# jbrandmeyer

Members

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## Reputation Activity

1.
When I see this topic headline:

littering and....

2. jbrandmeyer got a reaction from RapidReload in dice odds
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".
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.
3. jbrandmeyer got a reaction from CaribbeanNinja in dice odds
Light a man a fire, he'll be warm for the night.  Set a man on fire, he'll be warm for a lifetime.
Wait, that's not how that goes...
4. jbrandmeyer got a reaction from Rimsen in dice odds
Light a man a fire, he'll be warm for the night.  Set a man on fire, he'll be warm for a lifetime.
Wait, that's not how that goes...
5. jbrandmeyer got a reaction from Rimsen in dice odds
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.
6. jbrandmeyer got a reaction from Gilarius in dice odds
Light a man a fire, he'll be warm for the night.  Set a man on fire, he'll be warm for a lifetime.
Wait, that's not how that goes...
7. jbrandmeyer got a reaction from Gilarius in dice odds
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.
8. jbrandmeyer got a reaction from boow in dice odds
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.
9. jbrandmeyer got a reaction from The Jabbawookie in dice odds
Light a man a fire, he'll be warm for the night.  Set a man on fire, he'll be warm for a lifetime.
Wait, that's not how that goes...
10. jbrandmeyer got a reaction from Muelmuel in dice odds
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.
11. jbrandmeyer got a reaction from thestag in dice odds
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.
12. jbrandmeyer got a reaction from The Jabbawookie in dice odds
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.
13. jbrandmeyer reacted to Grumbleduke in dice odds
Close, but I think you also need to account for the different ordering of dice.
If we are going for 9 damage we want 3 doubles and 3 hits/crits, or 4 doubles and only 1 hit/crit.
If we want at least 9 damage, we can have 3 doubles and 3 hits, 4 doubles and at least 1 hit, or 5 doubles, or 6 doubles. This is going to be a pain, so let's stick with just 9 damage (although at least 9 would be more informative).
4 doubles, 1 hit/crit, 1 blank: (1/8)^4 * 4/8 * 3/8 (taking an accuracy to be a blank). But there are 30 different ways we can get that combination in a single roll (6 choices for which dice is the hit/crit, then for each 5 choices for which is the blank/accuracy). Which gives us: 0.00137329101 3 doubles, 3 hits/crits: (1/8)^3 * (4/8)^3. The number of combinations would be a bit harder to work out by hand, but we have a handy Choose function; we have 6 things and we can choose 3 to be hits (or doubles), 6C3 gives us 20 combinations. For a total probability of: 0.0048828125 Add them together and we get about 0.6%, or almost exactly 1 in 160 chance of getting 9 damage from 6 red dice.
Because I'm curious and have been a little maths-deprived lately, let's do the "at least 9 damage calculation as well.
6 doubles: 1/8^6 = 0.00000381469 Only 5 doubles: 1/8^5 * 7/8 * 6 = 0.00016021728 4 doubles and 2 hits: 1/8^4 * 4/8^2 * 6C2 = 0.00091552734 Add that on to our answer to the first part and we get: 0.73% or about 1 in 136 chance of getting at least 9 damage from 6 red dice.

14. jbrandmeyer reacted to Ginkapo in Table too small
Its not well known to be fair. We all know asbestos fibres are dangerous, but less know that breathing in any fibres isnt great. Not usually a major problem, just not something to encourage. Insulation is meant for the walls and ceiling behind plaster, lets leave it to that use.
15. jbrandmeyer got a reaction from Joker Two in Nadiri Starhawk Article
You can see the designers are running into trouble with adding new game mechanics that aren't broken in the the presence of previously printed upgrade cards and special effects.  Salvo: No crit effects.   "friendly non-unique squadrons without strategic."  The specific list of upgrades associated with Shriv.  "without the printed heavy keyword", "using its printed rear battery armament", and so on.

16. jbrandmeyer got a reaction from Thraug in Nadiri Starhawk Article
You can see the designers are running into trouble with adding new game mechanics that aren't broken in the the presence of previously printed upgrade cards and special effects.  Salvo: No crit effects.   "friendly non-unique squadrons without strategic."  The specific list of upgrades associated with Shriv.  "without the printed heavy keyword", "using its printed rear battery armament", and so on.

