if you offer free ice cream, people will still complain.
.

Soooo damn true

I've complained once cause I'm lactose intolerant and the end result was let us say not pretty....