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I'm not the math wiz you are looking for, but I know a guy...

I have some followup questions:

1) Is this on a single roll, or with re-roll(s). If so, how many dice get re-rolled?

2) Are we looking for exactly 9 hits, or 9 or more hits?

3) Hits only? Or are crits considered hits for the purpose of this exercise?

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Well, you are going to need three doubles, and two hits on the other three dice.

So, something like... I don't know. Not high.

And they'll evade and redirect, so it isn't going to help.

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1 hour ago, LTD said:

Well, you are going to need three doubles, and two hits on the other three dice.

So, something like... I don't know. Not high.

And they'll evade and redirect, so it isn't going to help.

*3 hits or crits.  Or 4 doubles, a hit or crit and a blank.

@boow Are you okay with getting more than 9 hits in that calculation, or does it need to be exactly 9?

Because if it's 9 and up, we have all results with 4 doubles + at least another damage and all results with 3 doubles and 3 hits or crits (0.125^3 *0.5^3 = 0.00024).

The chance of getting exactly 4 double hits on 6 reds is  (0.125^4 * 15 * [1-0.125]^[6-4]= 0.0028).

I *think* you just multiply that by the odds your other 2 reds do any damage at all (0.625^2).

So that would be 0.00024 + (0.0028 * 0.625^2) = 0.00133, or a 1.33% chance of doing 9 or more damage.

 

...Now for someone who actually understands probability to come along and correct this.

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Thanks for all the attempts and follow up questions. 
 

The actual roll I got was three doubles, two singles, and a crit. 
 

Context: My opponent had a corvette carrying Admiral Raddus, mc75 set aside (with a profundity hammerhead set aside). Also had two flotillas and a ton of squads. 
 

He chose first player and my rift assault objective. I plopped down a Cymoon ISD directly across from his corvette, made his corvette do a speed-1 maneuver through the objective and made his speed go up to 4. First turn he had to move his corvette before my ISD and got into long range. I had set a CF dial and had a 6 red dice shot at long range. Rolled the 9 hits, he evaded a double, burned a redirect, soaked 4 through shields and took 3 damage. Incidentally I had xx-9 turbolasers for two faceup cards, but didn't trigger structural damage. 
 

I actually had some 2 gauntlets on my fleet for some raid shenanigans, and both swooped in with 4 speed rogue moves and took shots on the corvette to KO it round 1, ending the game before he could Raddus out the rest of his fleet. 
 

So it was a lot of small seemingly inconsequential placement decisions and luck, but it was quite satisfying to get a win in that way.
 

But in true Star Wars fashion I’ll want to know the odds so I can throw some 3po odds trash talk his way. 

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13 minutes ago, boow said:

Thanks for all the attempts and follow up questions. 
 

The actual roll I got was three doubles, two singles, and a crit. 
 

Context: My opponent had a corvette carrying Admiral Raddus, mc75 set aside (with a profundity hammerhead set aside). Also had two flotillas and a ton of squads. 
 

He chose first player and my rift assault objective. I plopped down a Cymoon ISD directly across from his corvette, made his corvette do a speed-1 maneuver through the objective and made his speed go up to 4. First turn he had to move his corvette before my ISD and got into long range. I had set a CF dial and had a 6 red dice shot at long range. Rolled the 9 hits, he evaded a double, burned a redirect, soaked 4 through shields and took 3 damage. Incidentally I had xx-9 turbolasers for two faceup cards, but didn't trigger structural damage. 
 

I actually had some 2 gauntlets on my fleet for some raid shenanigans, and both swooped in with 4 speed rogue moves and took shots on the corvette to KO it round 1, ending the game before he could Raddus out the rest of his fleet. 
 

So it was a lot of small seemingly inconsequential placement decisions and luck, but it was quite satisfying to get a win in that way.
 

But in true Star Wars fashion I’ll want to know the odds so I can throw some 3po odds trash talk his way. 

I would say a 50% as it either happens or it doesn't

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Posted (edited)
7 hours ago, The Jabbawookie said:

...Now for someone who actually understands probability to come along and correct this.

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.

 

Edited by Grumbleduke
Google messed up the brackets in one of my numbers

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1 hour ago, boow said:

Thanks for all the attempts and follow up questions. 
 

The actual roll I got was three doubles, two singles, and a crit. 
 

Context: My opponent had a corvette carrying Admiral Raddus, mc75 set aside (with a profundity hammerhead set aside). Also had two flotillas and a ton of squads. 
 

He chose first player and my rift assault objective. I plopped down a Cymoon ISD directly across from his corvette, made his corvette do a speed-1 maneuver through the objective and made his speed go up to 4. First turn he had to move his corvette before my ISD and got into long range. I had set a CF dial and had a 6 red dice shot at long range. Rolled the 9 hits, he evaded a double, burned a redirect, soaked 4 through shields and took 3 damage. Incidentally I had xx-9 turbolasers for two faceup cards, but didn't trigger structural damage. 
 

