Guides/Probability - Revision history

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en>Vygder: heading cleanup

2010-12-07T08:50:02Z

heading cleanup

← Older revision		Revision as of 10:50, 7 December 2010
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	=Probability guide=		{{FoCW Header\|Probability guide}}

	This page intends to provide some basic information on how the mathematics behind [http://www.eternal-lands.com Eternal Lands] basically work, what probability certain event have, say make an enriched essence in a certain number of essences, and what to expect when playing.		This page intends to provide some basic information on how the mathematics behind [http://www.eternal-lands.com Eternal Lands] basically work, what probability certain event have, say make an enriched essence in a certain number of essences, and what to expect when playing.
	We aim to skip the unnecessary stuff and present the results with some hopefully helpful examples.		We aim to skip the unnecessary stuff and present the results with some hopefully helpful examples.

	==Basic considerations==		== Basic considerations ==
	There underlying problem in determining probabilities and related values is classically an [http://en.wikipedia.org/wiki/Urn_problem Urn problem]: Let's say, the chance for an item to break is 1 in 300 (uses, hits). Imagine an urn with 299 white marbles and 1 black marble. Take a marble from the urn, look at it and put it back in. Now, the chance to pick the black marble is obviously 1 in 300, thus the urn models our problem.		There underlying problem in determining probabilities and related values is classically an [http://en.wikipedia.org/wiki/Urn_problem Urn problem]: Let's say, the chance for an item to break is 1 in 300 (uses, hits). Imagine an urn with 299 white marbles and 1 black marble. Take a marble from the urn, look at it and put it back in. Now, the chance to pick the black marble is obviously 1 in 300, thus the urn models our problem.

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	Now, you'll probably want to use your items more than once which leads us to the next section:		Now, you'll probably want to use your items more than once which leads us to the next section:

	==Trying fate ''n'' times==		== Trying fate ''n'' times ==
	First off: ''n'' stands for a natural number, such as 1,2,3,4,... and so on. Another common abbreviation in probability theory is to denote the chance of success with ''p'' and the chance to fail (no success) with ''q'', obviously		First off: ''n'' stands for a natural number, such as 1,2,3,4,... and so on. Another common abbreviation in probability theory is to denote the chance of success with ''p'' and the chance to fail (no success) with ''q'', obviously
	q = 1 - p.		q = 1 - p.
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	When looking at break rates, do we want to know the probability for our items to break at a certain time or do we want to know how many spare items to buy or bring? Now, the first option certainly has interesting applications (say 2 [[Enriched Fire Essence\|EFEs]] right after another?) but mostly we'll be concerned with the second option.		When looking at break rates, do we want to know the probability for our items to break at a certain time or do we want to know how many spare items to buy or bring? Now, the first option certainly has interesting applications (say 2 [[Enriched Fire Essence\|EFEs]] right after another?) but mostly we'll be concerned with the second option.

	===I don't care when but ''k'' in ''n''===		=== I don't care when but ''k'' in ''n'' ===
	where ''k'' is another natural number, less or equal to ''n''. In the example above we find 3 possible ways to get a series of colored marbles with one black and two white marbles (we don't really get to see it as we have to put the marbles back), each of them has the chance of (299/300)(299/300)(1/300), so the probability to get one of those series is 3 times that. Now remember, we had p = 1/300 and q = 299/300, so we can write the probability as		where ''k'' is another natural number, less or equal to ''n''. In the example above we find 3 possible ways to get a series of colored marbles with one black and two white marbles (we don't really get to see it as we have to put the marbles back), each of them has the chance of (299/300)(299/300)(1/300), so the probability to get one of those series is 3 times that. Now remember, we had p = 1/300 and q = 299/300, so we can write the probability as
	3 * p^1 * q^2.		3 * p^1 * q^2.
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	Putting this to use in our series of 300 marbles example we see that the variance is q (as n*p=1), which means we shouldn't really be expecting to have series with more than 2 black marbles in them (0 is possible of course). However if we did series of 600, we'd expect 2 black marbles per series on average and as much as 4 and as low as 0.		Putting this to use in our series of 300 marbles example we see that the variance is q (as n*p=1), which means we shouldn't really be expecting to have series with more than 2 black marbles in them (0 is possible of course). However if we did series of 600, we'd expect 2 black marbles per series on average and as much as 4 and as low as 0.

