**KNOWN AW ISSUE**
Please be aware, there is a known issue with Saga badging when observing the AW map.
The team have found the source of the issue and will be updating with our next build.
We apologize for the inconvenience.
Please be aware, there is a known issue with Saga badging when observing the AW map.
The team have found the source of the issue and will be updating with our next build.
We apologize for the inconvenience.
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It's the same pool of possible pulls each time.
I’d give it a 4/10 overall
What you’re talking about is calling your shot before opening the first crystal. Your math is correct if, before opening either crystal, you’re calculating the odds of pulling and immediately duping Hercules. But that’s not the point OP was trying to make. OP was saying that the frequency of back to back pulls *of any champion* is too high, not specific champions.
Please don’t make me pull out the marbles again.
"Dependent events: Two events are dependent when the outcome of the first event influences the outcome of the second event. The probability of two dependent events is the product of the probability of X and the probability of Y AFTER X occurs.
P(XandY)=P(X)⋅P(Yafterx)"
https://www.mathplanet.com/education/pre-algebra/probability-and-statistic/probability-of-events#:~:text=Dependent events: Two events are,of Y AFTER X occurs.
The second one is will you pull *that same champion.* That’s where it becomes this 1/250 because now it matters which *specific* champion you pull, whereas it did not originally.
Therefore 1 x 1/250 = 1/250.
We have 2 crystals that we open. The chance of getting a champ on the first crystal is 100%.
Whatever the champ may be, the chance of getting that champ now that we have a specific champ in mind, is 1/250.
Point to note is that no matter which champ you want in the second crystal , the chance of getting any specific champ is 1/250
Remember statistics is the art of making the numbers say what you want them to say.
So we are actually both correct in our understandings of the situation.
Assuming behind the three doors, you have two Iron Men and one Ghost, you pick door 1. Here are the possible scenarios:
Scenario one: Door 1 - Ghost, Door 2 - IM, Door 3 - IM
Scenario two: Door 1 - IM, Door 2 - Ghost, Door 3 - IM
Scenario three: Door 1 - IM, Door 2 - IM, Door 3 - Ghost
So the odds of you picking the right door is 33.3%. Here is where the scenario changes, Kabam Montii will always open one of the doors with Iron Man. In scenario one, you picked correctly, so he could open either other door. In the other two scenarios, he only can open the door with the other Iron Man (since one is behind the door you picked, door 1). If in each of these three scenarios, you were to stick with your original choice, you would still be correct 33% of the time. If in each of these three scenarios you always switch to the other door, you will move your choice from Ghost to Iron Man 33.3% of the time, but you will correctly move from Iron Man to Ghost 66.6% of the time.
The key for me is to not look at it as two individual sets of choices, but as one big choice where the odds don't get "locked in" but more information is given so that you can make a better overall decision. The whole locked-in thing is fine and all, but it doesn't really explain why the odds don't shift, it just pre-supposes that you understand how the giving of information changes the system. I think most people get tripped up because they don't realize that Montii can't actually open the door you chose, and what that does to the probabilities.
Many of us here do know what the numbers are.
bottom line:chances are very very low. ANd that is just the point.
The mistake you are making again and again is that you are calculating the probability of pulling a particular champ twice in row, while what you should be calculating is probability of pulling any champ twice in a row.
Further your claim to have studied advance maths and then your disbelief that the chance of pulling the same champ again and the chance of pulling any champ, is the SAME, mildly amuses me. If you had studied just a tiny bit of probability theory or random processes then you would know that after pulling a certain champ, you have the same chance of pulling that champ again as you have the chance of pulling any other particular champ. Thats just how random processes work.
I would suggest you read probability theory by SL Ross instead of quoting random **** from the internet.
The two crystals are independent: the game doesn't remember what you pulled and use that information to influence the next crystal. The reason why the odds of pulling a specific champ twice in a row is 1/250 x 1/250 (in this example) is not because the second event is dependent on the first one, it is that for the event you are interested to occur, two separate independent events must happen concurrently. Statistical arithmetic says if these two events are independent you can multiply their odds. If they were dependent then multiplying would not be valid.
The reason is if the two events are independent, then no matter what happens with the first crystal the set possibilities for the second crystal is identical - that's what independent means. That means the sum total of all possibilities will just be the product of the possibilities of each crystal. But if the second one is dependent on the first, that means the set of all possibilities is *not* the same for all possible results of the first crystal. That's what dependent means: the first one influences the second one. In that case, you cannot multiply: you must now sum over all combinations directly.
I think what you mean is that the event you're interested in - pulling Nebula twice - is dependent on both events happening. That's not the same thing as saying the second event is dependent on the first. Saying the second is dependent on the first is semantically important, because it describes a situation where the multiplication short cut doesn't work.
If I say a coin has a 50% chance of landing heads when you flip it (let's not get into whether coin flips are perfectly even for now) it is obvious what I mean. But when I say there's a 25% chance of flipping two heads in a row, what do I mean? I mean that if I flip a coin twice, there's four possibilities: HH, HT, TH, TT. And one of them is two heads. So there's a one in four chance of getting that possibility.
