The G-Meister
Giga Slime
This post is far too long for the forums, so I've split it up into 2 extra replies. Unfortunately the forum play havoc with the display of equations, so anything in square brackets at the end of an equation is the equation number.
Contents
1. The Introduction
....1.1. A Quick Disclaimer
....1.2. Reader's Guidance
2. The Current System (Geometric Distributions)
....2.1. The Context
....2.2. The Maths
........2.2.1. How Probability of Card Drop Varies with Number of Kills
........2.2.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
........2.2.3. The Aftermath
....2.3. The Graphs
3. The Other End of the Scale (The Unit Step)
....3.1. The Maths
........3.1.1. How Probability of Card Drop Varies with Number of Kills
........3.1.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
....3.2. The Graphs
4. And Everything in Between (Weighted Geometric Distribution)
....4.1. The Introduction
....4.2. The Maths and Graphs
........4.2.1. How Probability of Card Drop Varies with Number of Kills
........4.2.2. Approximation to Unit Step
........4.2.3. Approximation to Geometric
........4.2.4. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
........4.2.5. The Scaling Factor
........4.2.6. The Remaining Graphs
....4.3. The Context
....4.4. The Implementation
5. How to Argue About This Topic
....5.1. I prefer the way the game currently works
....5.2. I would prefer the unit step
....5.3. I would prefer minimal weighting
....5.4. I would prefer a little weighting
....5.5. I would prefer moderate weighting
....5.6. I would prefer significant weighting
....5.7. My name is Own and 1% should be added to the drop rate after 1000 kills
6. A Note to the Developers
7. The Software Tool
8. The Epilogue
1. The Introduction
I hope no-one reading this is vegetarian or vegan, as you're going to be insulted at how meaty this article is. Jokes aside, actually take a look at the scroll bar, and actually go and grab a meal, because you'll probably need one at some point while reading. Despite the length, I would still advise reading the whole article, no matter what your level of experience in mathematics. While I tell certain people to dip out at certain sections, I use similar processes throughout, so understanding the ones at the start will give context to the ones elsewhere.
1.1. A Quick Disclaimer
If you've come here to express your views on what kind of solution you would like to see, I have just the section for you! Having said that, while you could quite easily scroll down to section 5, pick a group and go, without context it's a little fruitless. I'd still recommend reading the whole article, though if you can read and make sense of the graphs and equations, you should be set.
1.2. Reader's Guidance.
Our process here is gonna go like so. First, we'll explain how the game currently drops cards in words. We'll then translate that into simple maths, with some helpful graphs. We'll then move onto how this maths can be adapted to change how the RNG works, briefly dipping in to some higher-level maths, but providing yet more graphs and pictures for the layman. Finally, we round off by translating it back into words.
And the last thing before we move onto serious stuff, I believe there is limit of 10 files that can be uploaded to a single post. I assumed I would go well over that, and as such many of the images have been compacted together into single, larger images. Don't worry though, I make sure to reference them all throughout the text and in the right order.
2. The Current System (Geometric Distributions)
A lot of the information in this section can be found on the equivalent Wikipedia page, but here I have tried my best to put it both in normal people's words and in context, in the interest of readability.
2.1. The Context
Card drops in the game currently follow a geometric distribution. Keep that word in your head, we're gonna be using it a lot. For a geometric distribution to be obeyed, there are 3 criteria. Firstly, all trials (instances where a single, random "test" is carried out) must be independent and affect each other in no way. Secondly, there can only be two possible outcomes of these trials, often designated a success or failure. Thirdly, the probability of this success must be constant. If all three of these are true, the geometric distribution shows how likely it is to get the first success on any given trial.
This might sound like total jargon to anyone not vaguely familiar with statistics, but actually it's very easy to translate into the game's card drop mechanics. Each "trial" is a "kill", the "success" is the drop of a card, and cards only drop once, meaning we only consider the first "success". With that in mind, the three criteria are obeyed. The outcome of one kill (being card drop or no card drop) does not affect the outcome of any other, cards can either drop or not drop, and there is mountains of evidence that the probability of a card drop is constant.
2.2. The Maths
I tried my best to explain the graphs without the maths. Unfortunately I could not, so you'll get them later There are lots of letters from this point onwards, and I’m not just speaking about the multiple thousand words of explanation. All variables being used as part of the maths should be in italics. If you spot one that isn’t, or a sentence that reads weird, pop a reply down.