	===I don't care how, I had ''k'' just now===		=== I don't care how, I had ''k'' just now ===

	==Profit==		== Profit ==
	===Amount of items from a fixed set of ingredients===
			=== Amount of items from a fixed set of ingredients ===
	Here, we'll look at how many items you can get from a fixed set of ingredients, say for 100 or ''n'' items if you try until you run out of ingredients.<br />		Here, we'll look at how many items you can get from a fixed set of ingredients, say for 100 or ''n'' items if you try until you run out of ingredients.<br />
	We're going to calculate this in steps:		We're going to calculate this in steps:
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	See below for [[#Examples\|examples]].		See below for [[#Examples\|examples]].

	====Estimate of the necessary attempts====		==== Estimate of the necessary attempts ====
	As there are no fractions of items, say a half [[Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made		As there are no fractions of items, say a half [[Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made
	n * p * (1 - (2/3)^k(1 - p)^(k+1)) / (1 - 2/3 * (1 - p)) items.		n * p * (1 - (2/3)^k(1 - p)^(k+1)) / (1 - 2/3 * (1 - p)) items.
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	See below for [[#Examples\|examples]].		See below for [[#Examples\|examples]].

	==Astrological considerations==		== Astrological considerations ==
	We're only interested in the fail and degrade modifiers here, the exact way that the fail modifier works is unknown (to the author at this point). The degrade modifier works as follows: it's added to the "10k base", this means if ''dm'' is the value of your degrade modifier and the chance for an item to break is ''b'' in 10k, then the actual chance to break is ''b'' in (10k+''dm''). If you have the [[No More Tears Perk]] (possibly by using the cape) the modified value is multiplied by 4. Now, the degrade modifier varies from -900 to 900, looking at the best case, an item with a break rate of 1 in 10k we see that for those items the difference of 1 in 10k to 1 in 10,900 or 9,100 is not very large (0.01% compared to 0.009% or 0.011%). Looking at a rather bad case, say [[Mortar & Pestle\|mortars and pestles]], with a chance to break of 1 in 300, so roughly 33 in 10k, comparing the probabilities we see that we'll either break 3 more or less depending on astrology. If you've used mortars and pestles on a "bad day" you will think this unlikely, statistically however it's actually true.		We're only interested in the fail and degrade modifiers here, the exact way that the fail modifier works is unknown (to the author at this point). The degrade modifier works as follows: it's added to the "10k base", this means if ''dm'' is the value of your degrade modifier and the chance for an item to break is ''b'' in 10k, then the actual chance to break is ''b'' in (10k+''dm''). If you have the [[No More Tears Perk]] (possibly by using the cape) the modified value is multiplied by 4. Now, the degrade modifier varies from -900 to 900, looking at the best case, an item with a break rate of 1 in 10k we see that for those items the difference of 1 in 10k to 1 in 10,900 or 9,100 is not very large (0.01% compared to 0.009% or 0.011%). Looking at a rather bad case, say [[Mortar & Pestle\|mortars and pestles]], with a chance to break of 1 in 300, so roughly 33 in 10k, comparing the probabilities we see that we'll either break 3 more or less depending on astrology. If you've used mortars and pestles on a "bad day" you will think this unlikely, statistically however it's actually true.

