A Few Questions/Observations From an Old Player

[Timo Pietilä]

Once we have the number of standard deviations away from the average we are, we can use that value plus knowledge of the distribution to determine how unlikely it is that an AMHD would have that many hitpoints.

OK, and how likely that is? How do I use that info about "standard deviation" to calculate that? To me that is entirely pointless number without any real use.

I don't feel like going back several decades to learn something like this again.

[Derakon]

As I said, the odds of a monster being two standard deviations above the mean are about 2.5%. The actual math is something I don't personally remember off the top of my head.

[half]

For some reason I thought there were 10 stars of health, not 5.

There are 10 stars, but you can sometimes knock off one with 19% of the monster's hp. It is because you can't knock off 1.9 stars -- it is a rounding issue.

[Patashu]

OK, and how likely that is? How do I use that info about "standard deviation" to calculate that? To me that is entirely pointless number without any real use.

I don't feel like going back several decades to learn something like this again.

If you know the mean and standard variation of your normal distribution, then
1) calculate how far from the mean the target value is
2) divide by the distribution's standard deviation, to get the number of SDs away from the mean the target value is
3) Use this image: http://www.cs.bilkent.edu.tr/~korpe/...2/z-table.jpeg to determine the probability of being between the mean and that number of SDs (which you can multiply by 2, invert, etc. to get the other probabilities, like the probability of being higher than that number of SDs, or the probabily of being higher than that number of SDs in both directions)

This works for any distribution as long as the distribution is similar to a normal distribution (= has a bulge in the middle). If it does not (for example a completely flat distribution, or an exponential/logarithmic distribution), then you need different math.

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OK, but how do you calculate standard deviation?

First, calculate variance:

Var = E[X^2] - E[X]^2
where E[something] is the expected (mean) value of something. so E[X^2] is the mean of the squares, while E[X]^2 is the mean, squared.
Standard deviation is then Sqrt(Var).

In addition, variance has a useful property: Var(X+Y) = Var(X) + Var(Y). So if you need to figure out the standard deviation of 2d4, you can figure out the variance of 1d4, double it then sqrt it to get the standard deviation.

[Timo Pietilä]

As I said, the odds of a monster being two standard deviations above the mean are about 2.5%. The actual math is something I don't personally remember off the top of my head.

And this "in our example we use an SD of 1/8th of the monster's average HP" is what determines the "standard deviation" in reality and not just as example (in case of angband)?

[Timo Pietilä]

If you know the mean and standard variation of your normal distribution, then

...you know already a lot more than just deviation. Which makes my point, standard deviation without additional info is just useless number.

[LostTemplar]

Standard deviation approximately is average value over square root of number of dice in case of NdM type distributions. It is more accurate for larger N. E.g. 100d100 have mean of 5050 and deviation of 505, so chance that actual value will be greater then 6060 is about 2.5%.

[Timo Pietilä]

Standard deviation approximately is average value over square root of number of dice in case of NdM type distributions. It is more accurate for larger N. E.g. 100d100 have mean of 5050 and deviation of 505, so chance that actual value will be greater then 6060 is about 2.5%.

How did you get 2.5% out of that and values of 505, 5050 and 6060? My question is how do I use that info in practical way. Do I have to look it from a table Patashu posted, or is there a easy way to use it in calculations? I hope answer doesn't include integrals or laplaces.

[LostTemplar]

2.5% is from Derakon's post. Probability to be above two standard deviations. There is just a table or graph of pre-calculated values of normal distribution.

[Timo Pietilä]

2.5% is from Derakon's post. Probability to be above two standard deviations. There is just a table or graph of pre-calculated values of normal distribution.

Frak, I was hoping to get something useful from that. Apparently it isn't useful number for any of the people that doesn't have that table.

[Derakon]

To use the standard deviation, you need to also know the mean and the distribution. Without those, it's useless. The distribution is the shape of the curve, so the tables we've been referring to are also useless unless you happen to be working with a "normal distribution", as we have been here.

For example, the distribution of a 1d20 roll would be very flat, since every value has an equal chance of occurring. So even though the mean is 10.5 and the standard deviation is (according to the Net; don't feel like doing the math) 5.77, that doesn't mean that you have lower odds of getting a 20 than a 11.

[Oramin]

So, did somebody have an answer to the Til-I-Arc question?

[Gorbad]

Frak, I was hoping to get something useful from that. Apparently it isn't useful number for any of the people that doesn't have that table.

There is a thing called the Empirical Rule, which is exactly for people that don't want a table.

Some prerequisite info:

A 'normal distribution' is a standard bell curve, so most of the results will be in the middle. Like was said earlier, imagine 2d6, the odds for 2 and 12 are much lower than for 7 to occur, this is a bell curve, or normal distribution.

Standard deviation is a way to calculate how much your entire set of rolls differs from the mean (the average, i.e. 7 for 2d6). To do that requires a simple formula, but well, wikipedia is your friend. basically you find how much each separate sample differs from the mean, square that difference, sum all those, divide by the number of samples and take the square root. What it does is make a meaningful 'average' of the differences from the mean. For 2d6 this is about 2.41.

Anyway... the Empirical Rule (a great rule of thumb) is that in a normal distribution about 68% of all results if you sample ("roll") often enough will be within one standard deviation of the mean, so between 4.6 and 9.4 (7 plus or minus 2.41), 95% will be within 2.18 and 11.8 (twice the standard deviation), and 99% will be within plus or minus 3 times the standard deviation.

[fizzix]

So, did somebody have an answer to the Til-I-Arc question?

Vulnerabilities are not applied to melee damage. Whether they should be is debatable. When the game decides a message it goes through all possible slays and only picks a new one if it's more damage than whatever earlier one it found. Fire is the first slay it checks, and since it applies and nothing is greater, it picks that.

[Oramin]

Ah, good to know that about the vulnerabilities. Thanks.