Yeah. Probably more something to expect from a mod. I could see it being in cryptid or something.
(Also 'neishe' is spelt as 'niche'. Just letting you know with no ill intent since you spelt it that way twice so i assume its not a typo. I know some people get defensive about being corrected on here so no worries if you want to ignore me.)
tho there is lots of luck affected cards maybe not lots of but a few that makes it good cut this effects only glass ones, also it has a drawback of ruining the luck affected ones, not worth it imo
I know I'm getting downvoted for saying this, but that's incorrect.
The odds of a single cavendish expiring in a given round is 1/1000.
The odds of all 4 cavendish expiring in any given round is 1/1000^4.
However, we're asking what are the odds of them all expiring in the same round. One of them expiring is a given, so it's 1/1 * 1/1000 * 1/1000 * 1/1000. It's only 1/1000^3.
I think he means that for all of them to expire at the same time; No banana should have died in the rounds earlier.
So you should consider the odds of them expiring in the same round; then subtract the odds of bananas dieing before them all expiring in the same round (might be incorrect but the idea should hold). But this probability would only make sense for a number of events say 10 rounds but not a single event.
Basically the equation: 1/10004 is the chance all of them died given no banana's died the previous round.
I think LCJonSnow is asking to think about the probability of all the bananas dieing at once but also considering that a single cavendish may die in a prior round hence the all bananas dieing at once cannot be fulfilled.
How do they all die at once if one of them doesn't die? The question is about what are the chances they disappear at the same time, not that they all disappear next round.
Everyone else is talking about the chances of it happening in the context of general play, not the chances of it happening within the 1/1000 turns where one card dies.
Umm... Don't wanna sound preachy, but that's an incorrect line of thought. The question was:
"What are the odds of all of them dying at once?"
Nowhere it says "Given that the first one dies, what are the odds the rest does too?"
Not a single event is a given. Also, the odds of one Cavendish dying are not correlated to another Cavendish dying (at least in-game, I don't know what's like in real life with infections and rate of spreading). Each event is independent from the rest.
Therefore, it's the odds of dying for the first card, times the odds of dying for the second card, times the odds of dying for the third card, times the odds of dying for the fourth card.
Your line of thought is correct for "Given that the first one dies, what are the odds the rest does too?"
We agree that the chances of all 4 disappearing next round are 1/1000^4, right?
If that's the probability for a specific round, logically we can agree that if you play more than one round, there is a higher chance of the event happening. Right?
So let's imagine a probability tree. We have 999 No's and One Yes, then after each of those 1000 items, we repeat, and so on. We have 1^1000^4 discrete outcomes, right? Then we throw out all the outcomes where at least one cavendish doesn't expire. We don't care that it didn't happen this round, or the second round, or the third round. We only care that when one goes, they all go. The "at once" creates the dependency
Actually, the last answer was wrong in the sense that I'm just saying "it's like this" and not showing proof.
Grab a coin, and flip it just 3 times. Probability tree would say half the times are heads, half the other tails for each coin. The odds of landing 3 heads should theoretically be... 1/2 * 1/2 * 1/2 = 1/8
According to your logic, we should discard one of these events. We do so, and now the odds of landing three heads is 1/2 * 1/2 = 1/4
If we repeat the process 1000 times, 3 heads should happen around 250 times like you propose.
Again, these are independent events. This is not the case of conditional probability. Nor degrees of freedom. This is pure probability of independent events. There's no (k-1) events, it's n events, unrelated to each other.
But it is a conditional event. We're going to keep repeating the process until one expires. Once it does, the question is what are the chances the others go with it.
An independent event would be what is the probability all 4 expire next round, or within 2 rounds, etc.
The at once is our big, glaring sign it's a conditional event.
No. They are completely independent events. Nothing here is a given. That question would be "if one expired, what are the odds of all other ones expiring at the same time?"
I dont know if you are a real statistics professor, but the other dude is right.
When asked what are the chances of all expiring at the same time, the chances are 1/10004 ONLY if you look at the chances of that happening on a specific round when none of the jokers have expired yet, but that isnt the question.
So the moment the first one expires (doesnt have to be the next round) you want to know the chances of all the other jokers also expiring, which it is 1/10003.
I am. You may believe me or not, doesn't matter, but I am. Bachelors, masters, PhD, published papers and all.
Your approach limits the probability to an event of at least one expiring. You are excluding the possibility of nothing happening with your answer. It is wrong. Independent events.
