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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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" When the chance is defined in Mathematics, it is called probability.'
One of my first comments were, To which you replied, I've tried but cannot get something that confirms this statement of yours.
As for the coin question. The link you have shared says that the probability of getting three heads in a row is 1/8. Not that the probability of getting a third head in a row is 1/8. Those two are completely different probabilities.
Per your vedantu link: "Probabilities can be simply described as the number of desired outcomes divided by the total number of all outcomes."
If your first 2 coin tosses where you get head weren't included in your experiment then they don't count to the desired outcome and the 3rd coin toss is an equal 50/50 because it's only 1 event. If your experiment includes 3 coin tosses and you get 2 heads then the 3rd head is 12,5%. KEY WORD IS EXPERIMENT: THIS IS SET BEFORE THE COIN TOSSES.
To that end, @DNA3000 there one aspect of the crystal openings that I've never understood. To be crystal clear, I'm not trying to float conspiracy theories here. I genuinely don't know how it's handled, and I assume you would know best.
Is the 1%, 8%, whatever percent chance at a given rarity factored before looking at the pool?
To be more specific, with a Paragon crystal having a 1% chance at a 7* champ, is it rolling the mythical hundred sided die, hopefully landing on the one spot that is a 7*, then rolling again to determine the specific 7* champ?
Or does a given roll factor the pool of champs at a given rarity. In other words, with the currently very small pool of 7* champs compared to 6* champs factor into the equation some how?
**Note** I know this gets even more strange with a featured crystal, where there's an increased chance at a specific champ. I'm only curious about a standard crystal.
Thanks Professor!
First, a random number between 0 and 1 is generated to determine the rarity
If that number is less than 0.5 (50% chance), it will be a 4* champion.
If between 0.5 and 0.91 (41% chance), a 5*
Between 0.91 and 0.99 (8% chance), a 6*
Anything 0.99 or greater (1% chance), a 7*
Once the rarity is determined, the specific champion can be selected using a similar method:
Give every champion available in the selected rarity's pool an ID, 0-250 or whatever the maximum is for that pool.
Generate another random number between 0 and 1.
Multiply the random number by the number of champions in the pool (now the number lies between 0 and the max ID + 1)
Floor the number (now the random number is an integer that must match the ID of a champion available)
Select the champion with that ID
There are other ways to achieve the same result, but usually it looks something like that
first one has 50% probability to be head.
then two heads in a row have a probability of 25%, and you dont get HH without second head which means you have 25% probability for it to be a head when looking at consecutive heads.
So the probability goes down with each tosses,
And probability for third toss to be a head in consecutive tosses goes down to 12.5%.
Now when you look at third one after first two being head, one would think that it has 12.5% chance to be head again.
Which is straight up wrong because what you have calculated is 12.5% probability, while chance of it being a head has never gone down from 50% chance, and it does not change regardless of what number of try you are on.
Entire point was that probability of getting same outcome changes with nth consecutive try while chance does not. Hence probability and chances are not same thing. And it is inaccurate to call it same thing. Of course I am including first two coin tosses, Why else would I even get 12.5% probability without 8 total possible outcomes?
And chance and probability is the same thing. The latter just has a mathematical number
"You're stuck in the gamblers fallacy"
Thats an assumption that you are making.
According to you if probability is just a mathematical number, how is chance not just a mathematical number?
If you say chance and probability are same thing, you should be able to
Prove that probabilty of getting 3 7stars in a row is same as chance of getting a 7star.
And for @mgj0630 , yes, in order for the game to account for different sized pools of each Star/Rarity within the overall pool contained in the crystal, that would indeed have to be taken into account.
Even for like a standard PC (2/3/4*), where for a long while the pool of available 2* was more limited compared to the totality of champs that have all been released as 3* or 4*.
In order for the overall stated odds between getting a 2* vs 3* vs 4*, it has to take the differing sizes of each into account too.
So the most likely method is to first roll a hundred sided die (for example) to determine Star/Rarity, and then another “X-sided” die (X, sized to how many champs are possible from the already determined Star/Rarity) to determine the exact champ.
Since we've flipped about 47,032 coins in this thread, I figured it was as good a time as any to ask if anyone knew definitively.
"Probability is measured by the ratio of the favourable cases to the whole number of cases possible."
and
"the two terms chance and probability have extreme similarities, there are many differences between them"
Are you going to ignore this?
Also, from the page
Chance and Probability are very similar to each other. Both of them have the numbers 0 and 1. The difference they share is that chance doesn't have any obviousness whereas probability exactly defines the ratio of how likely an event is to happen.
