[Necha54]

Ok so i know its more of a rarity to find three rare monster in episode 2... But there is a room in the temple with 2 lilies, 2 rappies, and a hildebear/hidelt. .... Ive found a love happy and a nar Lilly before.... But i was wondering what the odds are of finding all three

[zoulen]

It is possible, however it'll be very rare and would be a Kodak moment. I think the best rare monster run I've seen is a friend doing the Addicting Food quest and getting 3 Nar Lilly's in a single map. Unlucky for him though because none of them dropped anything.

If you're seeking a percentage of that happening, however it's impossible to actually figure out.

[Necha54]

Well i believe i can do it cuz i started a new ranger and my first runs through caves and my first lily was a nar lily.... Haha So....

[SStrikerR]

The belief is that on versions prior to BB the rare monster appearance rate is 1/512. So in that case, it can be calculated to 7.4505806 * 10^-9, which is 7,450,580,600. Obviously the more enemies there are the better your chances are, but...yeah good luck with that rate.

If you're talking BB, rare monsters work on an entirely different system that I'm not getting into.

Spoiler!

I'm extremely tired and used google to do that, so the math may be off. Sue me and do it yourself if it really matters.

[TrueChaos]

i once has 2 Nar Lillies in the same room in Caves once. saw i pic of someone who was in Temple and had an Ob Lilly and a Torr in the same room

[Link1275]

I approximate your chances of seeing all 3 at the exact same time are

Spoiler!

.1111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 1111111

[MAD1b]

I remember once while hunting the Lavis Cannon on my Redria in Addicting Food, I found two red slimes at the same time. If I was lucky enough to see that, why is it that I still don't have a Lavis Cannon? Haha.

[Mitz]

Had Del Rappy and Pazuzu at the same time once but never 3 rates at once.

[GCoffee]

If the chance of a rare enemy appearing is indeed 1/512, then the only other number we need is the amount of enemies (with a rare version) that spawn in said room, which in your case is 5; the rest is easy math:

The chance of a single rare enemy appearing is 5/512, logically.
The chance of two rare enemies appearing is (5/512)*(4/512) = 5/65536 = 0.0000762939453125
The chance of three rare enemies appearing is (5/512)*(4/512)*(3/512) = 15/33554432 = 4.470348358154297×10-7

And as bonus:

The chance of four rare enemies appearing is (5/512)*(4/512)*(3/512)*(2/512) = 15/8589934592 = 1.746229827404×10-9
and for five rare enemies appearing (5/512)*(4/512)*(3/512)*(2/512)*(1/512) = 15/4398046511104 = 3.410605131648481×10-12


...Happy hunting, lol

[T0m]

Your math's a little bit off.

Originally Posted by GCoffee

The chance of a single rare enemy appearing is 5/512, logically.

Yeah that's almost right. (It's a little more complicated and should be 1-(511/512)⁵ or 100% minus the chance of a normal appearing to the fifth. It works out as 0,00977 against 0,00973, so the difference is negligible).
But it's downhill from there

Originally Posted by GCoffee

and for five rare enemies appearing (5/512)*(4/512)*(3/512)*(2/512)*(1/512) = 15/4398046511104 = 3.410605131648481×10-12

The chance of all 5 of them being rare is (with a rare monster chance of 1/512) simply (1/512)⁵, and not 5!(1/512)⁵. Striker already posted the correct number.
The reason is that in the first case there are 5 different possible outcomes that give you 1 rare out of 5 monsters, but there is only one single possible outcome that gets you 5 out of 5 rare monsters.

Let's reduce it to 1/6 to make it easier: dice. With two dice the chance of rolling a six is almost but not exactly 2*1/6. With six dice, this logic would give a probability of 6 * 1/6=1 of getting a six. And clearly -to anyone ever playing with dice- this isn't right.

With two dice the probability is 1-(5/6)²=0,306, and with six dice that is 1-(5/6)⁶=0,665.


...What? I lost everyone's interest right off the bat with the word "math"?