Extraordinary and legendary items

Xero · Post by **Xero** » Tue Dec 25, 2012 2:26 am

In a gambling game of chance, let's say roulette. There are 38 numbers that may win. 1-36, 0, 00. the odds a single number will hit are 1:38. After playing the game for 3 rounds, 0 hit twice and 36 has only hit once. A history of the last 100 games reveals that black has occurred 60 times, red 35 and green (0, 00) 5.
what is the probability the next winning number will be 0? What is the probability it will be black?

I have flipped a coin 10 times, 8 times it landed on heads, twice on tails. What is the probability it will be heads again?

Post by **Pyrizzle** » Tue Dec 25, 2012 3:51 am

Xero wrote:In a gambling game of chance, let's say roulette. There are 38 numbers that may win. 1-36, 0, 00. the odds a single number will hit are 1:38. After playing the game for 3 rounds, 0 hit twice and 36 has only hit once. A history of the last 100 games reveals that black has occurred 60 times, red 35 and green (0, 00) 5.
what is the probability the next winning number will be 0? What is the probability it will be black?

I have flipped a coin 10 times, 8 times it landed on heads, twice on tails. What is the probability it will be heads again?

Roulette spin being 0/00: 2/38 (2 spaces green, 36 spaces black/red)

Coin being heads again: 50/50 (unless it lands on it's side

)

Very well put examples Xero. +1

Xero · Post by **Xero** » Tue Dec 25, 2012 6:32 am

That is correct. The coin does not have a memory. Neither does the roulette wheel. Using the mathematical law of averages in the wrong context can be hazardous. If, after 20k kills your legendary has not dropped, the next time you kill a mob the chance that it will drop is 1:10000.

SlipEternal · Post by **SlipEternal** » Wed Dec 26, 2012 2:16 pm

Xero wrote:In a gambling game of chance, let's say roulette. There are 38 numbers that may win. 1-36, 0, 00. the odds a single number will hit are 1:38. After playing the game for 3 rounds, 0 hit twice and 36 has only hit once. A history of the last 100 games reveals that black has occurred 60 times, red 35 and green (0, 00) 5.
what is the probability the next winning number will be 0? What is the probability it will be black?

I have flipped a coin 10 times, 8 times it landed on heads, twice on tails. What is the probability it will be heads again?

If you flipped a coin 10 times, 8 times it landed on heads, twice on tails, what is the probability it landed on heads 8 times? 100%. If you flip the coin another 10 times, what is the probability it will land on heads 8 times again? This is a two-outcome experiment repeated multiple times (ignoring any outcomes where the coin does not land on heads or tails, such as it lands on its side, or it is sucked into a dimensional vortex, or any other outcome that is neither heads nor tails). Let's consider all possible ways that it can land on heads 8 times and tails twice. I will write all possible sequences of ten characters where the characters are either H or T (for heads or tails respectively) that satisfy the condition that 8 of the characters are H and two of the characters are T:

I put it in spoiler tags because there are 10C2=10C8=45 different ways of writing that sequence (10C2 is the evaluation of the probability expression nCr where n=10 and r=2 or r=8). What is the probability that if we flip a coin ten times, we get one of those sequences? Let's look at the first one: HHHHHHHHTT. All we do is multiply the probabilities: (0.5)*(0.5)*(0.5)*...*(0.5)=(0.5)^10=0.0009765625. It is not possible that any flip of a coin will be simultaneously heads and tails (since the experiment already stated it ignores any outcome that is not either heads or tails, repeating the experiment as many times as is necessary to obtain 10 trials where the outcome is either heads or tails). Thus, these probabilities are disjoint. So, by the sum principle of probabilities, if we have multiple disjoint events, the probability that one of them will occur is the sum of their probabilities. We have 45 disjoint events, all with the same probability. So, the probability that we flip a coin ten times and we have exactly eight flips that are heads and exactly two flips that are tails is 45*0.0009765625=0.0439453125. So, what are all of the possibilities:

The following table is the number of tails flipped and the probability of achieving that result:

