| Author | Flooders Tenthouse |
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you cannot be more correct in saying that if something has a probability of 1/1000..then the probability of that happening at least once in 50 attempts (our case) is 1- (999/1000)^50..this the the first thing they teach to 16 year old kids
but this has nothing to do with what we're concerned with (see 4780) |
that is the odds of occurring ONCE. what we're dealing with is a poisson distribution.
Right, but we ARE interested in the probability of it occurring once, as I have mentioned before. I understand that you can make a Poisson model of this, but I don't see a particular use of that in calculating the desired probability. I feel like you are unnecessarily complicating this, as I have mentioned above, it's a very simple calculation that gives us our desired probability.
depend upon each other not directly..but the entire event of 2 arts can only take place when it's initiated..that is the basis of conditional probabilities..of course pulling one is independent in itself
Yes the "condition" here that the first one has already been pulled. Because as I mentioned, we are looking for the probability of two in a row, given 2nd happens within 50 attempts of the first, not BOTH 1st and 2nd happen within 50 attempts of starting. The number of attempts it takes you to reach to 1st success is arbitrary, it can take as long as it wants, we don't care, so it is not at all a part of the calculations.
Ok read this part alone if you don't wanna read the above, this will surely make it clear as day.
Imagine a simple case. What is the probability of rolling a 3 on a dice twice, such that the second 3 is within 10 rolls of the first 3?
It is 1-(1/6)^10. Which is exactly the same as the probability of rolling a 3 in 10 well defined attempts. The probability is the same, as in both cases you roll a 3 only once. The first case, the first 3 you roll is somewhat of an illusion, you don't count that because that itself is your well defined start point. This has nothing to do with something like Poisson or Gaussian distributions. Yes you can model it using that, but it is not required for the very simple calculation I made. |
ok i see what you're saying; with staring at a point where our first outcome has already occurred and looking at the probability of the same happening second time within 50 attempts of that..but dude.. this clarifies the difference in the answers both of us were looking for (and any bystanders that bothered to read)
This takes me back to the question we had at hand..you totally negated the "fortune" of pulling a rare art in the first place..made it the exact same as the probability of rolling a 3 in 10 well defined attempts
This is the sole reason conditional probabilities were figured out..to take the first success into account..
you made it seem that the person was as lucky as pulling 2 art in 50 boxes as 1..are u kidding me? |
for Pankaj_Kalra21:
Well in conditional probability, you typically assume an event has occurred, which in this case is the first loot.
you made it seem that the person was as lucky as pulling 2 art in 50 boxes as 1..are u kidding me?
I feel like you are still misunderstanding me. I didn't say that. And that's not really relevant. Why? Because the guy who got the 2nd art in say about 50 attempts, you don't know how many attempts he took to get the first art. So it could be 51 attempts to get two arts if he got the first art in 1 attempt(which is extremely lucky), or he took 1050 attempts(1000 for the first, which is quite unlucky). The point is... we don't know.
I will give you one more example, that happened to me. I have won 2 straight ups in roulette in a row a few times. What's the chance to win two straight ups in a row, given it took me any arbitrary number of attempts to get the first straight? (1/38)^2? No, it's only 1/38. So I am claiming that getting two straight ups is just as likely as getting one. Sounds like a paradox? Certainly not, because it will be (1/38)^2 if I say you get two straights in a row, however, your first straight win is your FIRST attempt. In this case there is a very well defined start point. However if you are asking what's the probability of getting 2 straights in a row in general, it is 1/38 no matter how you put it or calculate it, simply because there is no condition or bound on the number of attempts I am allowed to get the first straight up. |
You don't need to apply conditional probability here because we are simply calculating the chance that an event x with a certain low probability occurs within y tries.
Its a very beginner question but i can see that you have a complicated thought process :p |
we are simply calculating the chance that an event x with a certain low probability occurs within y tries.
Yeah but the only reason you can assume it's just one event x is because of conditional probability. There are actually 2 events happening, first loot, and second loot, and due to conditional probability you can ignore the first loot.
