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1 hour ago, jbrandmeyer said:

For more on this topic, search the archives for my posts on the statistics of Sloane Phantoms, and on Blount's Z-95's.

I bet you're a real hit at parties...

I kid. That was an awesome read. Thank you for sharing it with us.

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17 minutes ago, jbrandmeyer said:

Light a man a fire, he'll be warm for the night.  Set a man on fire, he'll be warm for a lifetime.

that one made my day🤣

the real question is, how could C3-PO have computed the odds of survival a close-up encounter with an ISD in a YT1300 to 1 in 3720?😋

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16 hours ago, Grumbleduke said:

Add them together and we get about 0.6%, or almost exactly 1 in 160 chance of getting 9 damage from 6 red dice.

 

2 hours ago, jbrandmeyer said:

Great shot kid, that was one in a million 136.32!

Now do it again with one time rerolls of all non hit and crit results 😛

Personally I prefer and recommend the Monte Carlo approach. Roll 6 red dice many Million times and count the results to get the ratios :-p.

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59 minutes ago, gounour said:

that one made my day🤣

the real question is, how could C3-PO have computed the odds of survival a close-up encounter with an ISD in a YT1300 to 1 in 3720?😋

Maybe he went by the official numbers on storm tropper shooting accuracy... The Kamino Cloners were probably about as truthful as our Earth companies.

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3 hours ago, jbrandmeyer said:

Great shot kid, that was one in a million 136.32!

Dice pools in Armada are pretty small.  My favorite tool for analyzing their statistics is to use polynomials to describe their probability.

For example, the distribution of damage results on one red die is (3/8)*d^0 + (4/8)*d^1 + (1/8)*d^2, where the power of d represents the amount of damage, and each coefficient is the probability of getting that facing.

To find out the damage distribution for a pool of dice, just multiply out that polynomial and read off the coefficients.  With GNU Octave (a mostly Matlab-compatible free software package), this is equivalent to performing convolution.  The `conv` function operates on vectors of coefficients, as it is commonly used for the analysis of signal processing filters.

So, lets do that.


octave:1> dmg = [3, 4, 1]/8
dmg =

   0.37500   0.50000   0.12500

octave:2> dmg2 = conv(dmg, dmg)
dmg2 =

   0.140625   0.375000   0.343750   0.125000   0.015625

octave:3> dmg4 = conv(dmg2, dmg2)
dmg4 =

 Columns 1 through 6:

   0.01977539   0.10546875   0.23730469   0.29296875   0.21630859   0.09765625

 Columns 7 through 9:

   0.02636719   0.00390625   0.00024414

octave:4> dmg6 = conv(dmg2, dmg4)
dmg6 =

 Columns 1 through 5:

   0.0027809143   0.0222473145   0.0797195435   0.1689147949   0.2353477478

 Columns 6 through 10:

   0.2268676758   0.1550140381   0.0756225586   0.0261497498   0.0062561035

 Columns 11 through 13:

   0.0009841919   0.0000915527   0.0000038147

So there you have it, the exact distribution of the initial pool of damage on the first throw of 6 red dice.

Note that Octave is counting columns on a one-index basis, where we are looking at damage on a zero-index basis.  ie, 6 dice give a maximum of 12 damage.

There are some sanity checks we can perform.  For example, the sum of all those probabilities must equal 1.  Also, the chance of getting 12 damage is 1/8^6 and the chance of a complete whiff ought to be (3/8)^6.


octave:5> sum(dmg6)
ans =  1
octave:6> (1/8)^6
ans =  0.0000038147
octave:7> (3/8)^6
ans =  0.0027809

That checks out.

So, what are the odds of getting exactly 9 damage?  About one in 160.  Better question: What are the odds of getting 9 or more damage?  Only marginally better at 1 in 136.  If you have a robust tournament of 14 players playing three rounds, every one of which is actually throwing pools that large, every one of which is putting dice on target 3-4 times per game, then you might see one throw that strong the entire day.


octave:8> 1/dmg6(10)
ans =  159.84

octave:12> 1/sum(dmg6(10:end))
ans =  136.32

This also agrees with Grumbleduke's method, but I find that computing the entire distribution in one shot is a little less error prone.

I'll also note that the odds of getting three or less damage from the same initial pool is dishearteningly high at one in 3.6 throws, or roughly once per game depending on engagement and other details.  This is why red dice controlling upgrades are so valuable.

For more on this topic, search the archives for my posts on the statistics of Sloane Phantoms, and on Blount's Z-95's.

Awesome! Now we have two threads written in German in the main page.

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8 hours ago, Muelmuel said:

@boow having tried lone CR90 Raddus before, I think your friend should have fully up-armoured the CR90 (ECM/Derlin/Lando/etc.) since it's the vehicle that must survive before delivery. Also equip ET and go slow (speed 1-2) and maintain a nav dial and token. What this does is give the CR90 alot of options without overly committing on a run. What I don't understand, is why the CR90 ran the rift rather than going around it? Speed control is important to a Corvette's survival.

