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jbrandmeyer

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  1. Like
    jbrandmeyer reacted to flatpackhamster in Rebuilding the Defiance   
    Are you tempted to swap a CR90 for a pair of GR75s?  It would allow you to do more commanding of your squads, maybe even toss some tokens around.  You could put Leia on one of them and keep it close to the MC80 to give you command flexibility.
  2. Like
    jbrandmeyer got a reaction from Snipafist in Cannot Get Your Gifts Out December 2020   
    Thank you so much for the updates!  I'm looking forward to playing in meatspace again in the coming year.
    > The Foresight has a great effect, allowing you to cancel 2 dice at long range or reroll 2 at Medium/Short.  Or with Mothma, cancel 2 at medium and reroll 2 at close. 
    I think you meant to say "reroll 3 at close."
    IMO, the value of dual defense tokens can be over-stated.  An ISD-II has 54% odds of generating two accuracies in the initial throw, before anything like H9 or leading shots have had a chance to make things worse for the defender.  The Liberty suffers from a similar deficit with its brace tokens.  That said, with great maneuvering speed-1 is enough for the enemy traverse from long range all the way into short range, which you covered in the "catching prey" section on black dice ships.
  3. Like
    jbrandmeyer reacted to Green Knight in Opinions on Swivel-Mount Batteries   
    LTT is super flexible. OC it's even better on muni, but still great on clams.
  4. Like
    jbrandmeyer got a reaction from geek19 in Cannot Get Your Gifts Out December 2020   
    Thank you so much for the updates!  I'm looking forward to playing in meatspace again in the coming year.
    > The Foresight has a great effect, allowing you to cancel 2 dice at long range or reroll 2 at Medium/Short.  Or with Mothma, cancel 2 at medium and reroll 2 at close. 
    I think you meant to say "reroll 3 at close."
    IMO, the value of dual defense tokens can be over-stated.  An ISD-II has 54% odds of generating two accuracies in the initial throw, before anything like H9 or leading shots have had a chance to make things worse for the defender.  The Liberty suffers from a similar deficit with its brace tokens.  That said, with great maneuvering speed-1 is enough for the enemy traverse from long range all the way into short range, which you covered in the "catching prey" section on black dice ships.
  5. Like
    jbrandmeyer reacted to geek19 in Cannot Get Your Gifts Out December 2020   
    https://cannotgetyourshipout.blogspot.com/2020/12/john-updates-1215.html
    Ackbar - removed my writeups of the individual ships and went with a vaguer assessment of what does/doesn't work for him. I also updated my DnD character in one of the jokey jokes (as my Dragonborn Paladin has retired); this is the content you came here for right?
    MC30 - title changes and viability, and I stripped down the "Scooty Puff Jr" build to its most basic elements.  Can you run them titleless? I don't hate the idea!
  6. Like
    jbrandmeyer reacted to Ginkapo in Green Squadron A-Wing VS. Generic A-Wing   
    Its not A wing vs Green
    Its Y wing vs Green
  7. Thanks
    jbrandmeyer reacted to Grumbleduke in Resupply refresh behavior   
    Firstly, we don't have the new Rules Reference yet, so we don't know the full rules.
    But from what has been said on stream and so on, for the cards that have the two bars on either side, the left bar is what is put on the card at the start of the game, and the right bar is what is needed to ready the card during the status phase (at the end of a round).
    Cards that need to be exhausted to use now come with two options. Some, like the new Assault Concussion Missiles ready as normal during the status phase (the double arrows in a square icon), while others like Munitions Resupply have a cost associated with readying them (the single arrow and bar on the right). But all that cost does is ready or unexhaust the card so you can use it again in a later round.
  8. Haha
    jbrandmeyer reacted to Packerman29 in Rapid Launch Bays and...   
    When I see this topic headline:
     
    littering and....

  9. Like
    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.
  10. Haha
    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...
  11. Haha
    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...
  12. Like
    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.
  13. Like
    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...
  14. Thanks
    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.
  15. Like
    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.
  16. Like
    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...
  17. Like
    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.
  18. Like
    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.
  19. Like
    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.
  20. Like
    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.
     
  21. Like
    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. 
  22. Like
    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.
     
  23. Like
    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.
     
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    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.
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    jbrandmeyer reacted to JauntyChapeau in Denver area tournament schedule   
    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
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