Also, I'm using conditional probability in a very very basic conceptual sense, not mathematical sense, so I don't think it's a complicated thought process. |
| You can simply ignore the first success because we do not care about how many boxes were opened. We are simply concerned about the probability of another success within 50 tries. And my message was addressed to Pankaj_Kalra because he was insisting on conditional probability. |
occurs within y tries
the complicated thought process comes from the fact that we do not have a definite sample size to judge..perhaps, in the real world it's wiser to choose one..Like in the span of 50 he got 2 (very lucky). on the contrary, assuming he's still opening crates and only getting windflowers, that makes him extremely unlucky..
Se to end it, we decide to deem how lucky he was in that particular span of 50 crates..and that should be simple to figure out |
4787 is just my reasoning ^^
Anyways forget about all of this lol... I don't know much about probability and statistics anyway so I was probably wrong at some point...
To change the topic, what do you people think of this?
https://www.youtube.com/watch?v=c7O91GDWGPU |
Like in the span of 50 he got 2 (very lucky). on the contrary, assuming he's still opening crates and only getting windflowers, that makes him extremely unlucky..
^this is a false statement if i understand the use of word "Span" correctly. He did not get 2 rares within a span of 50. He got one after an unknown number of tries and then he got second sometime within the next 50 tries. |
| 2 in 50 observed attempts..and 1 after an unobservable number of attempts and 1 more right after that both can be used to make statistics |
2Ck (0.001^2) (0.999^k-2)
where k is the number of crates he opened/we're observing is the best method to deem "luck" in the most literal way..other methods do provide statistics but can't tell how lucky the dude got..of course, for that we need a sample size..i kept it 50 just to exaggerate it |
2 in 50 observed attempts..and 1 after an unobservable number of attempts
that makes no sense whatsoever. To reach 2 rares he must have opened a much larger number of boxes. How are you observing 2 in 50 attempts? The only thing you have is the time he obtained 1 and 2 which tells us nothing about the total number of boxes he could have opened to get the two rares. And we are not deeming anything; The probability tells us how lucky he got with his second rare and it is quite unambiguously within 5% odds as long as it is safe to assume 2 minutes = 50 boxes.
Also, you seem to be misinterpreting the 5% as you called it high which is not the case at all because it does not mean 5 in 100 boxes. It means that there is only a ~5% chance that you will get a single rare in 50 boxes i.e. 5% chance for 1 success in 50 tries. |
To reach 2 rares he must have opened a much larger number of boxes.
arent those only assumptions we can make without a definite set of data? the 5% stat and the equation are the only 2 assumptions we can make without the said data..5% saying that maybe u opened an infinite number of crates before and the odds of u not getting a rare art (on the attempt in which u did) was infinitesimally small
The equation says that idk how many crates you opened before this but im not here to second guess here..Let's talk about the 50 crate period in which you actually got the art (not saying 1st and the 50th, just 50 in general).
Now it's only a matter of your personal perception of luck..my way just seems more intuitive to me |
randomm is trying to say what virtual said in 4790.
pankaj why do you say that the attempts before the first success are unobserved?they are pretty well observed to me.take for example the n attempt.if it is success you start count it as observed but if was unsuccessful you count it unobserved?i feel like you observe all attempts this way no matter that you dont care about their size until the first success happens.although ,to my knowlege, observation affects probabilities of particles in quantum level only. |
although ,to my knowlege, observation affects probabilities of particles in quantum level only.
Hahaha good one xD My brain is getting "entangled" to to all this confusion hehe :P |
| lol for that i all i can say is imagine he got 50 rare arts out of the 50 crates..but the entire universe inclined itself in a way such that they did not show up in the arts page..spooky effect, right? :P until we somehow observe those crates, they remain in an entangled/unobserved state :D |
to make things simpler.
i got infinite number of chests and you got only 50.
you will start opening chests at the same time with me but only after i get my first rare art.
So in the next 50 attempts who has a bigger probability to get a rare art?me or you?? |
| Do i need a fort for a t5 to show up on the char page? I got sphynx immortals but i cant change to them on the recruiting page nor do warriors show up in my army :p |
Not warriors, guardians**
The non upgraded unit |