Code: Select all

Number of Tails        Probability        Formula
---------------        -----------        -------
0                      0.0009765625       (10C0)*(0.5)^0*(0.5)^10
1                      0.009765625        (10C1)*(0.5)^1*(0.5)^9
2                      0.0439453125       (10C2)*(0.5)^2*(0.5)^8
3                      0.1171875          (10C3)*(0.5)^3*(0.5)^7
4                      0.205078125        (10C4)*(0.5)^4*(0.5)^6
5                      0.24609375         (10C5)*(0.5)^5*(0.5)^5
6                      0.205078125        (10C6)*(0.5)^6*(0.5)^4
7                      0.1171875          (10C7)*(0.5)^7*(0.5)^3
8                      0.0439453125       (10C8)*(0.5)^8*(0.5)^2
9                      0.009765625        (10C9)*(0.5)^9*(0.5)^1
10                     0.0009765625       (10C10)*(0.5)^10*(0.5)^0

If we add up the probabilities, we get a total of 1 (in other words, we exhausted all possibilities, and the number of tails we get must be a number from 0 to 10). Now, what is the probability that in those 10 flips, we get at least 1 tails? That is the probability that we don't get heads every flip, or 1-0.0009765625=0.9990234375. So, there is a 99.9% chance that if we flip a coin 10 times, we will get at least one heads.

We can create a similar table for success or failure of receiving legendary items during 10 kills:

Code: Select all

Number of Items        Probability                              Formula
---------------        -----------                              -------
0                      0.99900044988002099748020998800045       (10C0)*(0.0001)^0*(0.9999)^10
1                      0.00099910035991601259874008399640       (10C1)*(0.0001)^1*(0.9999)^9
2                      0.00000044964012597480314974801260       (10C2)*(0.0001)^2*(0.9999)^8
3                      0.00000000011991602519580041997480       (10C3)*(0.0001)^3*(0.9999)^7
4                      0.00000000000002098740314958003150       (10C4)*(0.0001)^4*(0.9999)^6
5                      0.00000000000000000251874025197480       (10C5)*(0.0001)^5*(0.9999)^5
6                      0.00000000000000000000020991601260       (10C6)*(0.0001)^6*(0.9999)^4
7                      0.00000000000000000000000001199640       (10C7)*(0.0001)^7*(0.9999)^3
8                      0.00000000000000000000000000000045       (10C8)*(0.0001)^8*(0.9999)^2
9                      0.00000000000000000000000000000000       (10C9)*(0.0001)^9*(0.9999)^1
10                     0.00000000000000000000000000000000       (10C10)*(0.0001)^10*(0.9999)^0

(Note: The last two probabilities are so small, their first significant digit is too far to the right to be accurately represented).
Now, if we want to calculate the probability of getting at least one item, we can add up the probability of getting exactly 1, exactly 2, exactly 3, ..., exactly 10, or we can just subtract the probability of getting 0 from the total probability of 1: 1-0.99900044988002099748020998800045, which is just less than 0.1%.

Let's say we calculate this same table for x kills (omitting the probability, keeping only the formula):

Code: Select all

Number of Items        Formula
---------------        -------
0                      (xC0)*(0.0001)^0*(0.9999)^x

Now, the probability of finding an item after x kills is 1-(xC0)*(0.0001)^0*(0.9999)^x=1-(0.9999)^x, which is the formula I used in my original post.

Post by **Sarumar** » Fri Dec 28, 2012 12:53 am

Intesting... One question for you all; how long (in real life time) it takes before player gets one ring ... and then rings (like Nyktos have

) ?

SlipEternal · Post by **SlipEternal** » Fri Dec 28, 2012 7:24 pm

I compiled a table that is a bit more complete (it includes the effects of the Magic Finder skill).

