Void Relic/Math

More elaborate explanations about the math behind some of the calculations on Void Relics page for documentation. Readers are expected to have basic knowledge on ratios, rates, & percentages; combinatorics (e.g. combinations and permutations); probability; and algebra. Some of this content shares the same principles as User blog:FINNER/Warframe Expected & Nearly Guaranteed Numbers - An Explanation.
==Drop Chances==
From official Drop Tables on https://www.warframe.com/droptables/
Relic Quality 	Void Traces Required	Common 	Uncommon 	Rare
Intact (default)	0 	76% (25⅓% for each item, expressed as 25.33%)	22% (11% for each item)	2%
Exceptional 	25 	70% (23⅓% for each item, expressed as 23.33%)	26% (13% for each item)	4%
Flawless 	50 	60% (20% for each item) 	34% (17% for each item)	6%
Radiant 	100 	50% (16⅔% for each item, expressed as 16.67%)	40% (20% for each item)	10%
*Official repo doesn't express repeating decimal values and just rounds them to the nearest 2 decimal places. Even if item chances in a single drop table don't add up to 100% exactly, there will never be a small chance where a player gets no rewards from a drop table due to the game either truncating or rounding RNG values so that there will always be a reward hit.
==Notes==
"Expected" refers to expected value; 4-player groups opening very large numbers of relics can expect to receive a rare item for each of this many relics used.
"90%" refers to confidence; a 4-player group looking for one rare item has a 90% chance to find it without using more than this many relics (but they might use fewer).
A group of four players with Radiant relics has a 5.23% chance to receive two or more rare items; however, each player can keep only one. Using one Radiant relic at a time removes this possibility and may be worthwhile for getting the most out of rare or vaulted relics, but it does increase the expected number of runs required to find the item.
In-game, the rarity of a relic drop is indicated by the text color of its name: bronze (common), silver (uncommon) or gold (rare).
==Assumptions==
The following assumptions need to be made before proper statistical analysis can be made. If one of these is not true, then most (if not all) of the math in this article is invalid.
*Each reward roll is independent from each other when playing in a squad. In other words, player 1's relic reward roll will not affect the outcome of player 4's relic reward roll.
**By extension, a player that is not opening a relic (i.e. does not have a relic selected when running Void fissures) will not affect the chances of other player's reward rolls.
*Each instance of picking a reward from opening relics are independent from each other. In other words, the rewards you rolled or picked from a previous relic opening will not affect future chances of getting a certain reward from a relic.
*The type of relic opened (e.g. vs. ) does not affect chances of getting a certain reward from a relic. In other words, all relics at the same refinement level will have its chances as listed in the official drop table repository.
*The number of players in a squad does not affect drop chances.
*The official drop chances provided by DE are pulled directly from in-game chances for getting rewards from Void Relics.
**On a related note, the RNG used to produce random numbers for chosing relic rewards is uniformly distributed.
**There is no manipulation of drop chances by external, unaccounted factors such as the concept of a "pity system".
**Because the drop chances are rounded to the nearest two decimal places in the official drop table repository and thus item chances do not add up to 100% exactly, assume that in-game chances are truly represented as floating-point values with high precision and that all item drop chances of any particular relic will add up to 100% exactly.
***By extension, an item is always rolled for a player with a Void Relic, regardless if anyone picks it or not. In other words, there is never an instance where a player that is opening their Void Relic will not roll an item.
*Players can pick one and only one reward out of all the rewards rolled from a single relic opening instance.
==Same Relic Refinement==
A group of 4 players with relics of the same refinement level has the following chances to obtain a rare item:
Quality 	Chance for at least one rare item	Expected relics/player	Expected traces/player	90% relics/player	90% traces/player
Intact 	7.76% 	12.9 	0 	28.5 	0
Exceptional	15.07% 	6.6 	166 	14.1 	353
Flawless 	21.93% 	4.6 	228 	9.3 	465
Radiant 	34.39% 	2.9 	291 	5.5 	546
===Chance for at least one rare item===
Given there are four players in squad, to calculate the number of outcomes in which there are exactly r number of rare drops we use combinations:
{
\begin{aligned}
{}_n C_r &= \frac{n}{1} \times \frac{n - 1}{2} \times \dotsb \times \frac{n - (r - 1)}{r} = \frac{n (n - 1) \dotsb (n - r + 1)}{r (r - 1) \dotsb 1} = \frac{n!}{r! \times (n - r)!} \quad \text{where 0 ≤ r ≤ n} \\
{}_4 C_r &= \frac{4!}{r! \times (4 - r)!} = \frac{4 (3) (2) (1)}{r! \times (4 - r)!} = \frac{24}{r! \times (4 - r)!} \quad (\text{when n=4})
\end{aligned}
}
In the scenario where all players in the squad has the same refinement level for their relic, the order in which a rare reward is rolled does not matter because players may only pick one reward out of n rolled at any given time. In other words, there is functionally no difference if player 1 rolls a rare reward or if player 4 rolls it instead since all players in a squad can only get one rare reward from that specific relic opening. Player 1's rare reward may be different from player 4's rare reward, but for the sake of analysis, we will be ignoring this difference since we only care whether or not a rare rarity item is being rolled.
Note that nCr can be read as "from n number of options, pick r number of options without replacement".
