Hear me out. I am specifically talking only about the mean. The average damage of a dice pool would be the mean result that would come from that dice pool. So a dice pool of 4 reds would have a mean of 3 damage. In my simple mind I imagine that adding one reroll would result in a mean of 3.75 damage(mean of 5 reds). Reason being that you would always choose to reroll bad rolls, so even if your maximum damage potential is that of 4 reds the average damage output of this setup with a reroll would still be 3.75 damage.

I understand the distribution and consistency changes with rerolls, but aside from those does the mean actually change?(If you do know how the distribution works though that would be awesome )

I believe this mean is also the same for Dual turbo turrets but the distribution and consistency would be different.

I think this average gets awry with two++ rerolls though?

Your logic would be true if it was guaranteed that you always roll a blank in the initial dice. But this isnt true.

Well, excluding using accuracies in certain attacks, if I did not roll a blank/acc I would definitely have at least 4 damage which is higher than my guess of 3.75. So when you mention that this isn’t true do you mean that the average/mean of reroll is > dice pool + 1?

Sorry if I am not clear. Perhaps an equation will clarify

Question: ave dmg of dice pool + reroll = ave dmg of dice pool +1 die ?

Or is it > / < ?

And why? J

But it is a good approximation if you throw lots of reds.

To get a distribution you would have to define a decision policy, when to reroll. Just blanks? Reroll single damage?

One can try to find an optimal decision policy, like reroll if the die is below average. But sometimes you want to gamble for a damage spike.

Simulate it (not Monte Carlo, you can count all possibilities and get precise probabilities).

I had a formula for TRC means actually, but rerolls are harder. Maybe I find the notes.

As the number of red dice increase, your simplification becomes more accurate.

The lower the number of red dice however, the worse it overestimates the damage.

For example with a single red dice and reroll, the expected damage is 0.75.

Expected damge with a reroll (assuming you reroll blacks and accuracy) is 1.03. (also 1 Acc 4.7% and miss 9.4%).

Expected damage from your system is 1.5.

Your system will always overestimate the damage. The reason is that sometimes you will have no need for the reroll, thus why would it add further damage?

Another example, you roll 3 red dice with all three results being hits. You would not want to spend the reroll on this (unless you wanted an accuracy, or it was no risk such as with DTT) as you would trade 1x dice with 1.0 damage, and reroll it for 0.75 expected damage.

With foresight and knowing that 100% of the time you definitely would roll at least 1 blank result, then the reroll is the same as +1 attack. Hense rolling more red dice will increase the chance of a miss result, and make your system closer to true.

In summary;

"Question: ave dmg of dice pool + reroll < ave dmg of dice pool +1 die"

This difference narrows as the number of rolled red dice increase.

The probability calculations are quite involved, and rely on a number of assumptions.

But the gist of it is:

-For large, single-color dice pools, a reroll (CF token) adds almost the same expected damage as an extra die (CF dial).

-For small or multi-color dice pools, an extra die increases expected damage substantially more than a reroll.

Ah, I see what you are getting at. In my mind I was thinking of a normal distribution for dice pool + reroll where the upper part of the graph gets chopped off due to 1 less die as compared to dice pool + 1. I thought the mean would remain where it was but yes there is more math involved as some of the higher results that contribute to getting the average are lost(if u know a nice sinulation freeware idm spending some time on it )

I suspect then that DTT is closer to what I imagine regarding the normal distribution, as I am basically throwing dice pool + 1 and then removing a die?

The probability calculations are quite involved, and rely on a number of assumptions.

But the gist of it is:

-For large, single-color dice pools, a reroll (CF token) adds almost the same expected damage as an extra die (CF dial).

-For small or multi-color dice pools, an extra die increases expected damage substantially more than a reroll.

I suspect then that DTT is closer to what I imagine regarding the normal distribution, as I am basically throwing dice pool + 1 and then removing a die?

That's exactly right.

The formula you would use for working out the exact value would be a minor adaptation of the "roll X dice, drop Y lowest" used in D&D character generation.

In practical terms, though, the value of DTT is largely equivalent to getting a free CF token per turn to be used on red dice only.

I'll also add-on that adding further dice modification abilities lose value. I.e If you already have TRC, your CF token has diminishing returns.

Therefore CR-90 corvettes with TRCs don't benifit much by holding a CF token. They do benifit quite alot by adding a CF dial though.

