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 megatrons2nd wrote:
I had a game where my opponent kept rolling 2 dice and picking the highest for fleet, but this might apply for other situations as well.

When you have an ability to roll a dice again to attempt for a higher number will this skew the average roll to a higher number if you did roll two dice and choose the highest. I understand that if you rolled a 1, and a 6 you probably would have rerolled the 1. But would you really reroll the 5 to try for the 6? How about that 4, and if you did what if that 3 was the actual second roll?

I am thinking it does create a higher than average result, though my opponent did not believe so.

That is correct, and here is some math to show why you never want to let someone roll their rerolls as if they were + extra dice. (Mathtastic!)
WARNING! MATH CONTENT in a wall of text (If you want a long story short, skip to the results below)

*Disclaimer: There is a very good chance that this is all screwed up, but I don't think so.*

Example Situation: you are shooting two twin-linked brightlances at a pesky tank. For reasons not important, your are BS 2, so you need a 5+ to hit.

In excel I generated two columns of random numbers (=randbetween(1,6)) of one thousand rows each.
Basically I rolled each die 1,000 times. The number of 5+ rolls for each die was 330, and 331, for a total of 661 hits.
661/2000=0.3305, or 33.05% of those first rolls hit. (you divide by 2000 because you roll 2 dice 1000 times each. Math!)

If you roll your twin-links with the first to hit roll (rolling four dice 1000 times each) you end up with the results of the first roll, plus the second ones:
360 and 327 for the third and fourth die, respectively, which is another 687 hits.
687/2000=0.3435, or 34.35% of those "simultaneous rerolls" hit.

This gives us a total of 67.4% of all the rolls hit when you simultaneous roll your rerolls.

But wait, what if instead we did it properly? What if we did rerolls instead of simultaneous rolling? Lets math this out and see what happens.
So, I created a new column and gave it the following formula (=IF(A6<5,RANDBETWEEN(1,6),""). Scary! Really all it is saying, is that if the first roll was less than 5, it rerolls the die. If the first roll was 5 or 6, it puts nothing in the cell.

I had it run for all 1000 rolls for each of the first two dice, and something amazing happened. MATH!
Turns out, it didn't reroll all 1000 times. it only rerolled the 1-4 rolls, so it actually only made 670 rerolls of the first die, and 669 rerolls of the second die, substantially fewer dice were rolled than the simultaneous roll. Therefore, there were fewer hits as well.
Out of the actual rerolls, there were 214 hits with the first die, and only 216 hits with the second die, for a total of 430 hits (smaller pool of dice rolled, smaller number of hits rolled.) Therefore:
430/2000=0.215, or 21.5% of the actual rerolls hit.

This gives us a total of 54.55% of all the rolls hit when we actually rerolled instead of simultaneous rolled them.

RESULTS (as long as my math is right):
By simultaneous rolling rather than rerolling on a 5+ to hit, I was able to increase my number of hits by 12.85%!

This got even worse when I simultaneous roll 4+ hits, which increased the number of hits by over 26%!
on the 4+ simultaneous roll I ended up with 2015 hits, which is 15 more hits than were originally shot!

I didn't have the heart to try 3+, but I'm assuming they'd be much worse.

Lesson learned: simultaneous rolls are not the same as re-rolls, and people who know the probabilities and stats of this and do it anyways are manipulating the dice (i.e. cheating).

 rookshunter wrote:
RESULTS (as long as my math is right):

Either your math is wrong, your random number generator isn't sufficiently random, or you've made a typo somewhere in implementing your math. There is absolutely no difference between rolling a die and then re-rolling it if it fails vs. rolling two dice and calling it a success if one die succeeds. They're exactly equivalent because you always re-roll the failure so it doesn't matter when you roll the second die, you aren't gaining any new information from it.

