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September 27[edit]

Dirichlet eta function[edit]

Is there inverse function for eta function. Which gives value of s. — Preceding unsigned comment added by 151.236.160.86 (talk) 08:45, 27 September 2015 (UTC)[reply]

No, the eta function is not one-to-one. Sławomir
Biały 14:04, 27 September 2015 (UTC)[reply]

Wow, déjà vu. --RDBury (talk) 18:45, 27 September 2015 (UTC)[reply]

Medians are to Symmedians what Heights are to ... ?[edit]

The reflections of medians with regard to angle bisectors are called symmedians, and all three are concurrent. Do the three concurrent reflections of heights with regard to angle bisectors also bear a special name ? — 86.125.198.73 (talk) 14:26, 27 September 2015 (UTC)[reply]

They meet in the circumcenter, the latter being the isogonal conjugate of the orthocenter. — 86.125.198.73 (talk) 15:05, 27 September 2015 (UTC)[reply]

Apparently the term symaltitude seems to have been used by some. — 86.125.198.73 (talk) 18:20, 27 September 2015 (UTC)[reply]

Is this kind of statistical problem, able to be solved?[edit]

Imagine we have a rpg where you roll ((4d100)/2) -1 to find some stat like strenght, and females get a -14 modifier, that means you roll the dice and do the math and then remove -14 from it, this means that how entire world human population strenght is distributed can be seen by looking at this formula.
Now imagine another rpg that you roll 3d6 to find the stats.
Anyway, is there a way to convert the -14 to the 3d6 distribution?201.79.91.249 (talk) 17:22, 27 September 2015 (UTC)[reply]

I hope that I'm not alone in not having the faintest idea what this is asking.→86.138.241.98 (talk) 18:25, 27 September 2015 (UTC)[reply]

(edit conflict)For those of us who have never been a dungeon master, the notation 4d100 means roll a 100 sided die 4 times and add the results. I'm thinking the way to approach this is to convert the -14 into standard deviations (s.d.), then convert this number of s.d. to points for the other stat. The variance of 100 sided die would be ~~816.7~~833.25, so for 4 die it would be 4 times that and when you divide by 2 the variance is divided by 4, so the variance of he stat is again ~~816.7~~833.25. The s.d. of the stat is then ~~√816.7 = 28.6~~√833.25 = 28.9, so -14 is about -1/2 of a s.d. For the second stat, the s.d. for a six sided die is 2.92 and for 3 die it's 8.75. The s.d. of the second stat is then 2.96. So the equivalent to -1/2 s.d in points for the second stat is about -1.5. Not sure what to do about the half point since you probably want stats which are whole numbers; flip a coin maybe. --RDBury (talk) 18:40, 27 September 2015 (UTC)[reply]

The variance of a 100-sided die is 833.25, right? Bo Jacoby (talk) 04:43, 28 September 2015 (UTC).[reply]

Looks like it:

E[X^{2}]-E[X]^{2}={\dfrac {1}{100}}\sum _{k=1}^{100}k^{2}-\left({\dfrac {1}{100}}\sum _{k=1}^{100}k\right)^{2}=833.25

. --Kinu ^t/_c 07:31, 28 September 2015 (UTC)[reply]

Yep, I misread the formula. It's in Variance#Fair die. --RDBury (talk) 11:19, 28 September 2015 (UTC)[reply]

The sd of 3d6 is ${\sqrt {3}}$ times the sd of 1d6, not 3 times. Also, it seems you've made an error in calculating the sd of 1d6 as well - it would be impossible for the sd to be 2.92 when the deviation cannot be more than 2.5. The correct sd of 3d6 is ${\sqrt {3}}\cdot {\sqrt {35/12}}\approx {\sqrt {3}}\cdot 1.70783\approx 2.95804.$ -- Meni Rosenfeld (talk) 18:05, 28 September 2015 (UTC)[reply]

In the sentence "the s.d. for a six sided die is 2.92 and for 3 die it's 8.75", should be "variance" instead of "s.d.". -- Meni Rosenfeld (talk) 18:21, 28 September 2015 (UTC)[reply]

If you are interested in the mathematics (rather than just getting an answer on a silver platter). You can read this website. [[[1]] 175.45.116.61 (talk) 05:58, 28 September 2015 (UTC)[reply]

The expression

{\dfrac {1}{100}}\sum _{k=1}^{100}k^{2}-\left({\dfrac {1}{100}}\sum _{k=1}^{100}k\right)^{2}

could be interpreted to mean

{\dfrac {1}{100}}\sum _{k=1}^{100}\left(k^{2}-\left({\dfrac {1}{100}}\sum _{k=1}^{100}k\right)^{2}\right)

.