It's very explicit what the difference is. Just that probability has a proper ratio associated with it.
If you're the author of the crystal odds tool chain, and you're clever, and you want to optimize the behavior of the game servers, you can reduce two random rolls into one, by taking the crystal reward tables the crystal designers make which are designed to be readable by humans and convert them into a flattened version that the game servers are perfectly fine with that have identical behavior but are faster to calculate.
Let's try a simple example with easy numbers. Suppose I have a crystal with a one in four chance for a 7* and three in four chance for 6*. And on top of that, there are actually only four different 7* champs and five different 6* champs in the crystal. The direct way to implement this is to make a two-tier crystal that contains just two rewards, a "7* pool" and a "6* pool" and then within each pool the desired champs would be added. The crystal would then be designed more or less with three reward tables: the primary table, and two secondary tables.
Primary reward table:
0 to 0.25 - 7* pool
0.25 to 1.0 - 6* pool
7* pool table
0 to 0.25 - Champ A
0.25 to 0.5 - Champ B
0.5 to 0.75 - Champ C
0.75 to 1.0 - Champ D
6* pool table
0 to 0.2 - Champ K
0.2 to 0.4 - Champ L
0.4 to 0.6 - Champ M
0.6 to 0.8 - Champ N
0.8 to 1.0 - Champ O
The game would then roll one random number to see which reward in the primary table you get, and since both rewards are themselves tables, the game would then roll a second random number to see which of those actual champs you ultimately get. However, random numbers are, in general, "computationally expensive" relatively speaking. It takes far longer to generate a random number than do a table look up, for example. So a clever game implementation could allow the developers to design this way, because it is easy to understand, then use math to convert those three tables into one table:
Reward table:
0 to 0.0625 Champ A
0.0625 to 0.125 Champ B
0.125 to 0.1875 Champ C
0.1875 to 0.25 Champ D
0.25 to 0.4 Champ K
0.4 to 0.55 Champ L
0.55 to 0.7 Champ M
0.7 to 0.85 Champ N
0.85 to 1.0 Champ O
This is mathematically identical. We take the odds of the first table - 0.25 or 25% - and we "scale" every entry in the 7* table by that factor. We do the same thing for the 6* table and its differential odds (0.75 or 75%). We do this under the hood, in a way the developers don't even need to see. But the net result is we get identical drop odds, but with only one random number roll.
Multiplied by millions of random crystal rolls per day, this can add up. But whether the game actually does this or not, I do not know. However, these are the two ways I've seen it done, with some small variations in how the odds are expressed (for example, some implementations use weights instead of percentages, but I suspect MCOC uses percentages, as the published odds tend to land on exact percentages too often to be weighted).
I'm going to use the word "chance" simply because it's shorter.
H = Heads 50%, T = Tails 50%
BEFORE you flip any coins, the chance of you getting HHH = 0.5 x 0.5 X 0.5 = 12.5%
What is the chance of you flipping heads AFTER you have flipped HH?
(HH) = 1.0 x 1.0 = 100%
Why is this 100% you ask? Well because it already happened, so it has a 100% chance of happening.
Last Flip
H = 50%
So, then the calculation becomes (HH)H = 1.0 x 1.0 x 0.5 = 50%
It is important at what point in time you ask the question of "what is the chance?" Because past events have a 100% chance of happening and therefore don't influence future events. i.e. multiplying anything by 1 produces itself.
Before we start let’s address this semantic silliness. Chance is just a lay way of expressing what probability rigorously defines. The “chance” that something occurs is equivalent to the “probability” that it occurs, all that matters is that you correctly characterize what is going on using math. I have never, in my life, heard someone actually try to meaningfully distinguish between these things.
The issue that you are having comes from incorrectly conflating the probabilities of independent events with the probability of seeing a particular sequence of events. You were correct in that probability has a property of accumulating. The probability of getting at least one 7 star in one crystal is different (and smaller) than the probability of getting at least one in 2 crystals. So what’s going wrong?
The probability that you have gone ‘n’ crystals without seeing a 7 star is simply (.99)^n; Im assuming a 1% drop rate but I don't recall what the actual drop rates are. It doesn't matter to illustrate the point. As n grows, this probability does in fact go to 0. Does this mean that for n>>1 (sufficiently large) that the probability that the next crystal gives a 7 star is guaranteed? Obviously no. The probability that any crystal at any point in the sequence gives you a 7 star is exactly the same no matter what n is: 1%.