The issue with fleet re-rolls is that you aren't required to re-roll the die, you have to choose whether to keep the first result or to re-roll it and try for a better one. If you roll two dice at once and take the highest you're giving yourself access to knowledge of what the result of the re-roll will be before committing to throwing away the first result.

 DeathReaper wrote:

 rookshunter wrote:

My knee-jerk reaction is to never ever allow twin-linked to roll simultaneous.

Why?

Rolling two dice for twin linked is actually the exact same outcome as rolling one die and re-rolling it if there is a miss.

You have the same chance to hit.

This is because Dice do not have memory.

With simultaneous rolling, how do you know that second hit wasn't the re-roll of the first hit you rolled?

it doesn't matter, as long as one of them hit. If it makes you feel better, if I were your opponent, I would let you choose which die was the original roll and which is the re-roll.

That same question comes up when you use different colored dice for the re-roll, how do you know that second hit you made wasn't a first hit's reroll?

the second colored die is the re-roll, I am not sure what you are getting at here.

For example, say we have a Twin Linked Lascannon from a Space marine with a BS 4.

We roll a green and a red die, we note the green die as the roll and the red die as the re-roll if needed. We roll both dice the green comes up a 3 and the red comes up a 2. we hit because the roll (The green die) is a 3+, which hits.

Or same scenario the green die is a 1 and the red die is a 5, we hit because the roll missed but the re-roll was a 5.

Simultaneous rolling vs. rerolls have different statistical averages which is why you do one vs the other, and why there are so many different rules that either give you re-rolls (Fleet) or additional, simultaneous dice (Armourbane). I've yet to see where it says these two types of rolls are interchangeable.

Only if you are adding the two dice together to get a number.

If you roll 2 dice there are 36 possible outcomes. These being:

Spoiler:

As you can see the lighter die can be 1-6 and the darker die can be 1-6, with 36 possible different combinations no two repeated.

Not to burst anyone's bubble, and I realize it seems like a friendly way to speed up a game (I thought about it for a bit myself) but I'd rank this as dice manipulation. It's more subtle than loaded dice as statistical math is more difficult to see for most people, but the same principle applies.

It really isn't dice manipulation because dice do not have memory and you are not adding the result of the two dice together. (See the chart, each die has a 1/6 chance of showing any given number).

This only works if you are rolling one TL and it's reroll at a time. Also, if there is a special ability activated from a specific number rolled, or it is something like fleet, I hope its obvious why you don't wanna let someone do that (as mentioned by the OP). As pretty as that chart is (I'm digging it's feng shui), it is irreverent and you seem to have missed the point.

As soon as you roll two or more TL simultaneous w/their rerolls, you don't know which hits were rolled on the first roll, so you don't actually know how many rerolls you were legitimately entitled to. you cannot tell me which reroll is for which original roll so you start skewing the results. You essentially give yourself more "rerolls" than you should have had (see my epic math post).
Colored dice don't help either. Say blue was first roll, red=rerolls. Technically, for every blue hit, that is one less red die you should be rolling. Tell me, which red dice you rolled now don't count? the hits? The misses? How many of each?
Again, Now that you know the numbers behind the why, you don't want to continue doing it. That is manipulating the dice to give you a higher percentage of hits than you should have. That is called cheating.

 rookshunter wrote:

 megatrons2nd wrote:
I had a game where my opponent kept rolling 2 dice and picking the highest for fleet, but this might apply for other situations as well.

When you have an ability to roll a dice again to attempt for a higher number will this skew the average roll to a higher number if you did roll two dice and choose the highest. I understand that if you rolled a 1, and a 6 you probably would have rerolled the 1. But would you really reroll the 5 to try for the 6? How about that 4, and if you did what if that 3 was the actual second roll?

I am thinking it does create a higher than average result, though my opponent did not believe so.

That is correct, and here is some math to show why you never want to let someone roll their rerolls as if they were + extra dice. (Mathtastic!)
WARNING! MATH CONTENT in a wall of text (If you want a long story short, skip to the results below)

*Disclaimer: There is a very good chance that this is all screwed up, but I don't think so.*

It is screwed up.