The letter we will be using for "number of kills" is x. This is because all of the graphs will have the number of kills on the x-axis (along the bottom). We'll use p for the fixed drop chance.
2.2.1. How Probability of Card Drop Varies with Number of Kills
Simple answer, it doesn't, exactly as we just said in section 2.1. above. You might be asking what the point of this section is. It is less that I want to explain how this works, but more that I want to include it for completeness before we change the maths in it later. Having said that, it does have an important consequence. If the probability of success is constant and we can only have a success or failure, by logical deduction, the probability of failure must also be constant. Assuming all probabilities add up to 1 (which is a general assumption used throughout all of the mathematical topic), that probability is 1 - p. We will be using this in a moment.
2.2.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
This is where the distribution part comes in. While the drop chance of a card is constant, the amount of kills you have to make to get it behaves a little differently. Let's take an average sample of people (and by that I mean they behave exactly as theory predicts) and run them through the card drop process.
Here is where mathematical statements start. I'm going to write down an equation and explain what it means in words underneath.
P(X = 1) = p [1]
This might look very confusing for a few of you. What it says is "The probability of people getting a card drop on the first kill is equal to probability of a card drop". That should be a lot easier to understand, and fairly self-explanatory from what we have explained already (as well as a "no shit sherlock" statement). Basically, the number after "X =" is the number of kills, and everything after the second equals sign is the probability. Everything else is mathematical terminology you can safely ignore. Onto our next kill.
P(X = 2) = (1 - p) * p [2]
This translates to "The probability of people getting a card drop on the second kill is equal to the probability of failing to get a card on the first kill, multiplied by the probability of a card drop". That should be a reasonable deduction when people can only get one card. The people who got a card on the first kill we don't want to be considering any more, as they would never reach this point in the distribution. If you've wrapped your head around that, we follow a similar process for the third kill.
P(X = 3) = (1 - p) * (1 - p) * p = (1 - p)^2 * p [3]
This translates to "The probability of people getting a card drop on the third kill is equal to the probability of failing to get a card on the first and second kill, multiplied by the probability of a card drop". If you understood the process for the second kill, we have only done the same thing here for the third. The only extra thing we have done is re-write the probability of failing to get a drop for all of the previous kills. This is nice simplification for the step below.
Here, you should have an idea of the process. The last thing to do is extrapolate said process over all kills.
P(X = x) = (1 - p)^(x - 1) * p [4]
This reads "The probability of people getting a card drop on the xth kill is equal to the probability of failing to get a card to the power of (number of kills - 1), multiplied by the probability of a card drop". If that seems like a bit of a jump, run it though backwards, like so.
P(X = 3) = (1 - p)^(3 - 1) * p = (1 - p)^2 * p [5]
P(X = 2) = (1 - p)^(2 - 1) * p = (1 - p)^1 * p = (1 - p) * p [6]
P(X = 1) = (1 - p)^(1 - 1) * p = (1 - p)^0 * p = 1 * p = p [7]
If those look confusing, we are replacing our variable x, the number of kills, with a specified number we'd actually like to find the probability for. Anything to the power of 1 is itself, used in Equation 6, and anything to the power 0 is 1, used in Equation 7.
2.2.3. The Aftermath
Pun intended. The equation we are left with, describing the geometric distribution as a whole, is Equation 4 above. This has some interesting properties. Firstly, by taking a weighted average of this distribution (check this Wikipedia page or Google it for more info) we find that the average person gets a card drop at 1 / p kills. To make things a little clearer, if the drop chance is 1 / 300, then the average person gets a card drop at 300 kills. We will be using this property later. Secondly, the only letters on the right hand side are p and x. As x is a variable we change to get distribution frequencies, p is the only parameter of the distribution. By parameter, we mean a variable that is defined when the distribution is made.
Together, these two things demonstrate an underlying property that defines a geometric distribution - the fact that it is completely random. The distribution and all of the properties associated with it depend only on the fixed drop chance p, and at no point is a card drop ever guaranteed. This means it is possible (though impossibly unlikely) that You reach up into the 10s of thousands of kills, completely empty handed.
The actual property that defines complete randomness is the card drop probability never reaching 1, or being certain, at any point. However in this case, the geometric is by far the simplest of these types of distribution.