I actually had some 2 gauntlets on my fleet for some raid shenanigans, and both swooped in with 4 speed rogue moves and took shots on the corvette to KO it round 1, ending the game before he could Raddus out the rest of his fleet. 
 

So it was a lot of small seemingly inconsequential placement decisions and luck, but it was quite satisfying to get a win in that way.
 

But in true Star Wars fashion I’ll want to know the odds so I can throw some 3po odds trash talk his way. 

Did he not have a navigate dial, so that he could at least slow down to 3?

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Posted (edited)
38 minutes ago, Grumbleduke said:

little maths-deprived lately

Don´  t we all know that feeling...

Some good math there!

Edited by Jansen007

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10 minutes ago, Bertie Wooster said:

Did he not have a navigate dial, so that he could at least slow down to 3?

He did actually slow down to 3. and I didn't check to see if he maximized the yaws or not, but where he decided to end, he was about half in and half out of range. because i had placed the gravity well pretty much in the middle, both his corvette and my ISD were right on an edge of the starting area, so there was only one direction he could really turn towards.

This was on vassal, so we never know how accurate the distances are, but it showed me as having the shot, so i took it! We both placed our ships right at the range 3 line of deployment as well, and then he had the extra 1 move... Not sure why he had the corvette going so fast. i think he had some plan to get towards the rear of my ships and then hyperspace the Mc75 in.

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2 hours ago, boow said:

He did actually slow down to 3. and I didn't check to see if he maximized the yaws or not, but where he decided to end, he was about half in and half out of range. because i had placed the gravity well pretty much in the middle, both his corvette and my ISD were right on an edge of the starting area, so there was only one direction he could really turn towards.

This was on vassal, so we never know how accurate the distances are, but it showed me as having the shot, so i took it! We both placed our ships right at the range 3 line of deployment as well, and then he had the extra 1 move... Not sure why he had the corvette going so fast. i think he had some plan to get towards the rear of my ships and then hyperspace the Mc75 in.

I really doubt it may be any less accurate in Vassal than IRL

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4 hours ago, Grumbleduke said:

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 = 0.00022888183

Add that on to our answer to the first part and we get: 0.66% or about 1 in 150 chance of getting at least 6 damage from 6 red dice.

 

So you're telling me its possible

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3 hours ago, boow said:

He did actually slow down to 3. and I didn't check to see if he maximized the yaws or not, but where he decided to end, he was about half in and half out of range. because i had placed the gravity well pretty much in the middle, both his corvette and my ISD were right on an edge of the starting area, so there was only one direction he could really turn towards.

This was on vassal, so we never know how accurate the distances are, but it showed me as having the shot, so i took it! We both placed our ships right at the range 3 line of deployment as well, and then he had the extra 1 move... Not sure why he had the corvette going so fast. i think he had some plan to get towards the rear of my ships and then hyperspace the Mc75 in.

Yeah, I just checked it in Vassal. With most possible placements you'd have the shot. With the Rift Ambush movement, a CR90 going speed 3 directly across from a Cymoon is pretty rough. I did find that by doing an inside turn and giving it a click at the first (nav dial) and second joints it is possible to avoid the red range. But your friend may not have found that possibility.

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17 minutes ago, Bertie Wooster said:

Yeah, I just checked it in Vassal. With most possible placements you'd have the shot. With the Rift Ambush movement, a CR90 going speed 3 directly across from a Cymoon is pretty rough. I did find that by doing an inside turn and giving it a click at the first (nav dial) and second joints it is possible to avoid the red range. But your friend may not have found that possibility.

yeah he had some flotillas in the area that he was trying to manuever around. just imagine if i had rolled those 9 hits and no accuracy against a flotilla. i wouldve smashed my laptop.

As regards to the math above, 160 to 1 doesn't sound all that impressive, but it was still a lot of fun, haha!

 

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Turns out one of my numbers was incorrect (I'd put the calculation into Google and it messed up the brackets). Getting at least 9 damage on 6 red dice with no re-rolls is about 1 in 136.

I spotted that because, due to boredom, I ran the numbers for damage output on 6 red dice. I treated criticals as damage, and accuracies as no damage.

Probability and cumulative probability graphs here.

You have a 1 in 260,000 chance of rolling natural doubles, 1 in 10,500 of getting at least 11 damage, 1 in 926 of getting at least 10. These probably happen to someone, but probably won't happen to any one player.

It is 1 in 136 for getting at least 9 damage (which is reasonable), 1 in 30 for getting at least 8 (should happen to most players), and 1 in 9 for getting at least 7.

... today I hit 8 weeks without being allowed out of the house.

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@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).

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Posted (edited)

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.

Edited by jbrandmeyer

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