Say you want to find the number of outcomes in which there are exactly 2 rare drops:
{
\begin{aligned}
{}_4 C_2 &= \frac{4!}{2! \times (4 - 2)!} \\
&= \frac{4 (4 - 1) (4 - 2) (4 - 3)}{2 (2 - 1)} \\
&= \frac{4 (3) (2) (1)}{4} \\
&= 6
\end{aligned}
}
This result can be visualized as a wikitable of binary results, with ✔️ representing a success (rare drop) and ❌ representing a failure (no rare drop):
Outcome #	Player 1	Player 2	Player 3	Player 4
1 	✔️ 	✔️ 	❌ 	❌
2 	✔️ 	❌ 	✔️ 	❌
3 	✔️ 	❌ 	❌ 	✔️
4 	❌ 	✔️ 	✔️ 	❌
5 	❌ 	✔️ 	❌ 	✔️
6 	❌ 	❌ 	✔️ 	✔️
We can treat the scenario of opening Void Relics of a specific refinement in a squad (1 to 4 players) to get a rare item (or any rarity for that matter) as multiple Bernoulli trials (aka a Bernoulli process). A success outcome is defined as a rare reward being rolled and a failure outcome is defined as no rare reward being rolled. The results can be modelled after a binomial distribution.
Say we define binomial random variable X as the number of rare item rolls out of n number of players in squad with the same relic refinement. In other words, the number of successes out of n independent trials.
Using binomial distribution to calculate the probability that exactly r rare item(s) is/are rolled based on n number of players in squad with the same relic refinement:
{
\begin{aligned}
P(X = r) &= {}_n C_r \times p^r \times (1 - p)^{n - r} \quad \text{where} \quad 0 \le r \le n; 1 \le n \le 4 \\
\text{Probability for exactly X rare item} &= \text{Num. ways to express r rare drops from n players} \times \text{Rare item chance}^r \times \text{No rare item chance}^{n - r} \\
&= {}_n C_r \times \text{Rare item chance}^r \times (1 - \text{Rare item chance})^{n - r}
\end{aligned}
}
Assuming there are four players with the same relic refinement in a squad and we want to know exactly one rare item will be available to be chosen:
{
\begin{aligned}
\text{Probability for exactly one rare item} &= {}_4 C_1 \times \text{Rare item chance}^1 \times \text{No rare item chance}^{4 - 1} \\
&= 4 \times \text{Rare item chance}^1 \times (1 - \text{Rare item chance})^{3}
\end{aligned}
}
Using binomial distribution to calculate the probability that at least one rare item is rolled (in other words, one, two, three, or four rare items rolled) based on n number of players in a squad:
{
\begin{aligned}
P(1 \le X \le 4) &= \sum_{x = 1}^{n} P(X = x) & \text{where} \quad 1 \le n \le 4 \\
\text{Probability for at least one rare item} &= P(X = 1) + ... + P(X = n)
\end{aligned}
}
Using binomial distribution to calculate the probability that at least one rare item is rolled in a full squad with the same relic refinement. In other words, it is the sum of the probabilities for exactly one, two, three, and four rare items to be chosen by players at any single time:
{
\begin{aligned}
\text{Probability for at least one rare item} &= \sum_{x = 1}^{4} P(X = x) \\
&= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\end{aligned}
}
For example, the probability that at least one rare item is rolled in a full squad with Radiant refinement relics is calculated as such:
{
\begin{aligned}
P(1 \le X \le 4) &= \sum_{x = 1}^{4} P(X = x) \\
&= ({}_4 C_1 \times 0.1^1 \times (1 - 0.1)^{3}) + ({}_4 C_2 \times 0.1^2 \times (1 - 0.1)^{2}) + ({}_4 C_3 \times 0.1^3 \times (1 - 0.1)^{1}) + ({}_4 C_4 \times 0.1^4 \times (1 - 0.1)^{0}) \\
&= (4 \times 0.1 \times 0.9^{3}) + (6 \times 0.1^2 \times 0.9^{2}) + (4 \times 0.1^3 \times 0.9^{1}) + (1 \times 0.1^4 \times 0.9^{0}) \\
&= 0.2916 + 0.0486 + 0.0036 + 0.0001 \\
&= 0.3439 \\
&= 34.39%
\end{aligned}
}
Alternatively, this can be calculated from the probability that zero rare items are rolled in a full squad of Radiant refinement relics:
{
\begin{aligned}
P(1 \le X \le 4) &= 1 - P(X = 0) = 100% - P(X = 0) \\
&= 1 - ({}_4 C_0 \times 0.1^0 \times (1 - 0.1)^{4}) \\
&= 1 - (1 \times 1 \times (0.9)^{4}) \\
&= 1 - 0.6561 \\
&= 0.3439 \\
&= 34.39%
\end{aligned}
}
====Visualization====
 Possible outcomes of a full squad opening Radiant relics
Outcome #	Player 1	Player 2	Player 3	Player 4	At least one rare reward?	Probability of outcome	As a percentage	As a ratio	Calculation
1 	❌ 	❌ 	❌ 	❌ 	No 	0.6561 	65.61% 	6561:10000	0.1^0 * 0.9^4
2 	✔️ 	❌ 	❌ 	❌ 	Yes! 	0.0729 	7.29% 	729:10000 	0.1^1 * 0.9^3
3 	❌ 	✔️ 	❌ 	❌ 	Yes! 	0.0729 	7.29% 	729:10000 	0.1^1 * 0.9^3
4 	❌ 	❌ 	✔️ 	❌ 	Yes! 	0.0729 	7.29% 	729:10000 	0.1^1 * 0.9^3
5 	❌ 	❌ 	❌ 	✔️ 	Yes! 	0.0729 	7.29% 	729:10000 	0.1^1 * 0.9^3
6 	✔️ 	✔️ 	❌ 	❌ 	Yes! 	0.0081 	0.81% 	81:10000 	0.1^2 * 0.9^2
7 	✔️ 	❌ 	✔️ 	❌ 	Yes! 	0.0081 	0.81% 	81:10000 	0.1^2 * 0.9^2