2.3. The Graphs
To round off the geometric distribution, let's give some graphs of its function. These basically put all the maths we did above into pictures. All graphs are using the same drop chance, simply displaying the data in different ways.
The first graph is one of the card drop probability against number of kills, explained in 2.2.1. As you can see, it is constant for all kills, as we have said already many times. For now this is just for completion, as we will be including similar graphs later.
The second graph is one of Equation 4 that we derived in 2.2.2., that being the probability of reaching a given number of kills for a first card drop, plotted against the number of kills. We have P(X = x) on the y-axis, and x on the x-axis (see, calling the number of kills x did come in useful!). If you are confused as to why this decreases with the number of kills, look at the two parts of Equation 4. We have a constant p, multiplied by a "failure factor" of (1 - p)^(x - 1). This "failure factor" takes all the people that did not get a card drop on previous kills, and is what is actually decreasing. As 1 - p will always be between 1 and 0, raising it to a positive power will always cause the number to decrease. If you're still confused, just ignore it and take a look at the explanation of the third graph below, it's not the end of the world.
The third graph hasn't been explained anywhere yet, but it should be simple to understand. All we have done is taken the probability for each kill on the second graph and added it to that of all the previous kills. Mathematically, this gives us a sum of all the successes up the that point, known as a "cumulative distribution". In context, this is everyone who has got a card up to that point. This increases with the number of kills, just as you'd expect. This should be easier to digest in general as well as being a graph that fits nicely with multiple lines, so I generally refer to this one. For now, we're done with the geometric distribution (finally!).
3. The Other End of the Scale (The Unit Step)
Before we get on to the distribution I have been designing, I'd like to pop a quick word in about the exact opposite of the geometric distribution. The opposite of something that is completely random is something that is completely certain. This means that there is no factor influenced by randomness included in any of the calculations. Instead of having a fixed drop chance, we instead have a fixed number of kills at which a drop is guaranteed, and impossible beforehand. Let's make this a variable, and call it m. Using this, we can follow the process we did for the geometric distribution. The thing we end up with is a function called a "unit step", where the graph hops from 0 (no chance of drop) to 1 (guaranteed drop) instantaneously. As soon as we have a 1, the rest of the distribution is irrelevant, as everyone will have their card.
3.1. The Maths
3.1.1. How Probability of Card Drop Varies with Number of Kills
In maths, this gives us two statements.
p(x) = 0 (1 ≤ x < m) [8]
p(x) = 1 (x = m) [9]
On the left of the first equals sign, the p(x) means the probability of a card drop as a function of x. In the equivalent section for the geometric distribution, 2.2.1., I didn't give this notation to avoid extra confusion, as it would've read "p(x) = p". Equation 8 states that "the probability of a card drop on the 1st to (m - 1)th kill is 0". Equation 9 states that "the probability of a card drop on the mth kill is 1".
3.1.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
The fantastic thing about the unit step is, when we consider how many kills people get their card drop at, it follows exactly the same pattern as in 3.1.1. above.
P(X = x) = 0 (1 ≤ x < m) [10]
P(X = x) = 1 (x = m) [11]
If the 1 in Equation 11 seems confusing, all it means is that 100% of people get their drop when the number of kills equals m, exactly as we intended.
3.2. The Graphs
And the helpful graphs return once again.
The first graph shows the probability of a card drop against the number of kills. Hey look, it's 1, or 100%, at the value of m we are using. The second graph shows how many kills people make for a card drop against the number of kills. Hey look, it's 100% at the value of m we are using. The third graph shows the sum of how many people have a card drop against the number of kills. Hey look, it's 100% at the value of m we are using.
You get the idea.
4. And Everything in Between (Weighted Geometric Distribution)
4.1. The Introduction
We have two ends of a scale now. We have complete randomness, and complete certainty. To fill in the gap between them, we need a system that has a bit of both of these factors. The question is, how do we make one?
The way we go about weighting a distribution, be it between random and certain or otherwise, is by changing the graph of the probability of a card drop. You do this by making the probability a function of the number of kills. This is true for all distributions that aren't the geometric distribution, including the unit step. If you want to be nicer to players than the geometric distribution is, you increase that probability with the number of kills. However, this is where we open up completely and can put almost any graph in. Instead of explaining what we should want out of this graph, I am instead going to propose the solution I have been designing and talk about its consequences afterwards.