It is safer to write

\left({\dfrac {1}{100}}\sum _{k=1}^{100}k^{2}\right)-\left({\dfrac {1}{100}}\sum _{k=1}^{100}k\right)^{2}

Bo Jacoby (talk) 13:57, 28 September 2015 (UTC).[reply]

Perhaps, but the context makes the latter a more plausible interpretation. --Kinu ^t/_c 17:14, 28 September 2015 (UTC)[reply]

The "alternative interpretation" uses the same variable in two nested quantifications, so it is extremely awkward at best, meaningless at worst. There's no real risk of confusion. -- Meni Rosenfeld (talk) 18:05, 28 September 2015 (UTC)[reply]

The expression

{\dfrac {1}{100}}\sum _{k=1}^{100}\left(k^{2}-\left({\dfrac {1}{100}}\sum _{k=1}^{100}k\right)^{2}\right)={\dfrac {1}{100}}\sum _{k=1}^{100}\left(k^{2}-50.5^{2}\right)=833.25

is perfectly well defined, and is a perfectly plausible interpretation. Bo Jacoby (talk) 06:05, 29 September 2015 (UTC).[reply]

There may be weird syntaxes which allow it, but basically, you shouldn't use k as a summation value inside a scope which already assigns meaning to k. You'll have two different variables, both called k, in the same expression, and you can't know which one is being referred to. (You can use the convention that it always refers to the innermost scope, but that's just a sure recipe for confusion and disaster).

Curiously, the two expressions actually have the same value, since putting the second term inside the sum simply sums it as a constant 100 times and then divides by 100. -- Meni Rosenfeld (talk) 14:04, 29 September 2015 (UTC)[reply]

It is a special case of E[(X–E[X])^2] = E[X^2]–E[X]^2 = E[X^2–E[X]^2]. Bo Jacoby (talk) 03:30, 1 October 2015 (UTC).[reply]

If I understand correctly, you want to apply some transformation on a strength (not "strenght" as you wrote) attribute obtained with the first method, so that it would be on the same scale as an attribute obtained from the second method.

Following RDBury's calculations, ((4d100)/2)-1 has a mean of 100 and sd of 28.8661, and 3d6 has a mean of 10.5 and sd of 2.95804. So if you take the original value, multiply it by 0.28661/2.95804 = 0.102475, and then add 0.2525, you'll get a value with mean and variance equal to the 3d6 distribution. So for example, 120 from the first method will be equivalent to 12.55 on the second method.

Alternatively, you can simply divide the value by 10 and get a very good approximation (and then add 0.5 for an even better one). In fact, I suspect that the reason ((4d100)/2)-1 was chosen is that it matches the distribution of the original 3d6, just with a factor of 10 which allows a finer resolution.

Now, if there is a modifier of -14, you simply multiply it by the same amount (0.102475, or approximately, divide by 10). So on the 3d6 scale you need to subtract 1.4 for females. If you don't want fractional values, you can roll a 1d10 and subtract 2 if you get 1-4, and 1 if you get 5-10. Or toss a coin and choose 1 or 2 based on the result. -- Meni Rosenfeld (talk) 18:05, 28 September 2015 (UTC)[reply]

Intuitively (admittedly, the intuition of a long-time roleplayer), the curves are going to be somewhat different. Larger numbers of dice generally produce a "smoother" curve, i.e. a better approximation to a bell curve, so 4d100 is going to give a different distribution of results to 3d6 no matter how carefully you scale it. [anydice.com] can be used to plot these probabilities (with all divisions rounded down). Unfortunately, it only takes whole numbers in formulae, but this gives a good comparison (from Meni's simpler formula above):

   output 3d6
   output (4d100+8)/20

(4d100+8)/20 = ((4d100/2)-1)/10 + 0.5. The biggest difference is that values of 0,1,2,19 and 20 are possible in the 4d100 distribution, which are impossible in the 3d6 distribution. MChesterMC (talk) 16:05, 29 September 2015 (UTC)[reply]

Right. There are two factors at play here:

Using d100 instead of d6. This allows a finer resolution, so you can have a distinction between, say, 103 and 100, which is impossible in the scaled down version.
Using 4d instead of 3d. The more dice you add up, the better the approximation to the normal curve. This can most easily be seen by comparing the kurtosis - 3d6 has 2.57714, 4d100 has 2.69994, the normal distribution has 3 (The kurtosis of 3d∞ and 4d∞ is 2.6 and 2.7). As you note, this won't change by a linear transformation.