8 	✔️ 	❌ 	❌ 	✔️ 	Yes! 	0.0081 	0.81% 	81:10000 	0.1^2 * 0.9^2
9 	❌ 	✔️ 	✔️ 	❌ 	Yes! 	0.0081 	0.81% 	81:10000 	0.1^2 * 0.9^2
10 	❌ 	✔️ 	❌ 	✔️ 	Yes! 	0.0081 	0.81% 	81:10000 	0.1^2 * 0.9^2
11 	❌ 	❌ 	✔️ 	✔️ 	Yes! 	0.0081 	0.81% 	81:10000 	0.1^2 * 0.9^2
12 	✔️ 	✔️ 	✔️ 	❌ 	Yes! 	0.0009 	0.09% 	9:10000 	0.1^3 * 0.9^1
13 	✔️ 	✔️ 	❌ 	✔️ 	Yes! 	0.0009 	0.09% 	9:10000 	0.1^3 * 0.9^1
14 	✔️ 	❌ 	✔️ 	✔️ 	Yes! 	0.0009 	0.09% 	9:10000 	0.1^3 * 0.9^1
15 	❌ 	✔️ 	✔️ 	✔️ 	Yes! 	0.0009 	0.09% 	9:10000 	0.1^3 * 0.9^1
16 	✔️ 	✔️ 	✔️ 	✔️ 	Yes! 	0.0001 	0.01% 	1:10000 	0.1^4 * 0.9^0
Total 	Total 	Total 	Total 	Total 	Total 	1 	100% 	1:1 	1
Since the order in which a rare drop is rolled does not matter, the above table can be reduced to:
 Probability that X number of rare drops rolled
Number of rare drops rolled	Notation	Probability of outcome	As a percentage	As a ratio	Calculation
0 	P(X=0) 	0.6561 	65.61% 	6561:10000	4C0 * 0.1^0 * 0.9^4
1 	P(X=1) 	0.2916 	29.16% 	729:2500 	4C1 * 0.1^1 * 0.9^3
2 	P(X=2) 	0.0486 	4.86% 	243:5000 	4C2 * 0.1^2 * 0.9^2
3 	P(X=3) 	0.0036 	0.36% 	9:2500 	4C3 * 0.1^3 * 0.9^1
4 	P(X=4) 	0.0001 	0.01% 	1:10000 	4C4 * 0.1^4 * 0.9^0
Total 	Total 	1 	100% 	1:1 	1
Because players may only pick one reward out of four rolled, assuming they always pick the rare reward, the above table can be reduced to:
 Probability that players can get a rare drop
Rare drop?	Notation 	Probability of outcome	As a percentage	As a ratio	Calculation
No 	P(X=0) 	0.6561 	65.61% 	6561:10000	4C0 * 0.1^0 * 0.9^4
Yes 	P(X≥1) or P(1≤X≤4)	0.3439 	34.39% 	3439:10000	1 - (4C0 * 0.1^0 * 0.9^4)
Total 	Total 	1 	100% 	1:1 	1
===Expected Relics opened per player===
Say we define geometric random variable Y as the number of runs by a full 4-player squad with the same relic refinement before seeing at least one rare drop rolled.
To get the expected (average) number of relics opened per player to get a rare item, we have to determine the distribution of the number of failures needed before getting one success. In other words, the average number of full squad runs without a rare drop before the player finally sees a rare drop. The resultant probability distribution will follow a geometric distribution.
{
\begin{aligned}
P(Y = k) &= p \times (1 - p)^{k - 1} \quad \text{where } k > 0 \\
&= \text{Chance of rolling at least one rare drop} \times (1 - \text{Chance of rolling at least one rare drop})^{k - 1} \\
&= \text{Chance of rolling at least one rare drop} \times (\text{Chance of NOT rolling at least one rare drop})^{k - 1} \\
&= \text{Chance of rolling at least one rare drop} \times (\text{Chance of rolling zero rare drops})^{k - 1}
\end{aligned}
}
 Possible number of full squad runs of Radiant relics before seeing a rare drop
Number of runs before getting a rare drop	Notation 	Probability of occurrence	As a percentage 	As a ratio 	Calculation
1 	P(Y=1) 	0.3439 	34.39% 	3439:10000 	0.3439 * (0.6561)^0
2 	P(Y=2) 	0.22563279 	22.563279% 	about 2256:10000 	0.3439 * (0.6561)^1
3 	P(Y=3) 	0.148037673519 	14.8037673519% 	about 1480:10000 	0.3439 * (0.6561)^2
4 	P(Y=4) 	0.0971275175958 	9.71275175958% 	about 971:10000 	0.3439 * (0.6561)^3
5 	P(Y=5) 	0.0637253642946 	6.37253642946% 	about 637:10000 	0.3439 * (0.6561)^4
... 	... 	... 	... 	... 	...
10 	P(Y=10) 	0.0077475166035 	0.77475166035% 	about 77:10000 	0.3439 * (0.6561)^9
... 	... 	... 	... 	... 	...
100 	P(Y=100) 	2.60895094784e-19 	2.60895094784e-17% 	about 261 out of one sextillion	0.3439 * (0.6561)^99
... 	... 	... 	... 	... 	...
1,000 	P(Y=1000)	4.89215147266e-184 	4.89215147266e-182%	about 489 out of 10^186 	0.3439 * (0.6561)^999
... 	... 	... 	... 	... 	...
∞ 	P(Y=∞) 	0 	0% 	0:1 	0.3439 * (0.6561)^∞ converges to 0
Total 	Total 	1 	100% 	1:1 	1
Using the probability that a player will get at least one rare item in a full squad of players with a certain relic refinement, one can calculate the expected number of relics that each player will open (i.e. number of full squad runs) by using its reciprocal:
{
\begin{aligned}
E[Y] &= \frac{1}{p} \\
\text{Expected Relics opened per player} &= \frac{1}{P(1 \le X \le 4)}
\end{aligned}
}
For example, the expected number of relics that each player will open in a full squad of Radiant relics to get at least one rare item is:
{
\begin{aligned}
\text{Expected Relics opened per player} = \frac{1}{0.3439} = 2.9078 \approx 2.9
\end{aligned}
}
===Expected Void Traces used per player===
Multiply the expected number of relics opened per player by the number of Void Traces needed to upgrade one relic from Intact to that refinement.