Clean your ears out MLG.
Contents
1. The Introduction
....1.1. A Quick Disclaimer
....1.2. Reader's Guidance
2. The Current System (Geometric Distributions)
....2.1. The Context
....2.2. The Maths
........2.2.1. How Probability of Card Drop Varies with Number of Kills
........2.2.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
........2.2.3. The Aftermath
....2.3. The Graphs
3. The Other End of the Scale (The Unit Step)
....3.1. The Maths
........3.1.1. How Probability of Card Drop Varies with Number of Kills
........3.1.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
....3.2. The Graphs
4. And Everything in Between (Weighted Geometric Distribution)
....4.1. The Introduction
....4.2. The Maths and Graphs
........4.2.1. How Probability of Card Drop Varies with Number of Kills
........4.2.2. Approximation to Unit Step
........4.2.3. Approximation to Geometric
........4.2.4. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
........4.2.5. The Scaling Factor
........4.2.6. The Remaining Graphs
....4.3. The Context
....4.4. The Implementation
5. How to Argue About This Topic
....5.1. I prefer the way the game currently works
....5.2. I would prefer the unit step
....5.3. I would prefer minimal weighting
....5.4. I would prefer a little weighting
....5.5. I would prefer moderate weighting
....5.6. I would prefer significant weighting
....5.7. My name is Own and 1% should be added to the drop rate after 1000 kills
6. A Note to the Developers
7. The Software Tool
8. The Epilogue
1. The Introduction
I hope no-one reading this is vegetarian or vegan, as you're going to be insulted at how meaty this article is. Jokes aside, actually take a look at the scroll bar, and actually go and grab a meal, because you'll probably need one at some point while reading. Despite the length, I would still advise reading the whole article, no matter what your level of experience in mathematics. While I tell certain people to dip out at certain sections, I use similar processes throughout, so understanding the ones at the start will give context to the ones elsewhere.
1.1. A Quick Disclaimer
If you've come here to express your views on what kind of solution you would like to see, I have just the section for you! Having said that, while you could quite easily scroll down to section 5, pick a group and go, without context it's a little fruitless. I'd still recommend reading the whole article, though if you can read and make sense of the graphs and equations, you should be set.
1.2. Reader's Guidance.
Our process here is gonna go like so. First, we'll explain how the game currently drops cards in words. We'll then translate that into simple maths, with some helpful graphs. We'll then move onto how this maths can be adapted to change how the RNG works, briefly dipping in to some higher-level maths, but providing yet more graphs and pictures for the layman. Finally, we round off by translating it back into words.
And the last thing before we move onto serious stuff, I believe there is limit of 10 files that can be uploaded to a single post. I assumed I would go well over that, and as such many of the images have been compacted together into single, larger images. Don't worry though, I make sure to reference them all throughout the text and in the right order.
2. The Current System (Geometric Distributions)
A lot of the information in this section can be found on the equivalent Wikipedia page, but here I have tried my best to put it both in normal people's words and in context, in the interest of readability.
2.1. The Context
Card drops in the game currently follow a geometric distribution. Keep that word in your head, we're gonna be using it a lot. For a geometric distribution to be obeyed, there are 3 criteria. Firstly, all trials (instances where a single, random "test" is carried out) must be independent and affect each other in no way. Secondly, there can only be two possible outcomes of these trials, often designated a success or failure. Thirdly, the probability of this success must be constant. If all three of these are true, the geometric distribution shows how likely it is to get the first success on any given trial.
This might sound like total jargon to anyone not vaguely familiar with statistics, but actually it's very easy to translate into the game's card drop mechanics. Each "trial" is a "kill", the "success" is the drop of a card, and cards only drop once, meaning we only consider the first "success". With that in mind, the three criteria are obeyed. The outcome of one kill (being card drop or no card drop) does not affect the outcome of any other, cards can either drop or not drop, and there is mountains of evidence that the probability of a card drop is constant.
2.2. The Maths
I tried my best to explain the graphs without the maths. Unfortunately I could not, so you'll get them later There are lots of letters from this point onwards, and I’m not just speaking about the multiple thousand words of explanation. All variables being used as part of the maths should be in italics. If you spot one that isn’t, or a sentence that reads weird, pop a reply down.