	==Other considerations==		== Other considerations ==
	Keep in mind that the server always adds a small random number to your chances, there is no level where p=1 for anything except for the [[Magic]] skill, for all other skills, you will always have to work/calculate with a p<1.		Keep in mind that the server always adds a small random number to your chances, there is no level where p=1 for anything except for the [[Magic]] skill, for all other skills, you will always have to work/calculate with a p<1.
	==Examples==
			== Examples ==
	* expected result when making [[Potion of Spirit Restoration\|Potions of Spirit Restoration]] from 100 ingredients at level 30: p = 30/40 = 3/4, (2/3) * (1 - p) = 1/6, 100 * (3/4) / (1 - 1/6) = 100 * (3/4) * 6/5 = 90 Potions of Spirit Restoration. Remember, the calculation involved works with expected values and fractions of items, you will most likely get a few less.		* expected result when making [[Potion of Spirit Restoration\|Potions of Spirit Restoration]] from 100 ingredients at level 30: p = 30/40 = 3/4, (2/3) * (1 - p) = 1/6, 100 * (3/4) / (1 - 1/6) = 100 * (3/4) * 6/5 = 90 Potions of Spirit Restoration. Remember, the calculation involved works with expected values and fractions of items, you will most likely get a few less.
	* using the above example, how many attempts will we have to make? p = 3/4, n = 100 gives us		* using the above example, how many attempts will we have to make? p = 3/4, n = 100 gives us

en>GoodDay2Die: /* Amount of items from a fixed set of ingredients */ Lose, not Loose

2009-01-09T12:03:52Z

Amount of items from a fixed set of ingredients: Lose, not Loose

← Older revision		Revision as of 14:03, 9 January 2009
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	* make ''n'' attempts to mix the item		* make ''n'' attempts to mix the item
	** get the expected number of items (see expected value above)		** get the expected number of items (see expected value above)
	** ~~loose~~ the expected number of items		** lose the expected number of items
	* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.		* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.
	The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success,		The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success,

en>Ermabwed: links, i should really work on this some more... later

2008-11-25T07:50:49Z

links, i should really work on this some more... later

en>Ermabwed: added category

2008-06-18T18:47:15Z

added category

← Older revision		Revision as of 20:47, 18 June 2008
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	* Again using the same example the number of attempts necessary to produce the potions is		* Again using the same example the number of attempts necessary to produce the potions is
	100 * (0.995/(5/6)) = 120.		100 * (0.995/(5/6)) = 120.

			[[Category:Guides\|Probability]]

en>Ermabwed: /* Estimate of the necessary attempts */ added a simplification

2008-06-17T21:48:21Z

Estimate of the necessary attempts: added a simplification

← Older revision		Revision as of 23:48, 17 June 2008
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	To determine the sum of steps taken, we figure that we made ''n'' attempts in the first round, and we had (2/3 * (1 - p))^k * n items left (k=1) after the ''k'' step, we attempt to make them in the second round, i.e. after the 2nd round we'll have made n + 2/3 * (1 - p) * n attempts, the same formula that we used above then gives us, that after ''k'' steps we will have made		To determine the sum of steps taken, we figure that we made ''n'' attempts in the first round, and we had (2/3 * (1 - p))^k * n items left (k=1) after the ''k'' step, we attempt to make them in the second round, i.e. after the 2nd round we'll have made n + 2/3 * (1 - p) * n attempts, the same formula that we used above then gives us, that after ''k'' steps we will have made
	n * ((1 - (2/3 * (1 - p))^k ) / (1 - 2/3 * (1 - p)) ) attempts.		n * ((1 - (2/3 * (1 - p))^k ) / (1 - 2/3 * (1 - p)) ) attempts.
			Since (1-p)^k tends to zero for large ''k'' we can simplify the last formula to
			n / (1 - 2/3 * (1 - p)) = number of produced items / p.