For example for a full squad of players with Radiant relics:
{
\begin{aligned}
\text{Expected Void Traces used per player} &= \text{Expected number of Relics opened per player} * \text{Void Traces needed} \\
&= 2.9078 \times 100 \\
&= 290.78 \\
&\approx 291
\end{aligned}
}
===Number of Relics opened per player at 90% confidence level===
To get the average number of Radiant relics that 90% of the WARFRAME's player base will have to open in a full squad of Radiant relics before seeing at least one rare reward in all their runs, we can use the cumulative distribution function (CDF) of a geometric distribution which is the summation of the probabilities of x number of Radiant relics that are opened before seeing at least one rare reward rolled.
The CDF of a geometric distribution is:
{
\begin{aligned}
P(Y \le k) &= \sum_{x=1}^{k} P(Y = x) \\
&= 100% - \text{Chance of k number of failures in a row} \\
&= 1 - (1 - p)^{k}
\end{aligned}
}
Set P(Y ≤ k) to 90% and solve for k, given that p is 34.39%, the chance for rolling at least one rare item from a full squad with Radiant relics:
{
\begin{aligned}
0.9 &= 1 - (1 - 0.3439)^{k} \\
0.9 &= 1 - (0.6561)^{k} \quad &\text{Simplify} \\
0.9 - 1 &= 1 - (0.6561)^{k} - 1 \quad &\text{Subtract 1 from both sides} \\
-0.1 &= -(0.6561)^{k} \\
\frac{-0.1}{-1} &= \frac{-(0.6561)^{k}}{-1} \quad &\text{Remove negative sign} \\
0.1 &= (0.6561)^{k} \\
\log_{0.6561}{0.1} &= \log_{0.6561}{(0.6561)^{k}} \quad &\text{Use log to solve for exponent} \\
\log_{0.6561}{0.1} &= k \\
5.4636 &\approx k
\end{aligned}
}
===Number of Void Traces used per player at 90% confidence level===
Multiply the average number of relics opened per player at 90% confidence by the number of Void Traces needed to upgrade one relic from Intact to that refinement.
For example for a full squad of players with Radiant relics:
{
\begin{aligned}
\text{Void Traces used per player} &= \text{Number of Relics opened per player} * \text{Void Traces needed} \\
&= 5.4636 \times 100 \\
&= 546.36 \\
&\approx 546
\end{aligned}
}
==Mixed Runs with Intact and Radiant Relic Refinements==
Most players' relics, especially in public missions, will be either Radiant (from Elite Sanctuary Onslaught or self-refinement, best for finding specific uncommon or rare items) or, more likely, Intact (from almost every other source, best for farming). This variant of the above table gives drop chances for most real squads.
Quality 	Chance for at least one rare item	Expected relics/player	Expected traces/player	90% relics/player	90% traces/player
0 Radiant, 4 Intact	7.76% 	12.9 	0 	28.5 	0
1 Radiant, 3 Intact	15.29% 	6.5 	163 	13.9 	347
2 Radiant, 2 Intact	22.21% 	4.5 	225 	9.2 	458
3 Radiant, 1 Intact	28.56% 	3.5 	263 	6.8 	514
4 Radiant, 0 Intact	34.39% 	2.9 	291 	5.5 	546
===Chance for at least one rare item===
Given there are four players in squad, to calculate the number of outcomes in which there are exactly r number of rare drops we use combinations:
{
\begin{aligned}
{}_n C_r &= \frac{n}{1} \times \frac{n - 1}{2} \times \dotsb \times \frac{n - (r - 1)}{r} = \frac{n (n - 1) \dotsb (n - r + 1)}{r (r - 1) \dotsb 1} = \frac{n!}{r! \times (n - r)!} \quad \text{where 0 ≤ r ≤ n} \\
{}_4 C_r &= \frac{4!}{r! \times (4 - r)!} = \frac{4 (3) (2) (1)}{r! \times (4 - r)!} = \frac{24}{r! \times (4 - r)!} \quad (\text{when n=4})
\end{aligned}
}
In the scenario where a full squad is running with relics with different refinements, the order in which a rare reward is rolled does matter because the chance of rolling a rare reward is dependent on the relic refinement. Player 1's rare reward may be different from player 4's rare reward, but for the sake of analysis, we will be ignoring this difference since we only care whether or not a rare rarity item is being rolled.
Note that nCr can be read as "from n number of options, pick r number of options without replacement".
Say you want to find the number of outcomes in which there are exactly 2 rare drops:
{
\begin{aligned}
{}_4 C_2 &= \frac{4!}{2! \times (4 - 2)!} \\
&= \frac{4 (4 - 1) (4 - 2) (4 - 3)}{2 (2 - 1)} \\
&= \frac{4 (3) (2) (1)}{4} \\
&= 6
\end{aligned}
}
This result can be visualized as a wikitable of binary results, with ✔️ representing a success (rare drop) and ❌ representing a failure (no rare drop):
Outcome #	Player 1(Radiant)	Player 2(Intact)	Player 3(Intact)	Player 4(Intact)
1 	✔️ 	✔️ 	❌ 	❌
2 	✔️ 	❌ 	✔️ 	❌
3 	✔️ 	❌ 	❌ 	✔️
4 	❌ 	✔️ 	✔️ 	❌
5 	❌ 	✔️ 	❌ 	✔️
6 	❌ 	❌ 	✔️ 	✔️
We can treat the scenario of opening Void Relics of a specific refinement in a squad (1 to 4 players) to get a rare item (or any rarity for that matter) as multiple Bernoulli trials (aka a Bernoulli process). A success outcome is defined as a rare reward being rolled and a failure outcome is defined as no rare reward being rolled. The results can be modelled after a binomial distribution.