Now why are we trying to figure out how to help your friend rather than helping you pwn him better? 😁

If your red and blue objectives also give you full placement of all obstacles, you could try Romodi to give u more red dice. Also recommend XI7 before XX9, since pushing that extra damage through to the hull is better against small ships than fishing for structural damage (if you have both that's even better).

Well my opponent didn't 'run the rift', but as I was second player and we both had three ships to deploy, i was able to wait for him to place the CR90, and then place my ISD directly opposite, and we were as far from the center-placed rift as possible. The objective lets the second player move one of the first player's ship a speed-1 maneuver right at the start. But yeah, there was a lot of things he could've done differently, i think it was more that no one expected the 136.32 odds to hit!

As for me, my fleet was mostly hodge podge stuff that i hadn't tried much before. I actually was flying Romodi for the extra dice on obstructed attacks, along with Relentless, XX-9 and Vader boarding party. I had an Insidious Gladiator-II going at speed-3 with engine techs that was coming from the far side of the rift to catapult around it and get to the rear arcs. (it would've taken me longer than i thought!) Also had a Raider-I with only Boarding troopers

For squads, I had 2 gauntlets and Gar Saxon to try out raids for a change, 2 phantoms and Whisper for squad movement shenanigans, and a Colonel Jendon for two sure shots or something better.

So it was mostly a friendly game to try out stuff, and it just ended abruptly, haha!

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10 hours ago, jbrandmeyer said:

Dice pools in Armada are pretty small.  My favorite tool for analyzing their statistics is to use polynomials to describe their probability.

I suspected there was an easier way of analysing this, but I slept through most of my probability lecture course (in my defence, by the end I was one of the few people who still turned up, and the lecturer was viewed as so bad they were replaced by the department's best, award-winning lecturer the year after).  But that talk of using polynomials is giving my half-remembered flashbacks...

Also, all these people saying we shouldn't know the odds? What are you - some kind of rule-breaking, reckless, rebel scum?

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17 minutes ago, Grumbleduke said:

I suspected there was an easier way of analyzing this,

Easier is in the eye of the beholder.  My mistakes with math and arithmetic are strongly biased towards bookkeeping errors, so for me any method that cuts down on the bookkeeping gets classified as "easier".

8 hours ago, RapidReload said:

Now do it again with one time rerolls of all non hit and crit results 😛

Personally I prefer and recommend the Monte Carlo approach. Roll 6 red dice many Million times and count the results to get the ratios :-p.

One-time re-roll of every die is easy to handle.  Since every die can be re-rolled, you can just manage that in the very beginning by assuming that you will re-roll them before rolling them in the first place.  For example, black dice go from a distribution of [1, 2, 1]/4 to [1, 10, 5]/16 with Ordnance Experts assuming conservative re-rolling only.

Yes, this also implies that OE black dice are mathematically 16-sided.  Red dice that you know a priori will be re-rolled are mathematically 64-sided.

Harder cases for my method are when you get to re-roll just one die.  Or you have two features that allow you to re-roll dice one at a time (say, LTT + an accuracy token).  Or you want to evaluate the impact of hard-rolling for a crit result.  Or you only needed one accuracy and re-roll the others hoping for more damage.  In these cases I switch out to multivariate polynomials and Maxima to manage them.  Sloane Phantoms needed `(1/4)*d + (1/8)*d^2 + (1/8)*a + (1/4)*c + 1/4` to describe All The Things (TM) that can happen.

I get the appeal of Monte Carlo.  Analyzing the new situations just becomes a simple matter of programming.  But I get enough of that at work, and wanted to try something new to get closed-form solutions.  Octave and Maxima are a little more accessible to non-programmers, so that's also part of why I advocate for them.

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1 hour ago, Grumbleduke said:

 (in my defence, by the end I was one of the few people who still turned up, and the lecturer was viewed as so bad they were replaced by the department's best, award-winning lecturer the year after). 

I'm not quite seeing how getting replaced by the best is a signifier that they were bad. Surely getting replaced by Michael Gove is a greater insult? 

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21 minutes ago, Ginkapo said:

I'm not quite seeing how getting replaced by the best is a signifier that they were bad. Surely getting replaced by Michael Gove is a greater insult? 

Yep, but generally you don't get your world-renowned, star professor to lecture a first-year-undergraduate course unless the previous results were so shockingly bad you need to take drastic action. They should be teaching something fancy and advanced that only they can teach, not wasting their time on something anyone should be able to do.

Although I don't think that was the worst course; there was another one where the lecturer giving the course was viewed as so bad at it (but was allowed to keep giving it due to some combination of tenure and big donations from his family) that another lecturer taught the same course, unofficially, outside the normal timetable. Aah, good times.

Anyway. Fortunately you can do most of this dice-probability calculations with school-level probability (and some brute force).

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