However, the mean and variance are the most important attributes of a random variable, and by just matching these you will get a result which is difficult to distinguish in practice - especially when the difference in kurtosis is so small (3 dice already get you fairly close to normal). -- Meni Rosenfeld (talk) 18:36, 29 September 2015 (UTC)[reply]

PS. I suspect that your intuition about the difference between few and many dice, is based mostly on the difference in variance. 1d10+2 has a much higher variance than 3d4, and that's easy to notice. But if you linearly transform to match variance, and you use 3+ dice which are already close to normal, I'll be surprised (and impressed) if you can actually notice the remaining differences in the distribution. -- Meni Rosenfeld (talk) 09:21, 30 September 2015 (UTC)[reply]

There's certainly some difference, going back to anydice.com, and inputting "output ndX/n" for constant x and various values of N, then going to the "graph" tab shows the changes. You probably can't tell the difference between 4d6/4 and 5d6/5, but you probably can between 4d6/4 and 20d6/20 (the former having about a 40% chance of a 3 (all results rounded down), the latter having about an 81% chance). MChesterMC (talk) 16:04, 30 September 2015 (UTC)[reply]

4d6/4 and 20d6/20 have very different variances (35/48 and 7/48). Variance is linear in the number of dice but quadratic in multiplication factors, so in general the variance of xd6/x will be inversely proportional to x. As I explained, that's the main cause of the difference you're seeing. If you want to roll 20d6 and scale it to similar mean and variance as 4d6/4, the correct transformation isn't 20d6/20, it's 20d6/8.94-4.326. Anydice won't allow decimal places, but if you try 20d6/9-17/4 you'll get results which are pretty close. To visually see the similarity in the graph, though, you'll have to note the change in the x-axis displayed, and focus on the range 1-6. And of course, the fact that we have to round down (and the probabilities depend on how many original results map to a given value) causes some disruption (since the transformation was computed with the assumption that we don't round). -- Meni Rosenfeld (talk) 17:19, 30 September 2015 (UTC)[reply]

Thanks - as a side note, inputting both dice rolls at once, and clicking the "graph" view, shows the two graphs overlaid on each other for easy comparison. — Preceding unsigned comment added by MChesterMC (talk • contribs) 08:09, 1 October 2015 (UTC)[reply]

Thanks, I finally understood that you meant we should include both output lines in the same page.

Playing around with it a bit more, and taking the rounding into account, better ways to match 3d6 with 4d100 are either (4d100+17)/20 or (5*4d100+67)/98.

As for 4d6/4 and which we've also discussed, if you want to match it with 20d6, you'll get an impressively good fit with (20d6-38)/9. When there's rounding involved, you get smoother results in cases that matching the variance just requires dividing by an integer. -- Meni Rosenfeld (talk) 09:04, 1 October 2015 (UTC)[reply]

PS. If you compare 4d6/4 and (20d6-38)/9, you may note that the latter can take values of 0 and even negative. Even if these values are meaningless in the game context, it's really less of an issue than it sounds. You can simply specify a rule that if you get results outside the legal range, you roll again. Since the probability of this happening is so low (1 in a 1000 in this case), you will probably never have to worry about it, and it won't change the statistical properties of the roll. -- Meni Rosenfeld (talk) 09:12, 1 October 2015 (UTC)[reply]

Thanks for all the info (and continue trying to find a better answer), anyway this gave me some eureka moment, a digital song is made of values (samples, that is, interger values that are positive, negative and 0), you could get a song A and then song B, and calculate the average and standard deviation of stuff and etc..., and then get one of the songs, get their samples and and convert to other song average and standard deviation, maybe would sounds cool. — Preceding unsigned comment added by 201.79.73.30 (talk) 14:12, 2 October 2015 (UTC)[reply]