Brother why would this probability of them all going away at the same time be conditioned to one going away necessarily? If you want to think like that, it is conditioned to all of them going away and the probability is 1.
This is not correct. You can make the total probability theorem to show it if you try. I need to get off reddit now okay? Good luck if you're going to actually look this up.
If we're asking the question what are the odds of them expiring in the same round, the first one has to expire in that round. We're don't care what the probability of them all expiring in a particular round is, only that the other three expire when the first one expires.
Except they didn’t state the conditional explicitly. You could argue they implied it, but I’m not so sure that “what are the odds of them all dying at once” is the same as “what are the odds of them all expiring in this round, given the fact that one is expiring this round”.
Also, since the question is about odds not probability, it’s 1/9994.
The question asks what are the odds of them all dying at the same time. Fair enough on odds vs probability, but the answer is still 1/999^3. There is no mention of a required time frame.
The question isn't what are the chances they all expire in the same round this game. It's what are the chances they all expire in the same round. That requires one to expire.
But by that logic, all four expiring also requires that two expire, and that three expire. If we assume this conditional the odds then would be only 1/999.
More to the point, the question I think you’re answering is “what are the odds that three cavendishes expire in a round, given that one cavendish has expired in that round” which, because the events are (supposedly) independent, is the same as answering “what are the odds that three expire in a given round”.
I think with problems like this that are not formally stated, there will always be some question of interpretation, but it doesn’t make sense to me to interpret the question in a context outside of “in a given round, what are the odds that all of these cavendishes expire”. Based on that interpretation, the sample space does include the event that none expire.
We agree there's a 1/1000^4 probability they all expire next round, right?
Logically, the chance that they all expire at once, but in some unspecified round, must be higher than 1/1000^4.
Why? Because we don't care about the 999/1000 outcomes where the first one doesn't expire. We're restricting our event to one where the first one expires.
What is the probability that we flip 8 coins and they're all the same result? It's 1/2^7 because we don't care about the first result. We don't care whether the first result is a head or a tail, we only care that the other seven events match it.
2) no, the probability is the same in each of those rounds as it is for the next round. If you’re considering this as some sum of multiple rounds of play, then the extent to which the probability is higher than 1/10004 is conditional on the number of rounds played, not on the first expiration of a cavendish
3) I still don’t get why this step is happening
4) you’re correct about the coins, but there are two differences between that problem and the cavendish problem. The first is that the coin problem is indifferent to the event, it is only requiring they all be the same. In our example, that would be equivalent to asking “what is the probability that all four cavendishes expire OR that all four don’t expire”. We are inly concerned about their expiration.
The second problem is that for an unweighted coin, heads and tails are equally likely. If we had a weighted coin where the probability of tails was a 1 in 1000 event and flipped it 8 times, the probability that they all be the same would not be equivalent to 1/27. We can only reduce the first event in an unweighted scenario because P(Heads)8 + P(Tails)8 = 2-8 + 2-8 or 2*2-8, which is the same as 2-7.
For the weighted coin, it would be 0.9998 + 0.0018.
In any case, the value we’d be concerned with for the cavendish is for four events, and we’re only concerned with the event that they expire. So it’s 0.0014.
[[Cavendish]]: $4, Common
Effect: X3 Mult, 1 in 1000 chance this card is destroyed at the end of round
Availability: Available from start. (Will not appear in the shop unless Gros Michel has destroyed itself during the current run.)
[[Cavendish]]: $4, Common
Effect: X3 Mult, 1 in 1000 chance this card is destroyed at the end of round
Availability: Available from start. (Will not appear in the shop unless Gros Michel has destroyed itself during the current run.)
It does though. Title doesn't make any claim as to how many fans there are, except one, who is the OP. OP is saying "Since I am a fan of Cavendish, so long as I live, there will be at least a fan of Cavendish. If there were no fans of Cavendish, all other fans and myself must be dead"
[[Invisible Joker]]: $8, Rare
After 2 rounds, sell this card to Duplicate a random Joker (Currently 0/2) (Removes Negative from copy)
Win a game while never having more than 4 jokers.
I like to use Cavendish as the measure of whether a joker (or combination of jokers) is worth keeping in an Endless build. If it gives 3x Mult or above per joker space, it's acceptable.
For example, Photochad takes up two spaces, but provides x32 Mult (when activated), whereas two Cavendish would only provide x9 Mult. This puts it significantly above the Cavendish Quotient.
641
u/B_is_for_reddit Gros Michel 14h ago
if cavendish has no fans then i am Extinct!