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When calculating the number of slashing attacks for a screamer unit, do you roll a D3 then multiply by the number of screamers, or roll a D3 for each screamer then add them up?

Roll a D3 for each screamer.

Statistically its the same either way, just with rolling for each seperately you achieve the median more often.

Eihnlazer,
I find it hard to picture the statistics being identical for the two methods discussed here, even without crunching the numbers. Given that Statistics are based entirely on probability and correlating points of data, the fact that one method produces three results while the other produce such a wide spread of results would be enough to make it improbably that the statistics are the same. Rolling and multiplying leaves one with just three possible outcomes, there is a 0% chance to end up with any other number then either the maximum result, the minimal result or dead center between the two. Rolling individually and adding them together means one could end up with any number between the minimal result and the maximum result.

One is statistically more likely to roll a 1 on a D3 then to roll a 10 on a 10D3.

JinxDragon wrote:
Eihnlazer,
I find it hard to picture the statistics being identical for the two methods discussed here, even without crunching the numbers. Given that Statistics are based entirely on probability and correlating points of data, the fact that one method produces three results while the other produce such a wide spread of results would be enough to make it improbably that the statistics are the same. Rolling and multiplying leaves one with just three possible outcomes, there is a 0% chance to end up with any other number then either the maximum result, the minimal result or dead center between the two. Rolling individually and adding them together means one could end up with any number between the minimal result and the maximum result.

One is statistically more likely to roll a 1 on a D3 then to roll a 10 on a 10D3.

you clearly don't understand how math works then

Then educate me:
From my understanding statistics are based on probabilities of an event occurring, calculated by looking at a collection of data and plotting trends.

How can something which produces only three outcomes be statistical 'the same' as something which produces an increasingly greater number of outcomes the larger X is?

In a nutshell it would be:

Probability of getting a 1 on a D3 = 1/3

Probability of getting 10 1's on rolling 10 D3 = 1/3 x 2/3 x 2/3 x2/3 x 2/3 x 2/3 x 2/3 x 2/3 x 2/3 x 2/3

I think that is correct, it's been a long time since I did any pure stats.



Automatically Appended Next Post:
As for the OP question there is a rule that says that when shooting you roll once for variable shots and multiply by the number of shooters and you MIGHT be able to apply this to this case, but RAW you roll X many D3

How is 1/3 the same as 512/59049 ( 1/59049 ? )

Also, why did the probability of rolling a 1 on a D3 increase to 2/3 for the additional rolls?

JinxDragon wrote:
How is 1/3 the same as 512/59049 ( 1/59049 ? )

Also, why did the probability of rolling a 1 on a D3 increase to 2/3 for the additional rolls?

It isn't, I totally agree that 1D3 x 10 does not have the same probabilities as 10D3

If memory serves the way to calculate the probability of a combination like this one is the probability of the first event multiplied by the successive probabilities of the event NOT occurring.
Perhaps a more recent Stats GURU can confirm, I lost my guru mantle many years ago.

Then why did someone state it was statistically the same?
Then someone accuse me of not knowing math when I stated I couldn't see the probabilities of being anywhere close to the same?
Which I thought was important to statistics?

Though it wasn't a bad accusation, I have had a computer system available to me for decades now and I have not felt the desire to calculate something more complicated then character sheets without....

JinxDragon wrote:
Then why did someone state it was statistically the same and someone accuse me of not knowing math when I stated I couldn't see the probabilities of being anywhere close to the same?

I have no idea.

It has the same average, thats what you're missing

Explain further.

The average output is the only important part because that's the average amount of damage they do. Both may do it differently but both come out to the same average (of 2 per d3).

You may be right that they have different probability but its like saying 1+1+1 and 3x1 are different. They technically are but it doesn't matter because they both equal the same thing.

I have some concerns over this, but 2 am in the morning is not best to be trying to formulate a way to describe those concerns.

If there are 9 screamers and you roll 1 D3 the you have an equal chance of rolling a 1 2 or 3. So that means the the probabilities of getting 9 18 or 27 attacks is the same, ie 1/3

If however you roll a d3 for each the chances of getting 27 attacks is much less, 1/3^9

Well without getting too bogged down in stats and probability:

If you had 9 screamers and rolled 1D3 you'd end up with only 3 possible results; 9, 18 or 27 attacks every time.

If you rolled separately and assume an average result of 2 attacks per screamer you'll end up closer to the median (i.e. 18 attacks) every time.

Doing it the second way basically makes it much less likely to get the maximum 3 attacks per screamer but also much less likely to get the minimum. But which is the right way to do it?

Isn't that what I just said?

The correct way is to roll them individually as there is no direction other than they get D3 attacks each.

Using Exorcists myself that fire D6 Shots, i would agree that rolling 1D6 for all of them and 3D6 for 3 yield very different results.
WH 40k was based upon dice rolls and their randomness.

Those who say "maths are the same":
I'll roll 1D6 at the beginning of my game, and that's my roll To Hit for the entire game. Statistically, it would be the exact same, i agree. In relevance to the game? i just don't think it would be too fun...

 CrownAxe wrote:
The average output is the only important part because that's the average amount of damage they do. Both may do it differently but both come out to the same average (of 2 per d3).

Generally in a dice game a winning strategy is to minimise the random effect of the dice - throwing more dice here gives the same average result, but is statistically way more likely to result in that average result of 18.*

Lootas would be "better" if you rolled a D3 per model, as they would be more reliable, meaning your planning in the game is made easier.