To be specific, we're assuming the coin doesn't have a memory. When it lands heads, there's still a 50% chance of it landing heads or tails on the next flip. And the next one. And the next one. So I can say that *if* the coin lands heads, there's a 50% chance of landing heads or tails on the next flip, which means *if* the coin lands heads, the two possibilities are HT and HH. *If* the coin landed tails first, the two possibilities are TH and TT. We know this because the first flip doesn't influence the second one.
Notice we have a short cut we can use. If there are two possibilities for the first flip, and each of those two possibilities has two possibilities for the second flip, then there are four possibilities because 2 x 2 = 4. Because we have two sets of two possibilities.
Going further, there's a 12.5% chance of getting three heads in a row, and there's a 1/1024 chance of getting ten heads in a row. We can imagine list listing all the possibilities out: HHHHHHHHHH, HHHHHHHHHT, etc. But it is obvious there will be 2 x 2 x 2 x 2 x ... x 2 = 1024 possibilities, and one of them is all heads. All of this is due to the principle of independent events. Since each event is independent of the past, we can assess the odds of a sequence of events happening before it happens with relatively simple math. We are really just counting possibilities, only without having to write them all out.
But what if you've already flipped nine heads in a row? What are the odds of flipping heads again? Still 50% You might think but the odds of flipping ten in a row are 1/1024, how can the odds of flipping heads now be 50%? Shouldn't it be super rare?
It is super rare to flip ten in a row, but you've already flipped nine in a row. The odds of flipping nine in a row are 1/512 IF you haven't flipped a coin yet. What are the odds of flipping nine in a row now, after you've flipped nine in a row? 100%. The odds of doing that must be 100%, because you actually did it. It is done. Proof: let's bet on whether you flipped nine heads in a row. I choose that you did. I'm going to win that bet, aren't I?
Probability measures the chance that a particular event will go a certain way, before that event occurs. *After* that event occurs, there's no more probability, there's only history. So while the odds of flipping ten heads in a row is 1/1024 before you start flipping, the odds of flipping heads ten times in a row after you've already flipped nine heads in a row is now 50%. Because the odds of flipping nine in a row at this moment is 100%, and the odds of flipping one more head is 50%. 100% x 50% = 50%.
Probability measures the future. When the future becomes history, probability no longer applies.
The odds of getting double 1s is 1 in 36, but the odds of getting a second of any number rolled first, is only 1 in 6.
We're not looking at getting a specific champ, we are only looking at getting whatever champ we just got one more time.
To break it down: Assume there are 250 champs in pool. To get Corvus twice you have to get him the first time at a 1/250 chance and then again at a 1/250 chance. To get any champ twice, you just have to get a champ, and since it doesnt matter who it is it is a 1 in 1 chance of getting a champ. To duplicate that champ it is a 1 in 250 chance.
When Monty opens one of the three doors, or 999,998 empty boxes, or whatever, it is important to remember (as you also mention) he isn't opening randomly. He specifically does not open the option with the prize. So he isn't free to choose any option to open. Equally important, he can't open your choice, because that's structured into the game.
So when Monty starts opening options, there's two options he's not allowed to touch: your choice, and the prize. If you actually chose the prize, then there's actually only one option he can't touch: your choice. But if you didn't choose the prize, then there's two options he can't touch: your choice, and the prize box. To put it another way, the prize protects a box from being opened, just as your choice protects a box from being opened.
The reason why one box's odds improve while the other box's options don't improve is because the last box standing when Monty opens all other options could have been opened. It is a survivor. Why did it survive? Either it was lucky, OR OR OR OR it was never about luck, because it has the prize. The prize box is "protected."
So when the dust settles, we have two options: your choice, and the other one. Yours survived by default: it was guaranteed to survive. So we don't know anything new about your choice. But the other one survived the culling process. We now know something new about the survivor. It *may* have survived because it has the prize, which guaranteed its survival.
Metaphorically, hidden knowledge "flowed" into the survivor box. It didn't flow into your choice box, because the fact that it survived the process tells us nothing.
And that's why one box's odds change, and the other box, the choice box, doesn't. At the end of the game, just before you are allowed to switch, we know more than we did at the start about the survivor box, but know nothing more than we did at the start about the choice box. The choice box's odds are in fact "locked in." The "survivor box's" odds start low, and get progressively higher as Monty opens options, because the more options Monty opens, the more likely it is that the survivor box survived not because it is lucky, but because it contains the prize.
TL;DR: One option survives the game, and survival tells us something about that option. Your choice doesn't really survive anything, and thus we learn nothing new. So the survivor box's odds increase, while the choice box's odds stay the same.
But they aren’t.
But if Y depends on X, that means P(Y) depends on X - depending on what X ends up being, the odds of Y happening will be different. So P(Y) is not just one number, it depends on X. So P(X) x P(Y) is meaningless. You have to instead figure out what P(Y) is given X, and then use that. Thus: P(X) x P(Y given X).
However, crystal openings are independent. That means the odds of any particular champion popping out of crystal Y are the same regardless of what pops out of crystal X. P(X) is constant, and P(Y) is constant.
Here's an example of a dependent probability. What are the odds of pulling Doom from a Nexus 6* crystal? Nexus crystals are specifically programmed to never generate duplicate results. So the odds of Doom showing up in the second slot depend on what shows up in the first slot. If the first slot isn't Doom the odds are one thing, but if Doom shows up in the first slot the odds are something else (namely: zero). Those are dependent probabilities. Calculating this is thus more tricky.
One in the morning & another in the evening