17. jbrandmeyer reacted to Fraggle_Rock in RitR: More questions
As stated above, you are correct. For reference, you will only find the "tabling" rules regarding flotillas in the Tournament Regulations. Look under "End of Round" and "One Player Defeated."
This rule states that "At the end of a game round, all of one player's ships that are not flotillas are destroyed. The player with at least one ship remaining earns a win and the opposing player receives a loss."
Having said all of that, in one campaign I am playing in we decided to play with the "tabling rule" and are using the following rules:
- Player that has been tabled loses the battle.
- Ships still on the table (including squadrons and flotillas) are not considered destroyed and therefore are not scarred.
- Score points for the winner as equivalent to the full fleet cost. If allies are included in the fight, the full 45 points are added.
18.
Update to the schedule:
Sorry for the late notice, but the October 19th tournament has been moved to November 16th.  Link to Facebook Event We are cancelling the February 15th event due to it being only a week later than the Denver Prime tournament. The December 14th tournament will be a normal 400 point list.  A sector fleet tournament might be fun, but we're going to help people be ready for the upcoming Prime!  Link to Facebook Event
19. jbrandmeyer reacted to player3691565 in Is armada.....
Really hard
since I’ve just been handed my *** by my 9 year old son flying 2 ISDs and a medium fight ball my conclusion is it’s clearly child’s play and not the complex adult game I aways thought it was.......stamps feet and stomps off into the distant to calls of “daddy are you being salty”.

20. jbrandmeyer got a reaction from grunnax93 in Rebellion in the Rim Shenanigans
You start with 2 in your initial fleet.  But if you play the campaign objective then you can get a unique squadron as the location reward.
21.
@DblVsdGuy
Something like this?
Name: Untitled Fleet
Faction: Rebel
Assault: Suprise Attack
Defense: Asteroid Tactics
MC80 Star Cruiser (96)
• Weapons Battery Techs (5)
• Auxiliary Shields Team (3)
• Heavy Ion Emplacements (9)
• Spinal Armament (9)
• Mon Karren (8)
= 141 Points
MC75 Armored Cruiser (104)
• Bail Organa (7)
• Boarding Troopers (3)
• Electronic Countermeasures (7)
• External Racks (3)
• SW-7 Ion Batteries (5)
• XX-9 Turbolasers (5)
= 164 Points
GR-75 Medium Transports (18)
• Comms Net (2)
= 20 Points
GR-75 Medium Transports (18)
• Comms Net (2)
= 20 Points
• Ketsu Onyo (22)
• Shara Bey (17)
• Tycho Celchu (16)
= 55 Points
Total Points: 400

22. jbrandmeyer reacted to Mad Cat in Are Z-95s Any Good?
They are ok for the points but you soon come up against squadron control limits when you decide to put lots of them in a list. You can start to spend a lot of points on Expanded hangars or officers like Raymus to get them all doing what you want where as a list with about half that number of X-wings would not need these upgrades to operate well.
Red dice are fickle and can give extreme results (admittedly either way). I remember using 4 Z95s in a game, over 2 turns I threw 24 red dice with 8 swarm rerolls and inflicted 3 damage (2 of them were scattered). Red dice also have half the accuracies of blue dice so polishing off scatter aces can be frustrating.
Speed is a small problem compared to the humble TIE fighter but a lot of rebel fighters are speed 3 so it isn't like they are slowing their comrades down. AFFM can be nice and an FCT on a Pelta, MC80 or Nebulon could shuffle things along.
3 Hull is weak and you are vulnerable to Flak and Mauler Mithel's ramming attacks or Fel's auto damage. However Rebels have the best Escorts in the game so if your fighters include Jan, 2-4 X-Wings and maybe Bigggs then a pair of Z95s can hang around this fight - shooting away with swarm rerolls and only occasionally getting attacked by things that ignore the escort rule. Reserve hangar bays can also recycle a Z95 during the game. 3 points for a damaged 7 point fighter isn't as good a ratio as the Imperials can manage (3 for 11 or 12) but could be useful all the same.
Massed Z95 lists suffer from fragile hull, decent enemy flak, lots of counter(2) squadron opposition and it has squadron control issues. The double swarm reroll from Blout will help but I still recommend Jan + 2-3 X-Wings/YT1300s as a central core for protection.