Check the picture I posted, and realize that a single die can roll anywhere from 1-6 inclusive.

2 dice will give a possible 36 results without any repeating thus the chance of success are the same if you roll a die and re-roll it, or roll two dice and see if one is a hit.

The die rolls are Independent Events, the outcome of the first roll has no bearing on the second roll, ergo the probability of a hit coming up on the first die (For your example needing a 5+ to hit) is 33.33% and the probability of a hit coming up on the re-roll is 33.33%

Here is the explanation for independent events and re-rolls.

Basics of Mathhammer wrote:

Spoiler:

Re-rolls

The last basic concept is that of re-rolls. If you think about it, what a re-roll does is allow you to try again, if the original event was not to your liking. This is really just two mutually exclusive (as it cannot both happen and not happen) events, so you're using the sum rule from above. It's also the combination of two independent events (the first roll failed, and the second roll succeeded). That means that you're combining the odds that the event happens (p) with the odds that (it didn't happen (1-p) and that the re-roll did happen (p)). The odds on the second roll are no different than the odds on the first roll, so that simplifies things, and the formula at the end is p + ((1-p)*p).

Another way to approach this problem is by determining your chance of failure, then subtracting that from 1 to get your chance of success. So the equation is 1 - (1 - p)^2, since the only way to fail is for each roll to fail, and each roll has a 1 - p chance of failing.

Example: Let's say you want to know the odds of hitting with a twin-linked BS3 gun. BS3 means you're hitting on a 4+, so the basic odds (p) are 1/2. Therefore, the odds of hitting with the re-roll are:

1/2 + ((1-1/2)*1/2)
1/2 + (1/2 * 1/2)
1/2+1/4
3/4 (or 75%)

And yes, I stated for a TL Lascannon, which has only 1 shot, so I was talking specifically about rolling one TL and it's reroll at a time.

 rookshunter wrote:

 DeathReaper wrote:

 rookshunter wrote:

My knee-jerk reaction is to never ever allow twin-linked to roll simultaneous.

Why?

Rolling two dice for twin linked is actually the exact same outcome as rolling one die and re-rolling it if there is a miss.

You have the same chance to hit.

This is because Dice do not have memory.

With simultaneous rolling, how do you know that second hit wasn't the re-roll of the first hit you rolled?

it doesn't matter, as long as one of them hit. If it makes you feel better, if I were your opponent, I would let you choose which die was the original roll and which is the re-roll.

That same question comes up when you use different colored dice for the re-roll, how do you know that second hit you made wasn't a first hit's reroll?

the second colored die is the re-roll, I am not sure what you are getting at here.

For example, say we have a Twin Linked Lascannon from a Space marine with a BS 4.

We roll a green and a red die, we note the green die as the roll and the red die as the re-roll if needed. We roll both dice the green comes up a 3 and the red comes up a 2. we hit because the roll (The green die) is a 3+, which hits.

Or same scenario the green die is a 1 and the red die is a 5, we hit because the roll missed but the re-roll was a 5.

Simultaneous rolling vs. rerolls have different statistical averages which is why you do one vs the other, and why there are so many different rules that either give you re-rolls (Fleet) or additional, simultaneous dice (Armourbane). I've yet to see where it says these two types of rolls are interchangeable.

Only if you are adding the two dice together to get a number.

If you roll 2 dice there are 36 possible outcomes. These being:

Spoiler:

As you can see the lighter die can be 1-6 and the darker die can be 1-6, with 36 possible different combinations no two repeated.

Not to burst anyone's bubble, and I realize it seems like a friendly way to speed up a game (I thought about it for a bit myself) but I'd rank this as dice manipulation. It's more subtle than loaded dice as statistical math is more difficult to see for most people, but the same principle applies.