Only if you do very basic math hammer would you ever say the two methods were equal just because they shared a median point. Your expected result (and expectation >>>>> basic probability trees for strategy purposes) is much more consistent with the way screamers work, i.e. rolling a D3 for each

(Chance of 10 "1"s on a D3 is 1/3x1/3x......x1/3 = 1/(3^10), as you go down the "1" branch of the tree 10 times, and the probability at each point of rolling a 1 is 1/3rd)

*Binomial approximation to the normal; the famous bell curve is formed from many discrete binomial events. Rolling just 1D3 instead results in a flat "table", as all results are equally likely

 CrownAxe wrote:


You may be right that they have different probability but its like saying 1+1+1 and 3x1 are different. They technically are but it doesn't matter because they both equal the same thing.

You are right in that the two methods have the same average. They are not the same though.

When rolling 10 dice how high is your chance to roll 30 slash attacks?
When rolling 1 dice how high is your chance to roll 30 slash attacks?

If these two propabilities arent equal how can you say the methods results are the same?

You cut out a lot of outcomes out of the bell curve.

The standard deviation of both methods is different.

So no. You cant roll 1d3 for all if the rule tells you otherwise. Not from a rules standpoint and not from a statistics standpoint.

I never said they are the same, im saying that the only part that matters is their average output which is the same

 CrownAxe wrote:
I never said they are the same, im saying that the only part that matters is their average output which is the same

I was stating that your "only part that matters" is wrong, statistically and strategically.

What matters in a dice game is controlling for randomness. You cannot control a single roll, but when you have many rolls you can get a good idea of your expected result.

With your Method 2 of the outcomes with the lowest probability of Happening suddenly have the Same chance to happen as the statistical average. Your idea of only the average is what counts is blatantly wrong

nosferatu1001 wrote:

 CrownAxe wrote:
I never said they are the same, im saying that the only part that matters is their average output which is the same

I was stating that your "only part that matters" is wrong, statistically and strategically.

What matters in a dice game is controlling for randomness. You cannot control a single roll, but when you have many rolls you can get a good idea of your expected result.

You don't do screamer's slashing attack only once in a single game and be done with it. You use it several times over several games. And over several instances it averages out to be the same as either method

So "Sorry mate it was OP as a get out this game but next game it will even out so I'll take the 1/3 chance of 27 attacks" ?

Talk about what goes on on the table mate not the long term project forecast of the combat investment deficit as reduced to seasonally adjusted norms.

Uptopdownunder wrote:
So "Sorry mate it was OP as a get out this game but next game it will even out so I'll take the 1/3 chance of 27 attacks" ?

Talk about what goes on on the table mate not the long term project forecast of the combat investment deficit as reduced to seasonally adjusted norms.

Exalted

You cannot claim it is right because the average is the same over 25 games...

JinxDragon wrote:
Then why did someone state it was statistically the same?
Then someone accuse me of not knowing math when I stated I couldn't see the probabilities of being anywhere close to the same?
Which I thought was important to statistics?

First off the only thing statistics has in common with math is that it uses numbers and it is calculated. IMHO

For what my reply is worth I too lost my Stat guru status many moons ago. I use only a small portion of it for work.

Law of Averages: The law of averages is an erroneous generalization of the law of large numbers, which states that the frequencies of events with the same likelihood of occurrence even out, given enough trials or instances. The law of averages is usually mentioned in reference to situations without enough outcomes to bring the law of large numbers into effect.

If I roll 1D3 and apply that result 10 times, and do it enough (typically 500-1000 trials, margin of error = 1/sqroot x, x being the number of trials) I will get an arverage total of aprox. 20 (plus or minus a variable amount) depending on number of trials. The more trials the tighter the variable amount. 1/3 of the results rolled will be a one.

If I roll 10D3 enough times I should come up with 20 (plus or minus a variable amount) faster typically 100-200 trials (cant remember which error test allows for this, it might be a power test) again the more trials the tighter the variable amount. this is where the probablity of 512/59049 or 1/115 apx comes into play you would potentially need to roll 115 times to get 10 ones in one roll.

Outliers: results of 10 and 30 are rare in example number 2, however are common in example 1. This is why example one is never used by most games and why more trials are required to get a average.

 CrownAxe wrote:

nosferatu1001 wrote:

 CrownAxe wrote:
I never said they are the same, im saying that the only part that matters is their average output which is the same

I was stating that your "only part that matters" is wrong, statistically and strategically.

What matters in a dice game is controlling for randomness. You cannot control a single roll, but when you have many rolls you can get a good idea of your expected result.

You don't do screamer's slashing attack only once in a single game and be done with it. You use it several times over several games. And over several instances it averages out to be the same as either method

6 turns is still neglible, and with the actual method (1D3 per screamer) this gets even more reliable.

Not only does your method not follow the rules, it isnt remotely "the same" in the only part that matters, as your definition of what "matters" is incorrect, by any objective standard.

An average however is a pretty poor reflection on the experiment.

If I have a stone that weighs 1kg and a Meteorite that weighs 1 000 000 kg the average weight of the two objects is 500 000.5 kg.

It's a pretty meaningless number as in small sets the average is not any sort of representation of any particular member of the sample.

Die rolling and 40k is all about probabilities not averages.