	See below for [[#Examples\|examples]].		See below for [[#Examples\|examples]].

en>Ermabwed: /* Examples */ fixed the example according to the new formula

2008-06-17T21:27:04Z

Examples: fixed the example according to the new formula

← Older revision		Revision as of 23:27, 17 June 2008
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	Keep in mind that the server always adds a small random number to your chances, there is no level where p=1 for anything except for the [[Magic/Index\|Magic]] skill, for all other skills, you will always have to work/calculate with a p<1.		Keep in mind that the server always adds a small random number to your chances, there is no level where p=1 for anything except for the [[Magic/Index\|Magic]] skill, for all other skills, you will always have to work/calculate with a p<1.
	==Examples==		==Examples==
	* expected result when making [[Potions/Spirit Restoration\|Potions of Spirit Restoration]] from 100 ingredients at level 30: p = 30/40 = 3/4, (2/3) * (1 - p) = 1/6, 100 * (~~(1 - 3/4)^2 * 2/3 +~~ 3/4) /(1 - 1/6) = (19/24) * 6/5 * 100 = 95 Potions of Spirit Restoration. Remember, the calculation involved works with expected values and fractions of items, you will most likely get a few less.		* expected result when making [[Potions/Spirit Restoration\|Potions of Spirit Restoration]] from 100 ingredients at level 30: p = 30/40 = 3/4, (2/3) * (1 - p) = 1/6, 100 * (3/4) / (1 - 1/6) = 100 * (3/4) * 6/5 = 90 Potions of Spirit Restoration. Remember, the calculation involved works with expected values and fractions of items, you will most likely get a few less.
	* using the above example, how many attempts will we have to make? p = 3/4, n = 100 gives us		* using the above example, how many attempts will we have to make? p = 3/4, n = 100 gives us
	k > 2.57		k > 2.57

en>Ermabwed: /* Amount of items from a fixed set of ingredients */ fixed the formula (I think)

2008-06-17T21:24:00Z

Amount of items from a fixed set of ingredients: fixed the formula (I think)

← Older revision		Revision as of 23:24, 17 June 2008
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	** loose the expected number of items		** loose the expected number of items
	* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.		* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.
	The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success, (your level)/(2 * recommended level), use 99% or some other value from your experience if your level is ~~same~~ or higher than 2 times the recommended level. The expected number of critical fails is 1/3 of the remaining ingredients leaving you with 2/3 of (n - np) items to attempt in the second run. In this second run we expect to get 2/3 (n - np)p items, lose 1/3 of the unsuccessful attempts and have 2/3 of 2/3 * (n - np) - 2/3 (n - np)p ingredients left, now we see that by induction that after the ''k''th repetition of our process we will have (2/3)^k * n * (1-p)^k ingredients left, we will lose 1/3 * (2/3)^(k-1) * n * (1 - p)^k, the total of items made is thus the sum from j=1 to k-1 of (2/3)^j * n * (1 - p)^j ~~plus the initial n * p~~. The theory of [http://en.wikipedia.org/wiki/Series_(mathematics) infinite series] tells us, that the total of items made after infinite attempts (don't worry, this won't be necessary as there are no fractions of items) you will have		The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success,
	n * p ~~+ n * 2/3 * (1 - p)~~ / (1 - (2/3 * (1-p)~~)) = n * ((1 - p)^2 * 2/3 + p) / (1 - 2/3 * (1 - p~~)) items.		p = (your level)/(2 * recommended level),
			use 99% or some other value from your experience if your level is equal to or higher than 2 times the recommended level. The expected number of critical fails is 1/3 of the remaining ingredients leaving you with 2/3 of (n - n*p) items to attempt in the second run. In this second run we expect to get
			2/3 * (n - np)p items,
			lose 1/3 of the unsuccessful attempts and have
			2/3 of 2/3 * (n - np) - 2/3 (n - np)p = (2/3)^2 * (1 - p)^2 * n
			ingredients left, now we see that by induction that after the ''k''th repetition of our process we will have (2/3)^k * n * (1-p)^k ingredients left, we will lose 1/3 * (2/3)^(k-1) * n * (1 - p)^k, the total of items made is thus the sum from j=0 to k-1 of (2/3)^j * n * p * (1 - p)^j. The theory of [http://en.wikipedia.org/wiki/Series_(mathematics) infinite series] tells us, that the total of items made after infinite attempts (don't worry, this won't be necessary as there are no fractions of items) you will have
			n * p / (1 - (2/3 * (1-p))) items.
	See below for [[#Examples\|examples]].		See below for [[#Examples\|examples]].

