Meanwhile Bisco is 18+ Ex

Back to general discussion guys.

Unless bisco's is bugged it works like every other source of IIQ.

You guys are aware the IIQ belt, boots, gloves, other amulets, rings and chests exist right.

Bisco's collar is like 60c in standard (around the correct price compared to all other MF Items) and 18ex+ in ladder.

I'm pretty sure the extra 15ex comes from ladder/streamer hype and extreme misunderstanding of people's options.

That and everyone is flipping the shit out of it.
Last edited by RagnarokChu on Jan 11, 2018 4:45:06 PM
"
Unless bisco's is bugged it works like every other source of IIQ.

You guys are aware the IIQ belt, boots, gloves, other amulets, rings and chests exist right.

Bisco's collar is like 60c in standard (around the correct price compared to all other MF Items) and 18ex+ in ladder.

I'm pretty sure the extra 15ex comes from ladder/streamer hype and extreme misunderstanding of people's options.

That and everyone is flipping the shit out of it.


Ye, I'm sure it has nothing to do with bisco's being common for 2 leagues and all that being dumped into standard. It must be the flipperzzzz. :D
"
Unless bisco's is bugged it works like every other source of IIQ.




This in fact is the question. If the wording on Bisco's is correct, it is not
working in the same way. The player's quantity (usually referred to as "Quantity of items found") is not the same as "Quantity of items dropped by slain enemies". The wording indicates the same mechanical difference as e.g. "15% increased fire damage" (for you, the player) and "15% increased fire damage taken" (by the mob). Such a mob-related quantity modifier did not exist in 0.912c, when maps were made to be not affected by the "increased quantity of items found" stat.

Whether or not the "increased quantity of items dropped by slain enemies" stat affects mapdrops is something wich nobody knows atm.
"I have to proof my rights to this other twelve-year-old!" RaizQT, 11.11.16
"
Completed 29 ChallengesMiská wrote:
"
Unless bisco's is bugged it works like every other source of IIQ.

You guys are aware the IIQ belt, boots, gloves, other amulets, rings and chests exist right.

Bisco's collar is like 60c in standard (around the correct price compared to all other MF Items) and 18ex+ in ladder.

I'm pretty sure the extra 15ex comes from ladder/streamer hype and extreme misunderstanding of people's options.

That and everyone is flipping the shit out of it.


Ye, I'm sure it has nothing to do with bisco's being common for 2 leagues and all that being dumped into standard. It must be the flipperzzzz. :D

Like every single other item in the market being dumped into standard which where around longer then 2 leagues, that had equal or changes to item power or rarity :V?

In terms of outlier Bisco's collar has a higher inflation rate then legacy standard items, but still available to drop.
Last edited by RagnarokChu on Jan 11, 2018 5:32:31 PM
By the way, to all those people that wanted the math posted THERE ISN'T ENOUGH DATA IN THE VIDEO TO DO REAL STATS ON IT, BUT THE SAMPLE SIZE IS PLENTY LARGE ENOUGH, ALSO I DID SOME MATH:

You only need 30 samples if you want a basic picture of what drop rates are like. There is talk that this 'minimum' should be raised for serious scientific work, but you can get a fairly small confidence interval with just 30 datapoints.

There are two ways to run this experiment: by doing it with no other variables besides Biscos, where you clear 30+ white, unmodified maps, without any extra forms of item find or whatever,

or by acting to the best of your ability to raise item drop rates in general, like the streamer in the video did. Use all the sextants, alch every map, wear a bunch of item rarity, etc.



I'm gonna show you the basic, highschool level of stats that could be applied here.
We call this Confidence Interval for Two Independent Samples.

First, what numbers do we have?

He did 60 maps each time, and got 108 total from the control set and 160 from the biscos set.

The population, or n, of our data sets are 60 and 60, for the maps run with and without biscos. Call them n1 and n2.

The Means of out dataset are 108/60 and 160/60, or 1.8 and 2.666..., call THEM u1 and u2.


Problem though, we also need a standard deviation for both of these sets. Standard deviation is a sort of measurement of randomness. You would normally get this by comparing each datapoint to the mean of the dataset, sort of measuring how far each point is from the average.

Since we don't have that, I'm going to make up two datasets, one being super random (where every map got dropped in the same run) and the other super average (where rngesus is a merciful god and you got 1-3 each run). Both of these are extremes, and don't represent real drop rates.

Maths for Standard deviations spoilered
Spoiler

crazy dataset 59*0 1*108
Standard deviation control crazy
= sqrt((59(0-1.8)^2 + (108-1.8)^2)/60)
= sqrt((191.16 + 11278.44)/60)
= sqrt(11469.6/60)
= sqrt(191.16)
= 13.826062346163494716385436639116

average dataset 48*2 12*1:
Standard deviation control average
= sqrt((48(2-1.8)^2 + 12(1-1.8)^2)/60)
= sqrt((1.92 + 7.68)/60)
= sqrt(9.6/60)
= sqrt(0.16)
= 0.4

crazy dataset 59*0 1*160:
Standard deviation Biscos crazy
= sqrt((59(0-2.666...)^2 + (160-2.666...)^2)/60)
= sqrt((419.555... + 24751.111...)/60)
= sqrt(25170.666.../60)
= sqrt(419.5111...)
= 20.481970391324930120879734151542

average dataset 40*3 20*2:
Standard deviation Biscos average
= sqrt((40(2-2.666...)^2 + 20(1-2.666...)^2)/60)
= sqrt((17.777... + 55.555...)/60)
= sqrt(73.333.../60)
= sqrt(1.222...)
= 1.1055415967851332830383109122236


Now that we have our standard deviations, we need to pool them together, since we're comparing two sets of data. We call this the pooled estimate of the common standard deviation, and it means we assume that the two datasets have similar variance.