It really isn't dice manipulation because dice do not have memory and you are not adding the result of the two dice together. (See the chart, each die has a 1/6 chance of showing any given number).

This only works if you are rolling one TL and it's reroll at a time. Also, if there is a special ability activated from a specific number rolled, or it is something like fleet, I hope its obvious why you don't wanna let someone do that (as mentioned by the OP). As pretty as that chart is (I'm digging it's feng shui), it is irreverent and you seem to have missed the point.

As soon as you roll two or more TL simultaneous w/their rerolls, you don't know which hits were rolled on the first roll, so you don't actually know how many rerolls you were legitimately entitled to. you cannot tell me which reroll is for which original roll so you start skewing the results. You essentially give yourself more "rerolls" than you should have had (see my epic math post).
Colored dice don't help either. Say blue was first roll, red=rerolls. Technically, for every blue hit, that is one less red die you should be rolling. Tell me, which red dice you rolled now don't count? the hits? The misses? How many of each?
Again, Now that you know the numbers behind the why, you don't want to continue doing it. That is manipulating the dice to give you a higher percentage of hits than you should have. That is called cheating.

So as i posted earlier, and i don't think anyone is arguing that this is a correct way of doing it, you can't roll 3 (or any number of) red dice (original) and 3 (or same as before number of) black dice (re-rolls) and pick from red and for any red misses pick from black. You can roll 2 red dice, 2 black dice, and 2 blue dice and pick out one hit (if any) from each color for a total of 3 shots (add more color dice for more shots) or just roll one shot at a time.

Also as i said that its really only fast dice for single shot Twin linked stuff because with multiple shots per gun or lots of guns firing it gets out of hand and becomes slower.

 rookshunter wrote:
With simultaneous rolling, how do you know that second hit wasn't the re-roll of the first hit you rolled?

It doesn't matter, because twin-linking only re-rolls misses. So long as at least one of the dice scores a hit, it makes absolutely no difference to the outcome whether the hit was on the first roll or the re-roll. Either way the end result is a hit.

DR _ rooks was using a situation where they were firing TWO TL Heavy 1 weapons using 4 dice

You cannot use 4 single coloured dice for this, as if you end up with 2 hits, 2 misses, you dont know if the 2 hits were from the one weapon, the 2 misses from the other - i.e. you got:

BL One: 5, 6 (2 "hits" from one shot)
BL Two: 3, 2 (2 "misses" from one shot)

As you have no way of knowing which is which. Its just easiest not to try to fast dice roll in this instance.

nosferatu1001 wrote:
DR _ rooks was using a situation where they were firing TWO TL Heavy 1 weapons using 4 dice

You cannot use 4 single coloured dice for this, as if you end up with 2 hits, 2 misses, you dont know if the 2 hits were from the one weapon, the 2 misses from the other - i.e. you got:

BL One: 5, 6 (2 "hits" from one shot)
BL Two: 3, 2 (2 "misses" from one shot)

As you have no way of knowing which is which. Its just easiest not to try to fast dice roll in this instance.

Correct. I believe I erred on my first post by not pointing out it was more than one TL rolled at a time, but the other posts i mentioned it (thanks KMK and nos for catching my intent).
Still, if I fire TL scatter lasers IMO its so much easier to roll 4 dice, then reroll misses, than to roll 4 pairs of different colored dice. Worse yet, cant imagine trying that with a bunch of jetbike TL Suriken Catapults. Too many colors!


Automatically Appended Next Post:

 Peregrine wrote:

 rookshunter wrote:
RESULTS (as long as my math is right):

Either your math is wrong, your random number generator isn't sufficiently random, or you've made a typo somewhere in implementing your math. There is absolutely no difference between rolling a die and then re-rolling it if it fails vs. rolling two dice and calling it a success if one die succeeds. They're exactly equivalent because you always re-roll the failure so it doesn't matter when you roll the second die, you aren't gaining any new information from it.