When I build lists, I like to pair a strategy for adding dice (armament modifications, CF dial spamming, Ackbar) with a dice modification upgrade/strategy (CF token generation, DTT, HomeOne, TRC, ordnance experts, etc). Dice modifications need more dice for maximal chance of a dud result in need of changing, and massed dice enjoy modifications for maximal damage.

But I do not stack 2 or more dice adding strategies without a means to modify them reliably, nor would I get multiple dice modification abilities if I don't have very large dice pools.

Black Dice Example

Rather than talk about 4-5 red dice lets choose a simpler example of 2 black dice as there are no accuracies to consider keeping. In this example the player will only reroll on a blank and will keep a single hit. Rerolling a single hit would in any case not change the mean damage as you sacrifice one damage for a mean expected 1 extra damage.

Mean damage for 2 black dice = 2

Mean damage for 3 black dice = 3 (i.e. 2 black + Concentrate fire command)

Mean damage for 2 black dice with one reroll = 2.4375 (i.e. 2 black + Concentrate fire token)

Adding a rerolled third dice get you 1 extra damage so first you have to work out the probability of rolling at least one blank in order to do your reroll.

For one dice to roll a blank is 25% so the chances of rolling some damage (defined as a hit or a hit+crit) = 75%. For 2 dice to roll at least one blank it is 1 minus the probability of not rolling a blank squared (as you have 2 dice). So 1-(.75 x .75) = 0.4375.

If you had 3 or 4 or 5 black dice to calculate the chances rather than squaring the 0.75 you cube it or ^4 or ^5. As the number of dice increases the chances of rolling at least one blank increase so you calculate that then add the expected damage of the extra dice to the chances of rolling at least a blank.

Red Dice Example

OK so I'll have a go at the expected damage from 4 red dice + 1 CF token granting a reroll with no desire to keep an accuracy on the first roll. Also assume the player will be happy with 4 single hits and not use the reroll to attempt to get a double hit or generate an accuracy.

1 Red dice does 0.75 damage so the Expected damage of 4 red = 4 x 0.75 = 3

Chances of 1 red rolling no damage (blank or accuracy) = 0.375

Chances of 1 red rolling some damage (hit, crit or double hit) = 0.625

Chances of rolling some damage on N dice = (0.625)^N so for 4 dice (0.625)^4 = 0.153 so chances of rolling at least one blank/acc = 1-0.153 = 0.847

We have an 84.7% chance of rolling at least one duff dice so that means an 84.7% chance of using the reroll to get an extra 0.75 damage. So 0.75 X 84.7% = 0.636 damage which we add to the original expected 3 damage = 3.636

We can compare

4 red dice with a reroll = 3.636

5 red dice = 3.75

I've worked out the generic formula

N = Number of Dice being rolled

D = Expected damage form 1 die

S = Probability of scoring some damage on one die

Expected damage including 1 reroll = (NxD)+(1-(S^N))xD

If you want to work out the probabilities of mixed dice volleys such as an ISD-I front arc at close range then... Errm... I mean...

Well it gets complicated.

I tried to paste in form a spread sheet but can't get it to work. You can type it in with for example:

• the word "Dice" in cell A1
• the word "Results" in Cell E1
• the formula =(D2*B2)+(1-(C2^D2))*B2 in cell E2

Row 1

Dice, D, S, N, Results

Row 2

Red, 0.75, 0.625, 4, =(D2*B2)+(1-(C2^D2))*B2

Row 3

Blue, 0.75, 0.75, 3, =(D3*B3)+1-(C3^D3)*B3

Row 4

Black, 1, 0.75, 2, =(D4*B4)+1-(C4^D4)*B4

Then play around to your heart's content by varying the value of N in column D

I've worked out the generic formula

N = Number of Dice being rolled

D = Expected damage form 1 die

S = Probability of scoring some damage on one die

Expected damage including 1 reroll = (NxD)+(1-(S^N))xD

If you want to work out the probabilities of mixed dice volleys such as an ISD-I front arc at close range then... Errm... I mean...

Well it gets complicated.

I've worked out the generic formula

N = Number of Dice being rolled

D = Expected damage form 1 die

S = Probability of scoring some damage on one die

Expected damage including 1 reroll = (NxD)+(1-(S^N))xD

The problem with using such formulas is that so many assumptions are necessary (the key being: all dice in the pool are of the same type, accuracies have 0 value, crits have the same value as hits, and you will only reroll blanks) that the exact values returned are essentially worthless in practice. They only confirm the general principles we already know, and are too restrictive and situational to give additional insight (e.g. "how many red dice per arc do you need before upgrade X becomes better than upgrade Y?").