Code: Select all

MFL     LD %       LD (Ex)     LD (50%)     LD (95%)     LD (1-LD %)     ED %      ED (Ex)     ED (50%)     ED (95%)     ED (1-ED %)
---     ------     -------     --------     --------     -----------     -----     -------     --------     --------     -----------
0       0.010%     10000       6931         29956        92099           0.10%     1000        693          2994         6904
1       0.015%     6667        4621         19970        58695           0.15%     667         462          1996         4332
2       0.020%     5000        3465         14977        42582           0.20%     500         346          1496         3104
3       0.025%     4000        2772         11981        33172           0.25%     400         277          1197         2394
4       0.030%     3333        2310         9984         27035           0.30%     333         231          997          1933
5       0.035%     2857        1980         8558         22732           0.35%     286         198          854          1613
6       0.040%     2500        1733         7488         19556           0.40%     250         173          747          1378
7       0.045%     2222        1540         6656         17121           0.45%     222         154          664          1198
8       0.050%     2000        1386         5990         15198           0.50%     200         138          598          1057
9       0.055%     1818        1260         5445         13643           0.55%     182         126          543          943
10      0.060%     1667        1155         4991         12361           0.60%     167         115          498          850
11      0.065%     1538        1066         4607         11286           0.65%     154         106          459          772
12      0.070%     1429        990          4278         10374           0.70%     143         99           426          706
13      0.075%     1333        924          3993         9590            0.75%     133         92           398          650
14      0.080%     1250        866          3743         8910            0.80%     125         86           373          601
15      0.085%     1176        815          3523         8314            0.85%     118         81           351          559
16      0.090%     1111        770          3327         7789            0.90%     111         77           331          521
17      0.095%     1053        729          3152         7322            0.95%     105         73           314          488
18      0.100%     1000        693          2994         6904            1.00%     100         69           298          458
19      0.105%     952         660          2852         6529            1.05%     95          66           284          432
20      0.110%     909         630          2722         6190            1.10%     91          63           271          408

Table Key:
MFS = Magic Finder Skill
LD % = The percent chance of a legendary item being dropped
LD (Ex) = The expected number of monsters one needs to kill prior to a legendary item dropping
LD (50%) = 50% of all people will kill this number of monsters before their first legendary item will drop
LD (95%) = 95% of all people will kill this number of monsters before their first legendary item will drop
LD (1-LD %) = The number of monsters you need to kill so that the chance of you NOT finding a legendary item equals the chance of you finding one in a single drop
ED % = The percent chance of an extraordinary item being dropped
ED (Ex) = The expected number of monsters one needs to kill prior to a legendary item dropping
ED (50%) = 50% of all people will kill this number of monsters before their first legendary item will drop
ED (95%) = 95% of all people will kill this number of monsters before their first legendary item will drop
ED (1-ED %) = The number of monsters you need to kill so that the chance of you NOT finding an extraordinary item equals the chance of you finding one in a single drop

Note: The expected value is the mean, the 50% mark is the median. The distribution is a binomial distribution. From these statistics, it is clear that the first level of Magic Finder is the most significant. After that, it quickly stagnates, and each skill increase beyond the first offers less of a gain.

slayer · Post by **slayer** » Sat Dec 29, 2012 11:30 pm

Well I finally got my ElyR. 2000 weak potions at 0 magic finder, and 1147 with 8 MF. That was the last piece for my legendary/extraordinary collection. And since I'm a glutton for punishment, time to go pick up a second RoLS.

I'm really looking forward to some new content.

Post by **Pyrizzle** » Sun Dec 30, 2012 12:21 am

slayer wrote:Well I finally got my ElyR. 2000 weak potions at 0 magic finder, and 1147 with 8 MF. That was the last piece for my legendary/extraordinary collection. And since I'm a glutton for punishment, time to go pick up a second RoLS.

I'm really looking forward to some new content.

Congratulations on your ELYR Slayer! & good luck on your rols hunt!

slayer · Post by **slayer** » Tue Jan 01, 2013 12:58 am

Well it didn't take too long. I got RoLS #2 today at lunch. Of course I used the clear and load method, which to me feels like cheating. But it works

And it cuts down on a long grind in an area that is pretty poor XP at level 123.

SlipEternal · Post by **SlipEternal** » Tue Jan 01, 2013 4:46 pm

One last post about statistics related to finding legendary and/or extraordinary items. Since Samuel posted about the "measure items" as a way to determine approximately when you might obtain an item, I determined the statistics relating when you have a certain number of "measure items" to when you have a certain probability of obtaining legendary/extraordinary items. For these formulas, they come from analysis of the hypergeometric distribution.

Notation:
p = probability of a measure item dropping after 1 kill
q = probability of a legendary/extraordinary item dropping after 1 kill
n = number of measure items found
x = probability of finding at least one legendary/extraordinary item after finding n measure items

Formula:
n = Log[1-x]/Log[(p-pq)/(p+q-pq)]

In other words, you need to find at least (n) measure items before you will have probability (x) of finding at least one legendary/extraordinary item. As a quick example, by the time you find 9,214 weak poisons, you have a 99.99% chance of finding at least one ElyR if your Magic Finder skill is at 0.

Andor's Trail

Extraordinary and legendary items

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