Say we define binomial random variable X as the number of rare item rolls out of n number of players in squad with the same relic refinement. In other words, the number of successes out of n independent trials.
Using binomial distribution to calculate the probability that exactly r rare item(s) is/are rolled based on n number of players in squad with the same relic refinement:
{
\begin{aligned}
P(X = r) &= {}_n C_r \times p^r \times (1 - p)^{n - r} \quad \text{where} \quad 0 \le r \le n; 1 \le n \le 4 \\
\text{Probability for exactly X rare item} &= \text{Num. ways to express r rare drops from n players} \times \text{Rare item chance}^r \times \text{No rare item chance}^{n - r} \\
&= {}_n C_r \times \text{Rare item chance}^r \times (1 - \text{Rare item chance})^{n - r}
\end{aligned}
}
Assuming there are four players in a squad, 1 Radiant, 3 Intact relics, and we want to know exactly one rare item will be available to be chosen, there are two unique scenarios where this can happen:
*Player with Radiant relic rolls a rare reward (10% chance); all other players don't roll a rare reward
*One of the three other players with Intact relics rolls a rare reward (2% chance each); all other players doesn't roll a rare reward
{
\begin{aligned}
\text{Probability for exactly one rare item} &= (\text{Radiant rare item chance}^1 \times \text{Intact no rare item chance}^{3}) + \\
& \qquad {}_3 C_1 \times (\text{Radiant no rare item chance}^1 \times \text{Intact rare item chance}^{1} \times \text{Intact no rare item chance}^{2}) \\
&= (0.1^1 \times 0.98^3) + 3 \times (0.9^1 \times 0.02^1 \times 0.98^2) \\
&= 0.0941192 + 0.0518616 \\
&= 0.1459808 \\
&\approx 14.6%
\end{aligned}
}
Using binomial distribution to calculate the probability that at least one rare item is rolled (in other words, one, two, three, or four rare items rolled) based on n number of players in a squad:
{
\begin{aligned}
P(1 \le X \le 4) &= \sum_{x = 1}^{n} P(X = x) & \text{where} \quad 1 \le n \le 4 \\
\text{Probability for at least one rare item} &= P(X = 1) + ... + P(X = n)
\end{aligned}
}
Using binomial distribution to calculate the probability that at least one rare item is rolled in a full squad with 1 Radiant, 3 Intact relics. In other words, it is the sum of the probabilities for exactly one, two, three, and four rare items to be chosen by players at any single time:
{
\begin{aligned}
\text{Probability for at least one rare item} &= \sum_{x = 1}^{4} P(X = x) \\
&= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\end{aligned}
}
For example, the probability that at least one rare item is rolled in a full squad, 1 Radiant, 3 Intact, is calculated as such:
{
\begin{aligned}
P(1 \le X \le 4) &= \sum_{x = 1}^{4} P(X = x) \\
&= (0.1^1 \times 0.98^3) + {}_3 C_1 \times (0.9^1 \times 0.02^1 \times 0.98^2) + \\
& \qquad {}_3 C_1 \times (0.1^1 \times 0.02^1 \times 0.98^2) + {}_3 C_2 \times (0.9^1 \times 0.02^2 \times 0.98^1) + \\
& \qquad {}_3 C_2 (0.1^1 \times 0.02^2 \times 0.98^1) + {}_3 C_3 \times (0.9^1 \times 0.02^3 \times 0.98^0) + \\
& \qquad (0.1^1 \times 0.02^3) \\
&= 0.0941192 + {}_3 C_1 \times 0.0172872 + \\
& \qquad {}_3 C_1 \times 0.0019208 + {}_3 C_2 \times 0.0003528 + \\
& \qquad {}_3 C_2 \times 0.0000392 + {}_3 C_3 \times 0.0000072 + \\
& \qquad 0.0000008 \\
&= 0.0941192 + 3 \times 0.0172872 + \\
& \qquad 3 \times 0.0019208 + 3 \times 0.0003528 + \\
& \qquad 3 \times 0.0000392 + 1 \times 0.0000072 + \\
& \qquad 0.0000008 \\
&= 0.0941192 + 0.0518616 + \\
& \qquad 0.0057624 + 0.0010584 + \\
& \qquad 0.0001176 + 0.0000072 + \\
& \qquad 0.0000008 \\
&= 0.1459808 + \\
& \qquad 0.0068208 + \\
& \qquad 0.0001248 + \\
& \qquad 0.0000008 \\
&= 0.1529272 \\
\end{aligned}
}
Alternatively, this can be calculated from the probability that zero rare items are rolled in a full squad of different refinement relics:
{
\begin{aligned}
P(1 \le X \le 4) &= 1 - P(X = 0) = 100% - P(X = 0) \\
&= 1 - (0.9^1 \times 0.98^3) \\ 
&= 0.1529272
\end{aligned}
}
====Visualization====
 Possible outcomes of a full squad opening relics with different refinements (Radiant and Intact, "staggered runs")