	====Estimate of the necessary attempts====		====Estimate of the necessary attempts====
	As there are no fractions of items, say a half [[Alchemy/Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made		As there are no fractions of items, say a half [[Alchemy/Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made
	n * p ~~+ n~~ * (1 - (2/3)^k(1 - p)^k) / (1 - 2/3 * (1 - p)) items.		n * p * (1 - (2/3)^k(1 - p)^(k+1)) / (1 - 2/3 * (1 - p)) items.
	~~(Note that you'd usually expect the exponent k+1, however our first step yields n * p and doesn't fit the sequence, so we have only ''k''-1 terms in the sum)~~

	To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if		To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if
	k > - ln(n)/ln(2/3 * (1 - p)).		k > - ln(n)/ln(2/3 * (1 - p)).

en>Ermabwed: added number of attempts calculation

2008-06-13T13:48:25Z

added number of attempts calculation

← Older revision		Revision as of 15:48, 13 June 2008
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	** loose the expected number of items		** loose the expected number of items
	* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.		* repeat with the left-over ingredients (from non-critical fails) until there are no more ingredients left.
	The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success, (your level)/(2 * recommended level), use 99% or some other value from your experience if your level is same or higher than 2 times the recommended level. The expected number of critical fails is 1/3 of the remaining ingredients leaving you with 2/3 of (n - np) items to attempt in the second run. In this second run we expect to get 2/3 (n - np)p items, lose 1/3 of the unsuccessful attempts and have 2/3 of 2/3 * (n - np) - 2/3 (n - np)p ingredients left, now we see that by induction that after the ''k''th repetition of our process we will have (2/3)^k * n * (1-p)^k ingredients left, we will lose 1/3 * (2/3)^(k-1) * n * (1 - p)^k, the total of items made is thus the sum from j=1 to k of (2/3)^j * n * (1 - p)^j plus the initial n * p. The theory of [http://en.wikipedia.org/wiki/Series_(mathematics) infinite series] tells us, that the total of items made after infinite attempts (don't worry, this won't be necessary as there are no fractions of items) you will have		The expected number of produced items in the first ''n'' attempts is ''n'' * p where p is your chance of success, (your level)/(2 * recommended level), use 99% or some other value from your experience if your level is same or higher than 2 times the recommended level. The expected number of critical fails is 1/3 of the remaining ingredients leaving you with 2/3 of (n - np) items to attempt in the second run. In this second run we expect to get 2/3 (n - np)p items, lose 1/3 of the unsuccessful attempts and have 2/3 of 2/3 * (n - np) - 2/3 (n - np)p ingredients left, now we see that by induction that after the ''k''th repetition of our process we will have (2/3)^k * n * (1-p)^k ingredients left, we will lose 1/3 * (2/3)^(k-1) * n * (1 - p)^k, the total of items made is thus the sum from j=1 to k-1 of (2/3)^j * n * (1 - p)^j plus the initial n * p. The theory of [http://en.wikipedia.org/wiki/Series_(mathematics) infinite series] tells us, that the total of items made after infinite attempts (don't worry, this won't be necessary as there are no fractions of items) you will have
	n * p + n * 2/3 * (1 - p) / (1 - (2/3 * (1-p))) = n * ((1 - p)^2 * 2/3 + p) / (1 - 2/3 * (1 - p)) items.		n * p + n * 2/3 * (1 - p) / (1 - (2/3 * (1-p))) = n * ((1 - p)^2 * 2/3 + p) / (1 - 2/3 * (1 - p)) items.
	See below for [[#Examples\|examples]].		See below for [[#Examples\|examples]].
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	====Estimate of the necessary attempts====		====Estimate of the necessary attempts====
	As there are no fractions of items, say a half [[Alchemy/Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made		As there are no fractions of items, say a half [[Alchemy/Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made
	n * p + n * (1 - (2/3)^(k~~+1)~~(1 - p)^(k~~+1)~~) / (1 - 2/3 * (1 - p)) items.		n * p + n * (1 - (2/3)^k(1 - p)^k) / (1 - 2/3 * (1 - p)) items.
			(Note that you'd usually expect the exponent k+1, however our first step yields n * p and doesn't fit the sequence, so we have only ''k''-1 terms in the sum)