The letter we will be using for "number of kills" is x. This is because all of the graphs will have the number of kills on the x-axis (along the bottom). We'll use p for the fixed drop chance.
2.2.1. How Probability of Card Drop Varies with Number of Kills
Simple answer, it doesn't, exactly as we just said in section 2.1. above. You might be asking what the point of this section is. It is less that I want to explain how this works, but more that I want to include it for completeness before we change the maths in it later. Having said that, it does have an important consequence. If the probability of success is constant and we can only have a success or failure, by logical deduction, the probability of failure must also be constant. Assuming all probabilities add up to 1 (which is a general assumption used throughout all of the mathematical topic), that probability is 1 - p. We will be using this in a moment.
2.2.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
This is where the distribution part comes in. While the drop chance of a card is constant, the amount of kills you have to make to get it behaves a little differently. Let's take an average sample of people (and by that I mean they behave exactly as theory predicts) and run them through the card drop process.
Here is where mathematical statements start. I'm going to write down an equation and explain what it means in words underneath.
P(X = 1) = p [1]
This might look very confusing for a few of you. What it says is "The probability of people getting a card drop on the first kill is equal to probability of a card drop". That should be a lot easier to understand, and fairly self-explanatory from what we have explained already (as well as a "no shit sherlock" statement). Basically, the number after "X =" is the number of kills, and everything after the second equals sign is the probability. Everything else is mathematical terminology you can safely ignore. Onto our next kill.
P(X = 2) = (1 - p) * p [2]
This translates to "The probability of people getting a card drop on the second kill is equal to the probability of failing to get a card on the first kill, multiplied by the probability of a card drop". That should be a reasonable deduction when people can only get one card. The people who got a card on the first kill we don't want to be considering any more, as they would never reach this point in the distribution. If you've wrapped your head around that, we follow a similar process for the third kill.
P(X = 3) = (1 - p) * (1 - p) * p = (1 - p)^2 * p [3]
This translates to "The probability of people getting a card drop on the third kill is equal to the probability of failing to get a card on the first and second kill, multiplied by the probability of a card drop". If you understood the process for the second kill, we have only done the same thing here for the third. The only extra thing we have done is re-write the probability of failing to get a drop for all of the previous kills. This is nice simplification for the step below.
Here, you should have an idea of the process. The last thing to do is extrapolate said process over all kills.
P(X = x) = (1 - p)^(x - 1) * p [4]
This reads "The probability of people getting a card drop on the xth kill is equal to the probability of failing to get a card to the power of (number of kills - 1), multiplied by the probability of a card drop". If that seems like a bit of a jump, run it though backwards, like so.
P(X = 3) = (1 - p)^(3 - 1) * p = (1 - p)^2 * p [5]
P(X = 2) = (1 - p)^(2 - 1) * p = (1 - p)^1 * p = (1 - p) * p [6]
P(X = 1) = (1 - p)^(1 - 1) * p = (1 - p)^0 * p = 1 * p = p [7]
If those look confusing, we are replacing our variable x, the number of kills, with a specified number we'd actually like to find the probability for. Anything to the power of 1 is itself, used in Equation 6, and anything to the power 0 is 1, used in Equation 7.
2.2.3. The Aftermath
Pun intended. The equation we are left with, describing the geometric distribution as a whole, is Equation 4 above. This has some interesting properties. Firstly, by taking a weighted average of this distribution (check this Wikipedia page or Google it for more info) we find that the average person gets a card drop at 1 / p kills. To make things a little clearer, if the drop chance is 1 / 300, then the average person gets a card drop at 300 kills. We will be using this property later. Secondly, the only letters on the right hand side are p and x. As x is a variable we change to get distribution frequencies, p is the only parameter of the distribution. By parameter, we mean a variable that is defined when the distribution is made.
Together, these two things demonstrate an underlying property that defines a geometric distribution - the fact that it is completely random. The distribution and all of the properties associated with it depend only on the fixed drop chance p, and at no point is a card drop ever guaranteed. This means it is possible (though impossibly unlikely) that You reach up into the 10s of thousands of kills, completely empty handed.
The actual property that defines complete randomness is the card drop probability never reaching 1, or being certain, at any point. However in this case, the geometric is by far the simplest of these types of distribution.
2.3. The Graphs
To round off the geometric distribution, let's give some graphs of its function. These basically put all the maths we did above into pictures. All graphs are using the same drop chance, simply displaying the data in different ways.