Outcome #	Player 1(Radiant, 10% Rare)	Player 2(Intact, 2% Rare)	Player 3(Intact, 2% Rare)	Player 4(Intact, 2% Rare)	At least one rare reward?	Probability of outcome	As a percentage	As a ratio 	Calculation
1 	❌ 	❌ 	❌ 	❌ 	No 	0.8470728 	~84.71% 	~8471:10000	0.1^0 * 0.9^1 * 0.02^0 * 0.98^3
2 	✔️ 	❌ 	❌ 	❌ 	Yes! 	0.0941192 	~9.41% 	~941:10000 	0.1^1 * 0.9^0 * 0.02^0 * 0.98^3
3 	❌ 	✔️ 	❌ 	❌ 	Yes! 	0.0172872 	~1.73% 	~173:10000 	0.1^0 * 0.9^1 * 0.02^1 * 0.98^2
4 	❌ 	❌ 	✔️ 	❌ 	Yes! 	0.0172872 	~1.73% 	~173:10000 	0.1^0 * 0.9^1 * 0.02^1 * 0.98^2
5 	❌ 	❌ 	❌ 	✔️ 	Yes! 	0.0172872 	~1.73% 	~173:10000 	0.1^0 * 0.9^1 * 0.02^1 * 0.98^2
6 	✔️ 	✔️ 	❌ 	❌ 	Yes! 	0.0019208 	~0.19% 	~19:10000 	0.1^1 * 0.9^0 * 0.02^1 * 0.98^2
7 	✔️ 	❌ 	✔️ 	❌ 	Yes! 	0.0019208 	~0.19% 	~19:10000 	0.1^1 * 0.9^0 * 0.02^1 * 0.98^2
8 	✔️ 	❌ 	❌ 	✔️ 	Yes! 	0.0019208 	~0.19% 	~19:10000 	0.1^1 * 0.9^0 * 0.02^1 * 0.98^2
9 	❌ 	✔️ 	✔️ 	❌ 	Yes! 	0.0003528 	0.03528% 	~4:10000 	0.1^0 * 0.9^1 * 0.02^2 * 0.98^1
10 	❌ 	✔️ 	❌ 	✔️ 	Yes! 	0.0003528 	0.03528% 	~4:10000 	0.1^0 * 0.9^1 * 0.02^2 * 0.98^1
11 	❌ 	❌ 	✔️ 	✔️ 	Yes! 	0.0003528 	0.03528% 	~4:10000 	0.1^0 * 0.9^1 * 0.02^2 * 0.98^1
12 	✔️ 	✔️ 	✔️ 	❌ 	Yes! 	0.0000392 	0.00392% 	~4:100000 	0.1^1 * 0.9^0 * 0.02^2 * 0.98^1
13 	✔️ 	✔️ 	❌ 	✔️ 	Yes! 	0.0000392 	0.00392% 	~4:100000 	0.1^1 * 0.9^0 * 0.02^2 * 0.98^1
14 	✔️ 	❌ 	✔️ 	✔️ 	Yes! 	0.0000392 	0.00392% 	~4:100000 	0.1^1 * 0.9^0 * 0.02^2 * 0.98^1
15 	❌ 	✔️ 	✔️ 	✔️ 	Yes! 	0.0000072 	0.00072% 	~7:1000000 	0.1^0 * 0.9^1 * 0.02^3 * 0.98^0
16 	✔️ 	✔️ 	✔️ 	✔️ 	Yes! 	0.0000008 	0.00008% 	~8:10000000	0.1^1 * 0.9^0 * 0.02^3 * 0.98^0
Total 	Total 	Total 	Total 	Total 	Total 	1 	100% 	1:1 	1
Since the order in which a rare drop is rolled does not matter, the above table can be reduced to:
 Probability that X number of rare drops rolled
Number of rare drops rolled	Notation	Probability of outcome	As a percentage	As a ratio 	Calculation
0 	P(X=0) 	0.8470728 	~84.71% 	~8471:10000	
1 	P(X=1) 	0.1459808 	~14.6% 	~1460:10000	
2 	P(X=2) 	0.0068208 	0.68208% 	~68:10000 	
3 	P(X=3) 	0.0001248 	0.01248% 	~1:10000 	
4 	P(X=4) 	0.0000008 	0.00008% 	~8:10000000	
Total 	Total 	1 	100% 	1:1 	1
Because players may only pick one reward out of four rolled, assuming they always pick the rare reward, the above table can be reduced to:
 Probability that players can get a rare drop
Rare drop?	Notation 	Probability of outcome	As a percentage	As a ratio 	Calculation
No 	P(X=0) 	0.8471728 	~84.71% 	~8471:10000	
Yes 	P(X≥1) or P(1≤X≤4)	0.1529272 	~15.29% 	~1529:10000	
Total 	Total 	1 	100% 	1:1 	1
===Expected Relics opened per player===
Say we define geometric random variable Y as the number of runs by a full 4-player squad with 1 Radiant, 3 Intact relics before seeing at least one rare drop rolled.
To get the expected (average) number of relics opened per player to get a rare item, we have to determine the distribution of the number of failures needed before getting one success. In other words, the average number of full squad runs without a rare drop before the player finally sees a rare drop. The resultant probability distribution will follow a geometric distribution.
{
\begin{aligned}
P(Y = k) &= p \times (1 - p)^{k - 1} \quad \text{where } k > 0 \\
&= \text{Chance of rolling at least one rare drop} \times (1 - \text{Chance of rolling at least one rare drop})^{k - 1} \\
&= \text{Chance of rolling at least one rare drop} \times (\text{Chance of NOT rolling at least one rare drop})^{k - 1} \\
&= \text{Chance of rolling at least one rare drop} \times (\text{Chance of rolling zero rare drops})^{k - 1}
\end{aligned}
}
 Possible number of full squad runs of 1 Radiant, 3 Intact relics before seeing a rare drop
Number of runs before getting a rare drop	Notation 	Probability of occurrence	As a percentage 	As a ratio 	Calculation
1 	P(Y=1) 	0.1529272 	~15.29% 	~1529:10000 	0.1529272 * (0.8471728)^0
2 	P(Y=2) 	0.12955576422 	~12.96% 	~1296:10000 	0.1529272 * (0.8471728)^1
3 	P(Y=3) 	0.109756119531 	~10.98% 	~1098:10000 	0.1529272 * (0.8471728)^2
4 	P(Y=4) 	0.0929823990998 	~9.3% 	~930:10000 	0.1529272 * (0.8471728)^3
5 	P(Y=5) 	0.0787721593961 	~7.88% 	~788:10000 	0.1529272 * (0.8471728)^4
... 	... 	... 	... 	... 	...
10 	P(Y=10) 	0.0343742133867 	~3.44% 	~344:10000 	0.1529272 * (0.8471728)^9
... 	... 	... 	... 	... 	...