The only thing that becomes less probable is the chance (note how I could have said probability WLOG) that you are on the one string of length ‘n’ that is only made up of non-7 star pulls. Thinking in binary if a 0 represents having pulled anything other than a 7 star, and a 1 represents having gotten a 7 star then, for any n-length sequence of crystal pulls the most probable string is the n-length string of all 0’s, because the probability of getting a 0 at each point is higher than that of a 1, BUT THERE IS ONLY ONE SUCH STRING. The total probability of all other strings which have at least a single 1 at some digit ALL represent having gotten at least one 7 star and, for large enough n, their additive probabilities are in fact larger than that of the probability of the 0 string. Note how the individual digits never changed their probability only the probability of the ENTIRE sequence.
With this in mind, you don’t have a smaller chance of getting heads if you have previously flipped HH. In fact, in this case, you dont even have a “most probable” sequence because they are all just (1/2)^n. The statement that you have a smaller chance of seeing heads based on the previous observation is false by simple Independence of events and even more-so by the detailed binary sequence expression above. The accumulation of probability has NOTHING to do with whether or not independent events’ probabilities
P(Flipping heads on throw 3 | You flipped heads twice)= P(Flipping heads)= .5
The sentence would read “given I have flipped heads twice, what is the probability that I flip another heads?”
You conflated this sentence with the JOINT probability of:
P(H-H-H) =
P(Flipping heads three times in a row) = (.5)^3
This sentence would read “what is the probability that I toss a coin 3 time and get heads all three times”
I am glad, finally we have somebody with proper background on this. I would like you to point out whether I am correct or incorrect on following points.
1. Probability of getting a 7star changes depending upon how many tries you are calculating it for.
2. Probability is used to predict an outcome and it can fail to determine results.
3. Drop rate of getting a 7star remains same regardless of how many crystal one opens.
4. Probability of getting a 7star changes depending of how many tries you calculate it for, while actual chance of getting a 7star remains same.
5. You can not get a specific 'n' for n number of crystals required to get a guaranteed 7star.
6. Have I at any point told that chance of getting a 7star increases as you open more crystals?
And since we had lot of back and forth here about whether Probability of getting a head on third consecutive try is same as the chance of getting a head on third try or not. Can you plz clarify it and put to an end to this.
You have to be dangerously precise about how you ask 1 and 3. Depending on how you do so these are the same question or completely different.
1) If the question is, “whats the probability that you get at least one 7 star given you are going to try ‘n’ times”, then this grows as n gets large.
P(At least one 7 star | I will open ‘n’ crystals) = 1 - (.99)^n. Again, this has nothing to do with the probability of any individual crystal only the event which incapsulates the SEQUENCE.
By independence
P(Next crystal gives a 7 star | I have opened n crystals so far with no 7 stars)= P(7 star)=.01. No dependency on n
2) Probabilities don’t predict anything. They map weights to events. For something more “predictive” you could use expectation but thats not precisely what that does either. As far as “failing to predict”, what it seems like youre getting at is that unlikely things can happen. That is true. That’s not a failure of predictive power, just a fact of probability. The only failure that could occur is in incorrectly assigning probabilities to events or believing that things with small probabilities can’t occur or “ought not to”.
3) See the bottom of 1. The conditioning argument explains why there’s no dependence on the number you open for the probability of an independent crystal
4) The event of not getting a 7 star in n tries changes as n changes. Any crystal has a static chance of giving you a 7 star. For explanations, see 1 and 3.
5) No ‘n’ guarantees a 7 star. Since n never reaches infinity the equation in 1 never reaches 0, only approaches. If such an n existed that would be the number whales would need to buy to guarantee a 7 star.
6) Here’s what I saw that made me think you though exactly that:
“Anyone with proper understanding of probability will tell you that the probability of getting a head third time in a row is 12.5%…”
The way this reads, you have observed 2 heads and are now determining the probability of seeing the third based on that. As discussed in a previous comment this is a conditional probability so the answer should have been .5 or 50% by independence.
You also said
“probability for third toss to be a head in consecutive tosses goes down to 12.5%”.
It does not. The statement you should write is that the probability of the sequence of three heads is less than that of the sequence of 2 heads, not that probability for a third toss being a heads goes down given we’ve observed a sequence of two. The second phrasing is a conditional event, the first phrasing is two joint events. You were saying something about how 12.5 was chance and 50 was probability or some variation of that, but, as DNA also explained at length, thats not what happened.
The correctness of what you say depends both on the math paired with the correctness of the description of the experiment. Probabilities are sensitive to the exact phrasing of a