If you want to work out the probabilities of mixed dice volleys such as an ISD-I front arc at close range then... Errm... I mean...

Well it gets complicated.

Yeah, you can say that again!

If you wish to keep accuracies then just change the value of S in the formula but yes it doesn't take into account all variables in a player's decisions. Some element of player skill is still required which is, I suppose, why we play these games.

A more general maxim then is that...

As the number of dice in a volley increases, the value of a concentrate fire token approaches (but never quite reaches) the value of a concentrate fire command.

A more general maxim then is that...

As the number of dice in a volley increases, the value of a concentrate fire token approaches (but never quite reaches) the value of a concentrate fire command.

My rule of thumb:

-Bigger ships benefit significantly from banked CF tokens (due to larger pools, higher command value, typically fewer manipulation upgrades).

-Small ships should be banking other more useful tokens, and using CF only from dials, especially if loaded with TRC, DTT, etc.

...but we already knew that.

Your logic would be true if it was guaranteed that you always roll a blank in the initial dice. But this isnt true.

As the chance of blank increases, the power of reroll increases.

Maybe I could have written this better first time.

1 reroll is not 1 die added even when is close to it.

You are talking about average. 4 damage is inside of the 3 damage average. Think in blue die cause it easier.

4 blue dice have 3 damage average. With a reroll they wouldn't have 3.75 damage average cause if you get 4 hits you won't reroll and you won't get the bonus. You will reroll the blanks and, maybe the accuracy so you are rerolling from a 25% of getting a blank/accuracy with a 75% of getting the hit you want. It means you have a 93.75% of getting your hit with a reroll in a blue die. The thing here is that you just have 1 reroll. If you would have a reroll per die then each blue die would have a 0.9375 damage average and with 4 dice you would have 3.75 but you don't. You have 3 "0.75" dice and 1 "0'9375" die so you have 3.1875 damage average.

As long as the potential damage of 5 red die affect to their average, a reroll that means you can't do 10 damage can't means the same average than 4 dice.

With the red and black dice I am not sure as their chances of get a hit are different than their average damage. Maybe I do the maths later.

Those are my thoughts but I am not an statistics guru

After some seconds I think it is as easy as average x chance of blank so:

Blue +0.1875 per reroll

Red +0.28125 per reroll

Black +0.25 per reroll

You have 3 "0.75" dice and 1 "0'9375" die so you have 3.1875 damage average.

That would only be true if only a specific die (decided in advance) could be re-rolled. But you can reroll any of the four - whichever is weakest. I can do the exact math for you if you want (later, I'm at work now), but the expected outcome will be significantly closer to 3.75, though still somewhat lower of course.

Remember: nobody is saying that a reroll is exactly as good as an extra die. Only that, the more dice you roll, the smaller the difference between adding and rerolling.

You have 3 "0.75" dice and 1 "0'9375" die so you have 3.1875 damage average.

That would only be true if only a specific die (decided in advance) could be re-rolled. But you can reroll any of the four - whichever is weakest. I can do the exact math for you if you want (later, I'm at work now), but the expected outcome will be significantly closer to 3.75, though still somewhat lower of course.

Remember: nobody is saying that a reroll is exactly as good as an extra die. Only that, the more dice you roll, the smaller the difference between adding and rerolling.

I am not sure if just 1 reroll change the odds if you roll more dice. What is clear is if you mix colours and you have just 1 reroll to determine the bonus seems impossible for me.

You have 3 "0.75" dice and 1 "0'9375" die so you have 3.1875 damage average.

That would only be true if only a specific die (decided in advance) could be re-rolled. But you can reroll any of the four - whichever is weakest. I can do the exact math for you if you want (later, I'm at work now), but the expected outcome will be significantly closer to 3.75, though still somewhat lower of course.

Remember: nobody is saying that a reroll is exactly as good as an extra die. Only that, the more dice you roll, the smaller the difference between adding and rerolling.

I am not sure if just 1 reroll change the odds if you roll more dice. What is clear is if you mix colours and you have just 1 reroll to determine the bonus seems impossible for me.

You're going to pick the weakest die to reroll.

Blank is pretty much always the weakest die.

As the total number of dice rolled increases, the odds of rolling at least one blank approaches 100%.