	To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if		To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if
	k > - ln(n)/ln(2/3 * (1 - p)).		k > - ln(n)/ln(2/3 * (1 - p)).
	The difference between number of attempts up to and including step ''k'' and ''n'' tells us how much extra food we'd have to buy if we want to use ''n'' ingredients. This number is, as it turns out, what we're really after, as we've been working with fractions of items and expected values the whole time and expecting some sort of guaranteed number of items would be neglecting the random nature of the underlying processes.		The difference between number of attempts up to and including step ''k'' and ''n'' tells us how much extra food we'd have to buy if we want to use ''n'' ingredients. This number is, as it turns out, what we're really after, as we've been working with fractions of items and expected values the whole time and expecting some sort of guaranteed number of items would be neglecting the random nature of the underlying processes.

	~~(Needs~~ the ~~calculation of attempts after ''k'' steps)~~		So, to be on the safe side, always work with conservative numbers here.

	So, to ~~be on~~ the ~~safe side~~, ~~always work with conservative numbers here~~.		To determine the sum of steps taken, we figure that we made ''n'' attempts in the first round, and we had (2/3 * (1 - p))^k * n items left (k=1) after the ''k'' step, we attempt to make them in the second round, i.e. after the 2nd round we'll have made n + 2/3 * (1 - p) * n attempts, the same formula that we used above then gives us, that after ''k'' steps we will have made
			n * ((1 - (2/3 * (1 - p))^k ) / (1 - 2/3 * (1 - p)) ) attempts.

	See below for [[#Examples\|examples]].		See below for [[#Examples\|examples]].
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	k > 2.57		k > 2.57
	i.e. we should be out of ingredients after 3 rounds.		i.e. we should be out of ingredients after 3 rounds.
			* Again using the same example the number of attempts necessary to produce the potions is
			100 * (0.995/(5/6)) = 120.

← Older revision		Revision as of 09:50, 25 November 2008
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	Obviously we'd expect (w,w,w) most of the time with a probability of (299/300) to the power of 3, but this highlights another problem:		Obviously we'd expect (w,w,w) most of the time with a probability of (299/300) to the power of 3, but this highlights another problem:

	When looking at break rates, do we want to know the probability for our items to break at a certain time or do we want to know how many spare items to buy or bring? Now, the first option certainly has interesting applications (say 2 [[~~Alchemy/~~Enriched Fire Essence\|EFEs]] right after another?) but mostly we'll be concerned with the second option.		When looking at break rates, do we want to know the probability for our items to break at a certain time or do we want to know how many spare items to buy or bring? Now, the first option certainly has interesting applications (say 2 [[Enriched Fire Essence\|EFEs]] right after another?) but mostly we'll be concerned with the second option.