100 	P(Y=100) 	1.1316524957e-8 	1.1316524957e-6% 	about 11 out of one billion	0.1529272 * (0.8471728)^99
... 	... 	... 	... 	... 	...
1,000 	P(Y=1000)	1.69245158744e-73 	1.69245158744e-71%	about 169 out of 10^75 	0.1529272 * (0.8471728)^999
... 	... 	... 	... 	... 	...
∞ 	P(Y=∞) 	0 	0% 	0:1 	0.1529272 * (0.8471728)^∞ converges to 0
Total 	Total 	1 	100% 	1:1 	1
Using the probability that a player will get at least one rare item in a full squad of players with a certain relic refinement, one can calculate the expected number of relics that each player will open (i.e. number of full squad runs) by using its reciprocal:
{
\begin{aligned}
E[Y] &= \frac{1}{p} \\
\text{Expected Relics opened per player} &= \frac{1}{P(1 \le X \le 4)}
\end{aligned}
}
For example, the expected number of relics that each player will open in a full squad of 1 Radiant, 3 Intact relics to get at least one rare item is:
{
\begin{aligned}
\text{Expected Relics opened per player} = \frac{1}{0.1529272} = 6.53905910786 \approx 6.5
\end{aligned}
}
===Expected Void Traces used per player===
Multiply the expected number of relics opened per player by the average number of Void Traces needed to upgrade each relic from Intact to that refinement.
For example for a full squad of players with 1 Radiant, 3 Intact relics:
{
\begin{aligned}
\text{Expected Void Traces used per player} &= \text{Expected number of Relics opened per player} * \text{Average Void Traces needed} \\
&= 6.53905910786 \times \frac{(100 + 0 + 0 + 0)}{4} \\
&= 6.53905910786 \times 25 \\
&= 163.476477697 \\
&\approx 163
\end{aligned}
}
===Number of Relics opened per player at 90% confidence level===
To get the average number of Radiant relics that 90% of the WARFRAME's player base will have to open in a full squad of 1 Radiant, 3 Intact relics before seeing at least one rare reward in all their runs, we can use the cumulative distribution function (CDF) of a geometric distribution which is the summation of the probabilities of x number of relics that are opened before seeing at least one rare reward rolled.
The CDF of a geometric distribution is:
{
\begin{aligned}
P(Y \le k) &= \sum_{x=1}^{k} P(Y = x) \\
&= 100% - \text{Chance of k number of failures in a row} \\
&= 1 - (1 - p)^{k}
\end{aligned}
}
Set P(Y ≤ k) to 90% and solve for k, given that p is 15.29272%, the chance for rolling at least one rare item from a full squad with 1 Radiant, 3 Intact relics:
{
\begin{aligned}
0.9 &= 1 - (1 - 0.1529272)^{k} \\
0.9 &= 1 - (0.8470728)^{k} \quad &\text{Simplify} \\
0.9 - 1 &= 1 - (0.8470728)^{k} - 1 \quad &\text{Subtract 1 from both sides} \\
-0.1 &= -(0.8470728)^{k} \\
\frac{-0.1}{-1} &= \frac{-(0.8470728)^{k}}{-1} \quad &\text{Remove negative sign} \\
0.1 &= (0.8470728)^{k} \\
\log_{0.8470728}{0.1} &= \log_{0.8470728}{(0.8470728)^{k}} \quad &\text{Use log to solve for exponent} \\
\log_{0.8470728}{0.1} &= k \\
13.873615679123 &= k \\
13.87 &\approx k
\end{aligned}
}
===Number of Void Traces used per player at 90% confidence level===
Multiply the average number of relics opened per player at 90% confidence by the average number of Void Traces needed to upgrade each relic from Intact to that refinement.
For example for a full squad of players with 1 Radiant, 3 Intact relics:
{
\begin{aligned}
\text{Void Traces used per player} &= \text{Number of Relics opened per player} * \text{Void Traces needed} \\
&= 13.873615679123 \times \frac{(100 + 0 + 0 + 0)}{4} \\
&= 13.873615679123 \times 25 \\
&= 346.840391978 \\
&\approx 347
\end{aligned}
}
==Mixed Runs With All Different Refinements==
A very unlikely scenario, but for the sake of completeness, here are the values for a hypothetical 4-player squad with all different refinements:
Quality 	Chance for at least one rare item	Expected relics/player	Expected traces/player	90% relics/player	90% traces/player
1 Intact, 1 Exceptional, 1 Flawless, 1 Radiant	20.41% 	4.9 	Varies
*0 Void Traces for the player always running Intact relics
*122.5 for the player always running Exceptional relics
*245 for the player always running Flawless relics
*490 for the player always running Radiant relics
*Average 214.375 Void Traces used per player	10.1 	Varies
