#add an achievement for having both gros michel and cavendish in your joker slots at once
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get a gros michel, duplicate it with ankh or invis, have one die, then find cavendish before the second michel dies
So difficult but not impossible
its just an achievement, it wouldn't unlock anything
i mean it could
Fair yeah
If a Gros Michel is copied, all copies go extinct at the same time
I already don't like that implementation (despite it making logical sense) but id love this achievement
can you not buy gros michel if you have cavendish?
or does it extincting make it disappear from the shop
this is false
Yes, and I'm disappointed that Cavendish can reappear in the shop after going extinct
this isnt false it does happen
there's nothing here that indicates it would. each gros michel is rolling the probability independently
its litterally happend to me twice
becuase i have done it twice
that's unlucky i guess
and its in the If psudorandom
it's not
you use the same pseudorandom key for analogous rolls and get different results every time
otherwise gros michel would either go extinct in one round or never at all
do you want me to test it to conferm that it does infact always trigger at the same time?
like i know its how it works
here you go (this is on an almost unmodified version of the base game with debug mode turned on)
all of them were safe except the rightmost one
after two more rounds
okay i just tested it
using the debug plus tool
and your right
wow
that just means i had microspocic odds then
or the debug is diffrent but i doubt that lovely and the tool would change anything
i mean i did the test on a non-steamodded version of the game so
yeah.. still the odds of it happaning for me twice is bonkers
... huh. Color me surprised š
Could've sworn Invisible Joker copied every property of an existing joker, since it also copies enhancements and Perishable timers
ikr
Or showman
Itās just 1/36
So like 3%
thats for it once
i had it happen twice sepretly
1/1296
or 0.0007%
Maybe Iām misinterpreting this but what Iām doing is looking at the round where one grow michel breaks, then seeing if the other does which gets to 1/36. 1/1296 applies to looking at a random round and then seeing if both break
two breaking at once is 1/36
Im assuming one of them breaking as a given
Idk if that approach is right though
well 1/6 each round for a gros michel
so you can multiply them together to 1/36 that both will break in the same round
i had it happen twice and so i multiplyed the 1/36's together making it a 1/1296
Itās a 1/6 for any given round for both. However if we look at the round where at least one breaks there would be a 5/6 chance for only one to break (which would be the proof that two gros michel can decide to break at separate times) this means we see a 1/6 chance for two to break. I believe this is the correct way to model this as you stated you believed this due to you seeing both break twice, this implies you went until both broke, which is why I am looking at specifically a round where at least one broke.
but like thats dependent on wither the first one breaks which is a 1/6, your saying that when one breaks the other one has a 1/6 chance of also going which is true
your framing the question diffrently than it is and so the number you reach doesnt awnser it correctly
i unfortunatly havent gotten to do statisitics in school yet (to a meaningfull degree) so i might be wrong
but your framing feels weird
alright yeah have a good day
it'd be 10/11 and 1/11 respectively
25/36: neither break
5/36: banana 1 breaks
5/36: banana 2 breaks
1/36: both bananas break
P(both bananas break | at least one banana breaks) = (1/36) / (5/36 + 5/36 + 1/36) = 1/11
so i guess there's a 1/121 chance that both times you have two bananas they break on the same round
not that rare considering that someone has to get these odds, like out of the 20k players actively, someone would get these odds
Oh
The goat
wait... so does cavendish spawn when the game tries to spawn gros after you broke it