	===I don't care when but ''k'' in ''n''===		===I don't care when but ''k'' in ''n''===
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	====Estimate of the necessary attempts====		====Estimate of the necessary attempts====
	As there are no fractions of items, say a half [[~~Alchemy/~~Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made		As there are no fractions of items, say a half [[Steel Bar\|steel bar]], we can abort the infinite series above at a certain ''k'' and we'll have made
	n * p * (1 - (2/3)^k(1 - p)^(k+1)) / (1 - 2/3 * (1 - p)) items.		n * p * (1 - (2/3)^k(1 - p)^(k+1)) / (1 - 2/3 * (1 - p)) items.
	To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if		To determine the step ''k'' at which to stop, we look for the ''k'' at which the left over ingredients are less than 1. Above we found that after ''k'' steps we have (2/3 * (1 - p))^k * n items left, which is less than one if
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	==Astrological considerations==		==Astrological considerations==
	We're only interested in the fail and degrade modifiers here, the exact way that the fail modifier works is unknown (to the author at this point). The degrade modifier works as follows: it's added to the "10k base", this means if ''dm'' is the value of your degrade modifier and the chance for an item to break is ''b'' in 10k, then the actual chance to break is ''b'' in (10k+''dm''). If you have the [[~~Perks/No More Tears\|~~No More Tears]] ~~perk~~ (possibly by using the cape) the modified value is multiplied by 4. Now, the degrade modifier varies from -900 to 900, looking at the best case, an item with a break rate of 1 in 10k we see that for those items the difference of 1 in 10k to 1 in 10,900 or 9,100 is not very large (0.01% compared to 0.009% or 0.011%). Looking at a rather bad case, say [[~~Items/~~Mortar Pestle\|mortars and pestles]], with a chance to break of 1 in 300, so roughly 33 in 10k, comparing the probabilities we see that we'll either break 3 more or less depending on astrology. If you've used mortars and pestles on a "bad day" you will think this unlikely, statistically however it's actually true.		We're only interested in the fail and degrade modifiers here, the exact way that the fail modifier works is unknown (to the author at this point). The degrade modifier works as follows: it's added to the "10k base", this means if ''dm'' is the value of your degrade modifier and the chance for an item to break is ''b'' in 10k, then the actual chance to break is ''b'' in (10k+''dm''). If you have the [[No More Tears Perk]] (possibly by using the cape) the modified value is multiplied by 4. Now, the degrade modifier varies from -900 to 900, looking at the best case, an item with a break rate of 1 in 10k we see that for those items the difference of 1 in 10k to 1 in 10,900 or 9,100 is not very large (0.01% compared to 0.009% or 0.011%). Looking at a rather bad case, say [[Mortar & Pestle\|mortars and pestles]], with a chance to break of 1 in 300, so roughly 33 in 10k, comparing the probabilities we see that we'll either break 3 more or less depending on astrology. If you've used mortars and pestles on a "bad day" you will think this unlikely, statistically however it's actually true.

	==Other considerations==		==Other considerations==
	Keep in mind that the server always adds a small random number to your chances, there is no level where p=1 for anything except for the [[~~Magic/Index\|~~Magic]] skill, for all other skills, you will always have to work/calculate with a p<1.		Keep in mind that the server always adds a small random number to your chances, there is no level where p=1 for anything except for the [[Magic]] skill, for all other skills, you will always have to work/calculate with a p<1.
	==Examples==		==Examples==
	* expected result when making [[~~Potions/~~Spirit Restoration\|Potions of Spirit Restoration]] from 100 ingredients at level 30: p = 30/40 = 3/4, (2/3) * (1 - p) = 1/6, 100 * (3/4) / (1 - 1/6) = 100 * (3/4) * 6/5 = 90 Potions of Spirit Restoration. Remember, the calculation involved works with expected values and fractions of items, you will most likely get a few less.		* expected result when making [[Potion of Spirit Restoration\|Potions of Spirit Restoration]] from 100 ingredients at level 30: p = 30/40 = 3/4, (2/3) * (1 - p) = 1/6, 100 * (3/4) / (1 - 1/6) = 100 * (3/4) * 6/5 = 90 Potions of Spirit Restoration. Remember, the calculation involved works with expected values and fractions of items, you will most likely get a few less.
	* using the above example, how many attempts will we have to make? p = 3/4, n = 100 gives us		* using the above example, how many attempts will we have to make? p = 3/4, n = 100 gives us
	k > 2.57		k > 2.57