The issue with fleet re-rolls is that you aren't required to re-roll the die, you have to choose whether to keep the first result or to re-roll it and try for a better one. If you roll two dice at once and take the highest you're giving yourself access to knowledge of what the result of the re-roll will be before committing to throwing away the first result.

In this case, I think, my math is correct. Read more carefully next time. I didn't say one TL. The scenario was rolling two TL Brightlances. That situation makes all the difference in the world for reroll vs simultaneous.
I might have avoided this confusion by being more clear on that point. Conversely, as you and others have pointed out, a single TL shot result was already discussed and obvious, so I shouldn't have to point that out. Then again, posting at 4 am w/little sleep isn't the smartest thing I've done lately either.


Automatically Appended Next Post:

 DeathReaper wrote:

It is screwed up.

Check the picture I posted, and realize that a single die can roll anywhere from 1-6 inclusive.

2 dice will give a possible 36 results without any repeating thus the chance of success are the same if you roll a die and re-roll it, or roll two dice and see if one is a hit.

The die rolls are Independent Events, the outcome of the first roll has no bearing on the second roll, ergo the probability of a hit coming up on the first die (For your example needing a 5+ to hit) is 33.33% and the probability of a hit coming up on the re-roll is 33.33%

Here is the explanation for independent events and re-rolls.

Basics of Mathhammer wrote:

Spoiler:

Re-rolls

The last basic concept is that of re-rolls. If you think about it, what a re-roll does is allow you to try again, if the original event was not to your liking. This is really just two mutually exclusive (as it cannot both happen and not happen) events, so you're using the sum rule from above. It's also the combination of two independent events (the first roll failed, and the second roll succeeded). That means that you're combining the odds that the event happens (p) with the odds that (it didn't happen (1-p) and that the re-roll did happen (p)). The odds on the second roll are no different than the odds on the first roll, so that simplifies things, and the formula at the end is p + ((1-p)*p).

Another way to approach this problem is by determining your chance of failure, then subtracting that from 1 to get your chance of success. So the equation is 1 - (1 - p)^2, since the only way to fail is for each roll to fail, and each roll has a 1 - p chance of failing.

Example: Let's say you want to know the odds of hitting with a twin-linked BS3 gun. BS3 means you're hitting on a 4+, so the basic odds (p) are 1/2. Therefore, the odds of hitting with the re-roll are:

1/2 + ((1-1/2)*1/2)
1/2 + (1/2 * 1/2)
1/2+1/4
3/4 (or 75%)

And yes, I stated for a TL Lascannon, which has only 1 shot, so I was talking specifically about rolling one TL and it's reroll at a time.

Its not screwed up, I believe you read it wrong. I stated two TL Brightlance shots in my scenario. There is a big difference from that and a single TL roll. Also, The dice on a reroll are not added up ever on a TL shot, so I'm not seeing the relavance of your 36 results statement. Your probability equations look correct though (from what I know).


Automatically Appended Next Post:

 insaniak wrote:

 rookshunter wrote:
With simultaneous rolling, how do you know that second hit wasn't the re-roll of the first hit you rolled?

It doesn't matter, because twin-linking only re-rolls misses. So long as at least one of the dice scores a hit, it makes absolutely no difference to the outcome whether the hit was on the first roll or the re-roll. Either way the end result is a hit.

Again, with respect to everyone, my example was two TL Brightlance shots, not one TL shot. Yes, there is a difference. A big one as many have pointed out.

So basicly only roll 2 dice together if you're firing a single TL shot or if you have enough uniquely colored dice to represent each shot seperately... you know what, I'm just going roll all my dice once and reroll misses, lol.

I roll two dice for TL all the time, but only for weapons that are one shot anyways, like a TL fusion blaster.

As soon as you get into multiple rerollable shots, the effort you have to make to ensure that you keep the order of shots straight, but also which is the original and which is the reroll means it's not worth the effort.