The first graph is one of the card drop probability against number of kills, explained in 2.2.1. As you can see, it is constant for all kills, as we have said already many times. For now this is just for completion, as we will be including similar graphs later.
The second graph is one of Equation 4 that we derived in 2.2.2., that being the probability of reaching a given number of kills for a first card drop, plotted against the number of kills. We have P(X = x) on the y-axis, and x on the x-axis (see, calling the number of kills x did come in useful!). If you are confused as to why this decreases with the number of kills, look at the two parts of Equation 4. We have a constant p, multiplied by a "failure factor" of (1 - p)^(x - 1). This "failure factor" takes all the people that did not get a card drop on previous kills, and is what is actually decreasing. As 1 - p will always be between 1 and 0, raising it to a positive power will always cause the number to decrease. If you're still confused, just ignore it and take a look at the explanation of the third graph below, it's not the end of the world.
The third graph hasn't been explained anywhere yet, but it should be simple to understand. All we have done is taken the probability for each kill on the second graph and added it to that of all the previous kills. Mathematically, this gives us a sum of all the successes up the that point, known as a "cumulative distribution". In context, this is everyone who has got a card up to that point. This increases with the number of kills, just as you'd expect. This should be easier to digest in general as well as being a graph that fits nicely with multiple lines, so I generally refer to this one. For now, we're done with the geometric distribution (finally!).
3. The Other End of the Scale (The Unit Step)
Before we get on to the distribution I have been designing, I'd like to pop a quick word in about the exact opposite of the geometric distribution. The opposite of something that is completely random is something that is completely certain. This means that there is no factor influenced by randomness included in any of the calculations. Instead of having a fixed drop chance, we instead have a fixed number of kills at which a drop is guaranteed, and impossible beforehand. Let's make this a variable, and call it m. Using this, we can follow the process we did for the geometric distribution. The thing we end up with is a function called a "unit step", where the graph hops from 0 (no chance of drop) to 1 (guaranteed drop) instantaneously. As soon as we have a 1, the rest of the distribution is irrelevant, as everyone will have their card.
3.1. The Maths
3.1.1. How Probability of Card Drop Varies with Number of Kills
In maths, this gives us two statements.
p(x) = 0 (1 ≤ x < m) [8]
p(x) = 1 (x = m) [9]
On the left of the first equals sign, the p(x) means the probability of a card drop as a function of x. In the equivalent section for the geometric distribution, 2.2.1., I didn't give this notation to avoid extra confusion, as it would've read "p(x) = p". Equation 8 states that "the probability of a card drop on the 1st to (m - 1)th kill is 0". Equation 9 states that "the probability of a card drop on the mth kill is 1".
3.1.2. How Probability of Reaching a Particular Number of Kills for a Card Drop Varies with Number of Kills
The fantastic thing about the unit step is, when we consider how many kills people get their card drop at, it follows exactly the same pattern as in 3.1.1. above.
P(X = x) = 0 (1 ≤ x < m) [10]
P(X = x) = 1 (x = m) [11]
If the 1 in Equation 11 seems confusing, all it means is that 100% of people get their drop when the number of kills equals m, exactly as we intended.
3.2. The Graphs
And the helpful graphs return once again.
The first graph shows the probability of a card drop against the number of kills. Hey look, it's 1, or 100%, at the value of m we are using. The second graph shows how many kills people make for a card drop against the number of kills. Hey look, it's 100% at the value of m we are using. The third graph shows the sum of how many people have a card drop against the number of kills. Hey look, it's 100% at the value of m we are using.
You get the idea.
4. And Everything in Between (Weighted Geometric Distribution)
4.1. The Introduction
We have two ends of a scale now. We have complete randomness, and complete certainty. To fill in the gap between them, we need a system that has a bit of both of these factors. The question is, how do we make one?
The way we go about weighting a distribution, be it between random and certain or otherwise, is by changing the graph of the probability of a card drop. You do this by making the probability a function of the number of kills. This is true for all distributions that aren't the geometric distribution, including the unit step. If you want to be nicer to players than the geometric distribution is, you increase that probability with the number of kills. However, this is where we open up completely and can put almost any graph in. Instead of explaining what we should want out of this graph, I am instead going to propose the solution I have been designing and talk about its consequences afterwards.
Clean your ears out MLG.
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