**Mastery Loadouts**
Due to issues related to the release of Mastery Loadouts, the "free swap" period will be extended.
The new end date will be May 1st.
Due to issues related to the release of Mastery Loadouts, the "free swap" period will be extended.
The new end date will be May 1st.
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What you are saying , @Glads is =
1/250 * 1/250 or 0.004 * 0.004.
There is a small mistake in your consideration here. At first glance I also thought you were correct but it's not the case.
The first event occurence is not 1/250, it's 1.
It's 1/250 if you are looking for ONE SPECIFIC CHAMP.
Example -
Probability of pulling DOOM AND THEN DUPING HIM = 1/250 * 1/250 = this is what you are calculating @Glads
But what we need is ----
Probability of pulling ANY CHAMP AND THEN DUPING 'THAT' ANY CHAMP
= 1 * 1/250 = 0.004
What we are looking for is the second situation where ANY champ can be pulled 2 times back to back in a row.
Here can also look for sample space of that -
Example -
DOOM , DOOM
STRYFE, STRYFE
ICEMAN, ICEMAN
etc....
These will be 250 such possibilities of these types of combinations out of
250x250 total combinations = 62,500
So its gonna be 250/62,500 = also 0.004
So 2 ways to look at this both correct.
I guess, I Win.
1st is a overall probabilistic assessment of multiple independent events happening over time,like what is the chance of heads landing in a million flips. That should of course converge toward 50% if the coin and flips are fair. This one is the argument for the inverse multiplied example 1/250^2
2nd is what is the independent probability alone of a single coin flip. Or to put it in the framework of the first argument what is the probability of the next coin flip being heads in a series of length 1. This is the argument that 1/250 being the answer.
So your both right, or both wrong depending on if which thought process your approaching the question.
I do question when you get a probability of one.
Each crystal carries 1/250 chances. If I get a good champ or a bad champ that's not a probability of 1, it's 1/250 as there are 250 options u get one.
Then on the second crystal you have another 250 options again you only get 1. Therefore the chance that this happen is the number I have been saying.
To have probability of 1 which you are referring to you would need 250 options.
I hope th8s explains my reasoning
According to you probability of pulling the same champ in a row must be higher than pulling just one champ.
In a way you are right but you are talking about pulling a PARTICULAR CHAMP, like doom then yes you are right it's 0.004 * 0.004.
If you pull a good champ or a bad champ, which means you were not targeting anything , right ?
If you are not targeting anything there is no desired outcome so it's probability is 1 here, which means you are getting atleast 1 6* or 5* or whatever champ it is. You are guaranteed to get A champ but it can be any champ so it's 250/250.
You are looking at this the wrong way. The formula is = desired outcomes/ total outcomes.
Here your desired outcomes is 250.
This leads to far more questions
Okay, here’s an issue. You are trying to run an experiment in which you pull the same champion twice. It does not matter which champion, just two times in a row. As such, the first pull of the crystal does not have limiting parameters on it. Any champion will do to set the parameters for the second crystal pull, which is the one that matters to see if you get the same champion again. The first crystal is what sets the conditions for the experiment. It does not matter for the purposes of the experiment if you first pull Kitty Pryde or Hercules or Beast or Cyclops, it only matters that you pull that same champion again on the second crystal.
As such, the first crystal spin has no probability associated with it for the purposes of the experiment because the result does not matter. It would only matter if you called your shot prior to opening it. But, because that is not what we are testing, it doesn’t matter. Only the second crystal matters. That is why everybody else is telling you the odds of pulling the same *any* champion back to back is 1/250 (it’s actually lower than that because we don’t have 250 champions in the pool, but I understand it’s just a placeholder value).
Finally, to the proposition that something is wrong as a result of this 1 in 250 chance occurring: if it just kept happening to a specific player, then yeah I’d agree with you. But spread out across a community of this size? With this many active players? 1 in 250 isn’t just plausible, it’s practically inevitable with a fair degree of frequency. It is weird for you specifically to get struck by lightning, but unavoidable that *somebody* will.
It feels wrong when it is the same champ twice in a row, but it is not. Why would the odds of pulling say, Herc and then Corvus, be any different than pulling Herc twice in a row? The odds for each pull do not change because of your previous pull. Each pull is independent and based solely on the number of champs in the pool.
While it is true that you here was a 1 in 10 chance of pulling this specific marble, that fact is not relevant to anything because you didn’t care about what marble you grabbed. You just wanted a marble.
Now you put that marble back in the bag and shake it up. You think to yourself, “what a fool I was! I miss my red marble friend!” Now you want to reach blindly back into the bag and hope you pull out the red marble. The odds of that are 1 in 10 because now you have a specific target in mind whereas previously you did not.
All of this is to say, I hope that you and your marbles are reunited.
Couldn’t not post this. #Illuminaticonfirmed
The odds of pulling 2 specific champs that you want is 1/40000 or about .0025%.
But the odds of pulling 2 in a row like they are complaining about is .5%.
Each ticket represents a champ and each box a crystal. You're welcome 😉
If you still don't get it, then google dice rolls, as its probablymore commonly explained. Rolling the same number on 2 dice is a 1/6 chance. Rolling 2 1s specifically is a 1/36 chance (1/6 x 1/6). In this example, getting any champ twice is a 1/250 chance, while getting a specific champ, like getting double 1s is 1/x^2 or 1/250 x 1/250 or 1/62500.
Here's a another simple example for anybody still not getting it 3 pages into this thing. We get two crystals that has 3 champs:
Herc
Mags
Doom
Glads would tell us that the odds of getting duplicate champions out of this crystal is (1/3)*(1/3), or 1/9. 1 out of 3 for the first crystal time 1 out of 3 for the second crystal.
You're excited, because these champs are dope, and you pop both crystals immediately. These are all of the possible outcomes of your two pulls:
Herc, Herc
Herc, Mags
Herc, Doom
Mags, Herc
Mags, Mags
Mags, Doom
Doom, Herc
Doom, Mags
Doom, Doom
How many total outcomes are there? 9
How many outcomes have consecutive champs? 3
Your odds of getting the same champ consecutively from this dope crystal is 3/9, or 1/3.
If you still don't get the math here, lay out all of the options again for opening 2 crystals with a 4 champion pool and see what the odds of consecutive champs are. Then do it with 2 crystals with a 5 champion pool, a 6 champion pool, and so on until you get to a 250ish champion pool. You'll get the point long before you get that high.
That's because if your concern is 'people pull the same champion back to back, the odds for that must be crazy', then it doesn't matter who they pull first: any champion will do - so the odds of pulling them is 1 in 1: you're going to get someone from that crystal*.
The chance to pull any champion back to back is therefore 1 X 1/250
The second stage is where probability comes into it: Having defined your champion, the odds of pulling that particular champion again are 1/250.
----
* If you want to pre-define the champion, then your maths is perfect: If you have two crystals and you're hoping to pull and Awaken Corvus in particular, then the odds of pulling Corvus back to back from two crystals is 1/250 X 1/250.