More maths
Spoiler


Sp = square root(((60-1)Sd1^2 + (60-1)Sd2^2)/(60+60+2))

Standard deviation pooled crazy:
= square root(((60-1)13.826062346163494716385436639116^2 +(60-1)20.481970391324930120879734151542^2)/(60+60+2))
= square root((59*191.15999999999999999999999999999 + 59*419.51111111111111111111111111112)/122)
= square root((11278.439999999999999999999999999 + 24751.155555555555555555555555556)/122)
= sqrt(36029.595555555555555555555555555/122)
= sqrt(300.24662962962962962962962962963)
= 17.327626197192436827317616612165

Standard deviation pooled average:
= square root(((60-1)0.4^2 +(60-1)1.1055415967851332830383109122236
^2)/(60+60+2))
= square root((59*0.16 + 59*1.222...)/122)
= square root((9.44 + 72.111...)/122)
= sqrt(81.551111111111111111111111111116/122)
= sqrt(0.66845173041894353369763205828784)
= 0.81758897400768776126962113154335



Alright, NOW we can do a confidence interval.

This amounts to taking the difference in the means, and comparing it to the pooled standard deviation modified by a T value and the populations.

The T value we just get from a table since this is high school maths. In this case we have 60 data points and want a 5% confidence, so we come up with 2.000, yay for easy math.

Maths the mathining
Spoiler

(mean1 - mean2) +/- t * Sp * sqrt(1/n1+1/n2)

Mean1-mean2 = 1.8-2.666 = -0.8666...

sqrt(1/n1+1/n2) = sqrt(1/60 + 1/60) = 0.1825741858350553711523232609336

Crazy interval:

-0.8666... +/- 2 * 17.327626197192436827317616612165 * 0.1825741858350553711523232609336
-0.8666... +/- 6.3271544908131715343934992178662

-7.1938211574798382010601658845329 to 5.4604878241465048677268325511995

Or: biscos gives between 546% fewer maps per run and 719% more maps per run.
Crazy dataset can't prove anything, biscos sucks.

average interval
-0.8666... +/- 2 * 0.81758897400768776126962113154335* 0.1825741858350553711523232609336
-0.8666... +/- 0.29854128255434368177662504462735

-1.165207949221010348443291711294 to -0.56812538411232298489004162203932

with the unnaturally average dataset, we can say with 95% confidence that biscos does have an effect on the drop rate of maps, giving between 116% and 56% more maps.



We don't actually have a standard deviation though, so none of this is actually more than an example.
The 'line' where we move from not confident to confident comes at a standard deviation of around 2.5, which is relatively low. If you have for or five runs with more than 5 maps that run, we probably won't be able to prove anything either way.
Last edited by Duodecimus on Jan 11, 2018 10:01:43 PM
i think its hit its max at 19-20ex.

gonna be falling now
Wait, people actually fell for the "Bisco affects map drops" meme? HAHA

"My favorite streamer said so so it must be true. He's so good at right-clicking his way through maps with his mouth open."
"
Completed 16 ChallengesDuodecimus wrote:
By the way, to all those people that wanted the math posted THERE ISN'T ENOUGH DATA IN THE VIDEO TO DO REAL STATS ON IT, BUT THE SAMPLE SIZE IS PLENTY LARGE ENOUGH, ALSO I DID SOME MATH:

You only need 30 samples if you want a basic picture of what drop rates are like. There is talk that this 'minimum' should be raised for serious scientific work, but you can get a fairly small confidence interval with just 30 datapoints.

There are two ways to run this experiment: by doing it with no other variables besides Biscos, where you clear 30+ white, unmodified maps, without any extra forms of item find or whatever,

or by acting to the best of your ability to raise item drop rates in general, like the streamer in the video did. Use all the sextants, alch every map, wear a bunch of item rarity, etc.



I'm gonna show you the basic, highschool level of stats that could be applied here.
We call this Confidence Interval for Two Independent Samples.

First, what numbers do we have?

He did 60 maps each time, and got 108 total from the control set and 160 from the biscos set.

The population, or n, of our data sets are 60 and 60, for the maps run with and without biscos. Call them n1 and n2.

The Means of out dataset are 108/60 and 160/60, or 1.8 and 2.666..., call THEM u1 and u2.


Problem though, we also need a standard deviation for both of these sets. Standard deviation is a sort of measurement of randomness. You would normally get this by comparing each datapoint to the mean of the dataset, sort of measuring how far each point is from the average.