*0 Void Traces for the player always running Intact relics
*252.5 for the player always running Exceptional relics
*505 for the player always running Flawless relics
*1010 for the player always running Radiant relics
*Average 441.875 Void Traces used per player
===Visualization===
 Possible outcomes of a full squad opening relics with different refinements
Outcome #	Player 1 (Intact, 2% Rare)	Player 2 (Exceptional, 4% Rare)	Player 3 (Flawless, 6% Rare)	Player 4 (Radiant, 10% Rare)	At least one rare reward?	Probability of outcome	As a percentage	As a ratio 	Calculation
1 	❌ 	❌ 	❌ 	❌ 	No 	0.7959168 	~79.59% 	~7959:10000	(0.02^0 * 0.98^1) * (0.04^0 * 0.96^1) * (0.06^0 * 0.94^1) * (0.1^0 * 0.9^1)
2 	✔️ 	❌ 	❌ 	❌ 	Yes! 	0.0162432 	~1.62% 	~162:10000 	(0.02^1 * 0.98^0) * (0.04^0 * 0.96^1) * (0.06^0 * 0.94^1) * (0.1^0 * 0.9^1)
3 	❌ 	✔️ 	❌ 	❌ 	Yes! 	0.0331632 	~3.32% 	~332:10000 	(0.02^0 * 0.98^1) * (0.04^1 * 0.96^0) * (0.06^0 * 0.94^1) * (0.1^0 * 0.9^1)
4 	❌ 	❌ 	✔️ 	❌ 	Yes! 	0.0508032 	~5.08% 	~508:10000 	(0.02^0 * 0.98^1) * (0.04^0 * 0.96^1) * (0.06^1 * 0.94^0) * (0.1^0 * 0.9^1)
5 	❌ 	❌ 	❌ 	✔️ 	Yes! 	0.0884352 	~8.84% 	~884:10000 	(0.02^0 * 0.98^1) * (0.04^0 * 0.96^1) * (0.06^0 * 0.94^1) * (0.1^1 * 0.9^0)
6 	✔️ 	✔️ 	❌ 	❌ 	Yes! 	0.0006768 	~0.07% 	~7:10000 	(0.02^1 * 0.98^0) * (0.04^1 * 0.96^0) * (0.06^0 * 0.94^1) * (0.1^0 * 0.9^1)
7 	✔️ 	❌ 	✔️ 	❌ 	Yes! 	0.0010368 	~0.10% 	~10:10000 	(0.02^1 * 0.98^0) * (0.04^0 * 0.96^1) * (0.06^1 * 0.94^0) * (0.1^0 * 0.9^1)
8 	✔️ 	❌ 	❌ 	✔️ 	Yes! 	0.0018048 	~0.18% 	~18:10000 	(0.02^1 * 0.98^0) * (0.04^0 * 0.96^1) * (0.06^0 * 0.94^1) * (0.1^1 * 0.9^0)
9 	❌ 	✔️ 	✔️ 	❌ 	Yes! 	0.0021168 	~0.21% 	~21:10000 	(0.02^0 * 0.98^1) * (0.04^1 * 0.96^0) * (0.06^1 * 0.94^0) * (0.1^0 * 0.9^1)
10 	❌ 	✔️ 	❌ 	✔️ 	Yes! 	0.0036848 	~0.37% 	~37:10000 	(0.02^0 * 0.98^1) * (0.04^1 * 0.96^0) * (0.06^0 * 0.94^1) * (0.1^1 * 0.9^0)
11 	❌ 	❌ 	✔️ 	✔️ 	Yes! 	0.0056448 	~0.56% 	~56:10000 	(0.02^0 * 0.98^1) * (0.04^0 * 0.96^1) * (0.06^1 * 0.94^0) * (0.1^1 * 0.9^0)
12 	✔️ 	✔️ 	✔️ 	❌ 	Yes! 	0.0000432 	0.00432% 	~4:100000 	(0.02^1 * 0.98^0) * (0.04^1 * 0.96^0) * (0.06^1 * 0.94^0) * (0.1^0 * 0.9^1)
13 	✔️ 	✔️ 	❌ 	✔️ 	Yes! 	0.0000752 	0.00752% 	~8:100000 	(0.02^1 * 0.98^0) * (0.04^1 * 0.96^0) * (0.06^0 * 0.94^1) * (0.1^1 * 0.9^0)
14 	✔️ 	❌ 	✔️ 	✔️ 	Yes! 	0.0001152 	0.01152% 	~1:10000 	(0.02^1 * 0.98^0) * (0.04^0 * 0.96^1) * (0.06^1 * 0.94^0) * (0.1^1 * 0.9^0)
15 	❌ 	✔️ 	✔️ 	✔️ 	Yes! 	0.0002352 	0.02352% 	~2:10000 	(0.02^0 * 0.98^1) * (0.04^1 * 0.96^0) * (0.06^1 * 0.94^0) * (0.1^1 * 0.9^0)
16 	✔️ 	✔️ 	✔️ 	✔️ 	Yes! 	0.0000048 	0.00048% 	~5:1000000 	(0.02^1 * 0.98^0) * (0.04^1 * 0.96^0) * (0.06^1 * 0.94^0) * (0.1^1 * 0.9^0)
Total 	Total 	Total 	Total 	Total 	Total 	1 	100% 	1:1 	1
Since the order in which a rare drop is rolled does not matter, the above table can be reduced to:
 Probability that X number of rare drops rolled
Number of rare drops rolled	Notation	Probability of outcome	As a percentage	As a ratio 	Calculation
0 	P(X=0) 	0.7959168 	~79.59% 	~7959:10000	
1 	P(X=1) 	0.1886448 	~18.86% 	~1886:10000	
2 	P(X=2) 	0.0149648 	~1.5% 	~150:10000 	
3 	P(X=3) 	0.0004688 	0.04688% 	~47:100000 	
4 	P(X=4) 	0.0000048 	0.00048% 	~5:1000000 	
Total 	Total 	1 	100% 	1:1 	1
Because players may only pick one reward out of four rolled, assuming they always pick the rare reward, the above table can be reduced to:
 Probability that players can get a rare drop
Rare drop?	Notation 	Probability of outcome	As a percentage	As a ratio 	Calculation
No 	P(X=0) 	0.7959168 	~79.59% 	~7959:10000	
Yes 	P(X≥1) or P(1≤X≤4)	0.2040832 	~20.41% 	~2041:10000	
Total 	Total 	1 	100% 	1:1 	1
==Sanity Checks==
To confirm that the math behind these probabilities are correct (as well as the listed drop chances of items according to DE), one can perform controlled experiments based on the results of their full-squad Void Relic openings.
===Chi-squared test===
A chi-squared test can be used to determine if there is any statistically significant difference between observed results and expected results.
==Interactive Graph==
==External Links==
*3.2: Probability Mass Functions (PMFs) and Cumulative Distribution Functions (CDFs) for Discrete Random Variables
Category:Real-World References