Therefore, as the number of dice rolled increases, the likelihood that you will gain the maximum benefit out of a single reroll increases.

Hmm I understand that point but if you are assuming that as your roll is bigger enough to guarantee a blank don't have you to remove the 0.75 that blank die add to your average? Not for bothering! I am really trying to understand the arcane secrets of statistics.

I would like to see the maths behind in the "for-dumb'est" way you could

EDIT: Okay, I was thinking on it and I figured out the reason. As I said you work over the blank odds but what you are trying to explain me is that is not the blank odds of the die rather than the blank odds of the roll. Isn't it?

Hmm I understand that point but if you are assuming that as your roll is bigger enough to guarantee a blank don't have you to remove the 0.75 that blank die add to your average? Not for bothering! I am really trying to understand the arcane secrets of statistics.

I would like to see the maths behind in the "for-dumb'est" way you could

EDIT: Okay, I was thinking on it and I figured out the reason. As I said you work over the blank odds but what you are trying to explain me is that is not the blank odds of the die rather than the blank odds of the roll. Isn't it?

WARNING: LONG ASSPOST WITH LOTS OF MATH AHEAD

Yeah, it's a separate thing. I'm sorry, I'm not great at explaining statistics...

So, we know that the average damage of a red die is .75.

This is found by finding the sum of all sides and dividing it by the number of sides of the die.

Say you're going to roll 8 dice. Your average damage is 8*(.75)=6. This is found by multiplying the average damage of each die.

Let's say your roll is one of each face, including the ones that are the same value, so 2 hits, 2 crits, a double, an acc, and 2 blanks. Your total damage is exactly in line with what the average predicted.

The average damage on each of your dice was .75. This is calculated by taking the total damage dealt and dividing it by the number of dice that you rolled, so 6/8 = .75.

Now, after you've rolled, suppose you can select one of those dice to reroll. You will of course choose a blank. Notice how you did not decrease your total damage dealt, because you removed a blank, so you haven't decreased the average damage per die associated with the initial dice roll. You rolled 8 dice; you have 6 damage.

Then, you prepare to roll the die the second time. At this point, not knowing what its result will be, you can estimate that you'll get .75 damage out of it.

So you can now back up and say that, assuming you rolled that blank in your initial pool, your average damage on 8 dice with one selective reroll was 6.75, or (average damage of the initial pool)+(average damage of the rerolled die).

But we can remove even that assumption by identifying what the likelihood of it happening is.

What is the likelihood that you'll roll a blank in your initial pool?

Well, that's pretty easy to calculate. The likelihood of rolling a blank on one (let's call acc a blank for simplicity) is 3/8 = .375. The likelihood of not rolling a blank is the inverse, or 1-(3/8) = 5/8 = .625.

The likelihood of rolling at least one blank can be redefined as the inverse of rolling no blanks. We can find the likelihood of finding NO blanks by just raising the non-blank odds to the power of the number of dice we're rolling, so (5/8)^8 = .023.

So this tells us we have a .023, or 2.3%, chance of rolling NO blanks on 8 reds. Now we redefine or problem back to the blanks: if there is at least one blank, the condition "are there no blanks" is not true, and vice versa. So, we can find the odds of rolling at least one blank by taking the inverse again: 1 - .023 = .977, or nearly 98% chance of rolling at least one blank.

Now we can put it back together.

If you have .98 chance of rolling a blank on those 8 dice, and then IF that happens you'll reroll for an average of .75 extra damage, you can find the average damage increase with (chance of the triggering event happening)*(average impact of the event being triggered). In this case, that's .977 * .75 = .73ish extra damage for a selective reroll on 8 dice.

If you actual read and followed that, you can see why your average damage increase improves for a single reroll as your total dice pool expands. The odds of rolling at least one blank increase, which increases the likelihood that you will use your reroll, which increases your average damage UP TO the average damage of the rerolled die.

Oh, also, caveat: the above is super duper simplified. It works within the constraints of rolling and rerolling a buttload of red Armada die, but don't try taking it to a stats class and expect it to be rigorously accurate. I'm an English Lit major who does cybersecurity for a living, I'm like the least qualified person in the room to explain this stuff...

Oh, also, caveat: the above is super duper simplified. It works within the constraints of rolling and rerolling a buttload of red Armada die, but don't try taking it to a stats class and expect it to be rigorously accurate. I'm an English Lit major who does cybersecurity for a living, I'm like the least qualified person in the room to explain this stuff...

But it is accurate. I followed it.