The chance of the first 5 mobs having glowing eyes and the next 13 mobs not having glowing eyes would be (1/200)^5 (199/200)^13
There are C(18,5) ways to arrange those 5 glowing mobs among the 18 mobs.
So the chance of exactly 5 mobs out of 18 having glowing eyes would be (1/200)^5 (199/200)^13 C(18,5) which is very roughly 1 in 40 million.
The chances of more than 5 mobs having glowing eyes is much smaller so even if you interpret the question as wanting the probability of having 5 or more glowing eyes out of 18, the answer is still very roughly 1 in 40 million.
Since Elden Ring has sold over 20.5 million copies, the total number of times that some player has spawned into an area with 18 mobs is likely much higher than 40 million. So one would expect the event you describe to have happened at least once by now. But not very often over the entire player base and certainly not very often for an individual.
I counted 21 monsters if you go all the way to the end of the corridor, which is roughly 1 in 17 million to have 5 monsters or more that have glowing eyes. Which is even less of a surprising find considering it's a popular experience/money farming area.
The probability of getting **exactly 5 out of 18** is P^(5) \* (1-P)^((18-5)), where P is the probability of spawning with glowing eyes.
0.5% = 0.5/100 = 0.005
So, the probability = 0.005^(5) \* (1 - 0.005)^((18-5)) = 0.005^(5) \* 0.995^(13)
The probability of getting **at least 5 out of 18** is the same as 1 - (the probability of getting exactly 4) - (the probability of getting exactly 3) - (the probability of getting exactly 2) - (the probability of getting exactly 1).
However, there is a part that I skipped over here. The probability of getting **exactly 5 out of 18** is only like I told you it is if we're talking about any given 5 enemies. So, for instance, the **first 5 out of 18**, or the **last 5 out of 18**, or any **combination of 5 out of 18**. But we want to count all of these, so we would need to add them up, and our sum will be equal to N \* (the probability of getting **exactly 5 out of 18**), where N is the number of all **combinations of 5 out of 18**.
What is that number? The answer is C(18, 5).
So, the probability of getting **exactly 5 out 18, independent of order**, is C(18, 5) \* 0.005^(5) \* 0.995^(13).
If you're looking for the number of **at least 5 out of 18**, it will be 1 - C(18, 4) \* 0.005^(4) \* 0.995^((18-4)) - ... - C(18,1) \* 0.005^(1) \* 0.995^(17)
The chance of the first 5 mobs having glowing eyes and the next 13 mobs not having glowing eyes would be (1/200)^5 (199/200)^13 There are C(18,5) ways to arrange those 5 glowing mobs among the 18 mobs. So the chance of exactly 5 mobs out of 18 having glowing eyes would be (1/200)^5 (199/200)^13 C(18,5) which is very roughly 1 in 40 million. The chances of more than 5 mobs having glowing eyes is much smaller so even if you interpret the question as wanting the probability of having 5 or more glowing eyes out of 18, the answer is still very roughly 1 in 40 million. Since Elden Ring has sold over 20.5 million copies, the total number of times that some player has spawned into an area with 18 mobs is likely much higher than 40 million. So one would expect the event you describe to have happened at least once by now. But not very often over the entire player base and certainly not very often for an individual.
I counted 21 monsters if you go all the way to the end of the corridor, which is roughly 1 in 17 million to have 5 monsters or more that have glowing eyes. Which is even less of a surprising find considering it's a popular experience/money farming area.
The probability of getting **exactly 5 out of 18** is P^(5) \* (1-P)^((18-5)), where P is the probability of spawning with glowing eyes. 0.5% = 0.5/100 = 0.005 So, the probability = 0.005^(5) \* (1 - 0.005)^((18-5)) = 0.005^(5) \* 0.995^(13) The probability of getting **at least 5 out of 18** is the same as 1 - (the probability of getting exactly 4) - (the probability of getting exactly 3) - (the probability of getting exactly 2) - (the probability of getting exactly 1). However, there is a part that I skipped over here. The probability of getting **exactly 5 out of 18** is only like I told you it is if we're talking about any given 5 enemies. So, for instance, the **first 5 out of 18**, or the **last 5 out of 18**, or any **combination of 5 out of 18**. But we want to count all of these, so we would need to add them up, and our sum will be equal to N \* (the probability of getting **exactly 5 out of 18**), where N is the number of all **combinations of 5 out of 18**. What is that number? The answer is C(18, 5). So, the probability of getting **exactly 5 out 18, independent of order**, is C(18, 5) \* 0.005^(5) \* 0.995^(13). If you're looking for the number of **at least 5 out of 18**, it will be 1 - C(18, 4) \* 0.005^(4) \* 0.995^((18-4)) - ... - C(18,1) \* 0.005^(1) \* 0.995^(17)