Since we don't have that, I'm going to make up two datasets, one being super random (where every map got dropped in the same run) and the other super average (where rngesus is a merciful god and you got 1-3 each run). Both of these are extremes, and don't represent real drop rates.

Maths for Standard deviations spoilered
Spoiler

crazy dataset 59*0 1*108
Standard deviation control crazy
= sqrt((59(0-1.8)^2 + (108-1.8)^2)/60)
= sqrt((191.16 + 11278.44)/60)
= sqrt(11469.6/60)
= sqrt(191.16)
= 13.826062346163494716385436639116

average dataset 48*2 12*1:
Standard deviation control average
= sqrt((48(2-1.8)^2 + 12(1-1.8)^2)/60)
= sqrt((1.92 + 7.68)/60)
= sqrt(9.6/60)
= sqrt(0.16)
= 0.4

crazy dataset 59*0 1*160:
Standard deviation Biscos crazy
= sqrt((59(0-2.666...)^2 + (160-2.666...)^2)/60)
= sqrt((419.555... + 24751.111...)/60)
= sqrt(25170.666.../60)
= sqrt(419.5111...)
= 20.481970391324930120879734151542

average dataset 40*3 20*2:
Standard deviation Biscos average
= sqrt((40(2-2.666...)^2 + 20(1-2.666...)^2)/60)
= sqrt((17.777... + 55.555...)/60)
= sqrt(73.333.../60)
= sqrt(1.222...)
= 1.1055415967851332830383109122236


Now that we have our standard deviations, we need to pool them together, since we're comparing two sets of data. We call this the pooled estimate of the common standard deviation, and it means we assume that the two datasets have similar variance.

More maths
Spoiler


Sp = square root(((60-1)Sd1^2 + (60-1)Sd2^2)/(60+60+2))

Standard deviation pooled crazy:
= square root(((60-1)13.826062346163494716385436639116^2 +(60-1)20.481970391324930120879734151542^2)/(60+60+2))
= square root((59*191.15999999999999999999999999999 + 59*419.51111111111111111111111111112)/122)
= square root((11278.439999999999999999999999999 + 24751.155555555555555555555555556)/122)
= sqrt(36029.595555555555555555555555555/122)
= sqrt(300.24662962962962962962962962963)
= 17.327626197192436827317616612165

Standard deviation pooled average:
= square root(((60-1)0.4^2 +(60-1)1.1055415967851332830383109122236
^2)/(60+60+2))
= square root((59*0.16 + 59*1.222...)/122)
= square root((9.44 + 72.111...)/122)
= sqrt(81.551111111111111111111111111116/122)
= sqrt(0.66845173041894353369763205828784)
= 0.81758897400768776126962113154335



Alright, NOW we can do a confidence interval.

This amounts to taking the difference in the means, and comparing it to the pooled standard deviation modified by a T value and the populations.

The T value we just get from a table since this is high school maths. In this case we have 60 data points and want a 5% confidence, so we come up with 2.000, yay for easy math.

Maths the mathining
Spoiler

(mean1 - mean2) +/- t * Sp * sqrt(1/n1+1/n2)

Mean1-mean2 = 1.8-2.666 = -0.8666...

sqrt(1/n1+1/n2) = sqrt(1/60 + 1/60) = 0.1825741858350553711523232609336

Crazy interval:

-0.8666... +/- 2 * 17.327626197192436827317616612165 * 0.1825741858350553711523232609336
-0.8666... +/- 6.3271544908131715343934992178662

-7.1938211574798382010601658845329 to 5.4604878241465048677268325511995

Or: biscos gives between 546% fewer maps per run and 719% more maps per run.
Crazy dataset can't prove anything, biscos sucks.

average interval
-0.8666... +/- 2 * 0.81758897400768776126962113154335* 0.1825741858350553711523232609336
-0.8666... +/- 0.29854128255434368177662504462735

-1.165207949221010348443291711294 to -0.56812538411232298489004162203932

with the unnaturally average dataset, we can say with 95% confidence that biscos does have an effect on the drop rate of maps, giving between 116% and 56% more maps.



We don't actually have a standard deviation though, so none of this is actually more than an example.
The 'line' where we move from not confident to confident comes at a standard deviation of around 2.5, which is relatively low. If you have for or five runs with more than 5 maps that run, we probably won't be able to prove anything either way.

Or maybe, just maybe he should have count all maps, not just strand. Or maybe just maybe he should have run all the maps white with no sextants and no rolls on maps. Also if he get same ammount of maps he would probably not post it at all, because publication bs, when you don't want to post something underwhelming. Maybe he got less maps with Bisco's like 3 times and posted only when he got sattisfied results? Who knows.I don't think there is need for such walla of text.
Last edited by SunL4D2 on Jan 12, 2018 1:48:32 AM
"
Completed 16 ChallengesDuodecimus wrote:
complex math



everyone posting his results...

you DO realize he only counted SHAPED STRANDS, right? he NEVER mentioned how many actual MAPS he dropped.

the entire argument based on his results is flawed from the beginning...

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