I have included Z95s when I need one extra squadron to keep the numbers up. In Sato lists I like 1-2 of them as you just need to keep some squads alive to get the dice colour changes. A Z95 on an obstacle can come out later in the game and help.
I also include them as additions to a Jan/X-Wing force where they are less vulnerable due to the escort.
23. jbrandmeyer got a reaction from Muelmuel in Are Z-95s Any Good?
If you are running a fleet that has some extra squadron commanding ability (say, an Ackbar assault frigate list), then you can build an SFC out of them that works.  I've run 3x X-wings and 3X Z-95's pretty well as a punch-forward alternative to Shara, Tycho and 2x A-wings.  Its also more punchy than Biggs and 3x X-Wings.
For the low, low cost of only 5 more points, you get +1 deployment, +4 hull, and 9 more attack dice on your turn. Where the A-wing crew works around the periphery to slow the enemy down and gum up the works, the XZ crew try to hit and kill one or two linchpin enemies, while sticking around long enough to slow down a non-Intel blob.
I personally find that the next upgrade for this particular SFC should be to upgrade an X-Wing to Biggs, and not a Z-95 to Blount.  Blount's cost includes his defense tokens, which aren't going to help. By the time the escorts are dead, the wing had better have already done its job.  Also, Blount's second re-roll only applies to his neighbors, not to himself.  So while Biggs ends up aiding everyone by making the X-wings better escorts, Blount only really aids 2 of his companions.
24. jbrandmeyer got a reaction from OgRib in Are Z-95s Any Good?
If you are running a fleet that has some extra squadron commanding ability (say, an Ackbar assault frigate list), then you can build an SFC out of them that works.  I've run 3x X-wings and 3X Z-95's pretty well as a punch-forward alternative to Shara, Tycho and 2x A-wings.  Its also more punchy than Biggs and 3x X-Wings.
For the low, low cost of only 5 more points, you get +1 deployment, +4 hull, and 9 more attack dice on your turn. Where the A-wing crew works around the periphery to slow the enemy down and gum up the works, the XZ crew try to hit and kill one or two linchpin enemies, while sticking around long enough to slow down a non-Intel blob.
I personally find that the next upgrade for this particular SFC should be to upgrade an X-Wing to Biggs, and not a Z-95 to Blount.  Blount's cost includes his defense tokens, which aren't going to help. By the time the escorts are dead, the wing had better have already done its job.  Also, Blount's second re-roll only applies to his neighbors, not to himself.  So while Biggs ends up aiding everyone by making the X-wings better escorts, Blount only really aids 2 of his companions.
25. jbrandmeyer got a reaction from JolliGreenGiant in Are Z-95s Any Good?
If you are running a fleet that has some extra squadron commanding ability (say, an Ackbar assault frigate list), then you can build an SFC out of them that works.  I've run 3x X-wings and 3X Z-95's pretty well as a punch-forward alternative to Shara, Tycho and 2x A-wings.  Its also more punchy than Biggs and 3x X-Wings.
For the low, low cost of only 5 more points, you get +1 deployment, +4 hull, and 9 more attack dice on your turn. Where the A-wing crew works around the periphery to slow the enemy down and gum up the works, the XZ crew try to hit and kill one or two linchpin enemies, while sticking around long enough to slow down a non-Intel blob.
I personally find that the next upgrade for this particular SFC should be to upgrade an X-Wing to Biggs, and not a Z-95 to Blount.  Blount's cost includes his defense tokens, which aren't going to help. By the time the escorts are dead, the wing had better have already done its job.  Also, Blount's second re-roll only applies to his neighbors, not to himself.  So while Biggs ends up aiding everyone by making the X-wings better escorts, Blount only really aids 2 of his companions.
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