Probability Theory
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- Queeb Salaron
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Probability Theory
So I came up with this new probability theory that breaks most things down into a 50-50 shot. (This is a joke theory, but bear with me.) Here's how it works:
Contemporary probability theory functions as a way to predict what will happen when a situation with multiple possible outcomes arises. Take, for example, rolling a single 6-sided die. There are 6 possible outcomes for a roll, and therefore, under modern theory, a 1:6 chance that any given number will appear when the die is rolled.
But let's turn the situation around a bit: Let's say that you wanted to know what the odds would be for rolling a 1 with that same 6-sided die. I submit that there are only two possible outcomes: Either you WILL roll a 1, or you WON'T roll a 1. Because there are only two possible outcomes, the ratio is increased to 1:2.
This theory applies to rolling multiple dice as well. Say you wanted to roll a five dice and wanted to predict the probability that all five of those dice will read "1." Well, it either WILL happen, or it WON'T happen, which makes it a 1:2 shot, as there are only two possible outcomes.
This theory encompasses relative probability, too. Say you wanted to predict what would happen when rolling a 6-sided die in relation to the number three. There are three possible outcomes in this case: You will either roll a number less than three, three, or a number greater than three. So the relative probability of rolling into any of those categories is 1:3.
I haven't found anyone who can disprove the theory without simply saying, "No, you're wrong."
Let's see those reasoning skills!!
Contemporary probability theory functions as a way to predict what will happen when a situation with multiple possible outcomes arises. Take, for example, rolling a single 6-sided die. There are 6 possible outcomes for a roll, and therefore, under modern theory, a 1:6 chance that any given number will appear when the die is rolled.
But let's turn the situation around a bit: Let's say that you wanted to know what the odds would be for rolling a 1 with that same 6-sided die. I submit that there are only two possible outcomes: Either you WILL roll a 1, or you WON'T roll a 1. Because there are only two possible outcomes, the ratio is increased to 1:2.
This theory applies to rolling multiple dice as well. Say you wanted to roll a five dice and wanted to predict the probability that all five of those dice will read "1." Well, it either WILL happen, or it WON'T happen, which makes it a 1:2 shot, as there are only two possible outcomes.
This theory encompasses relative probability, too. Say you wanted to predict what would happen when rolling a 6-sided die in relation to the number three. There are three possible outcomes in this case: You will either roll a number less than three, three, or a number greater than three. So the relative probability of rolling into any of those categories is 1:3.
I haven't found anyone who can disprove the theory without simply saying, "No, you're wrong."
Let's see those reasoning skills!!
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- Queeb Salaron
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Well yes, but all those different ways are still lumped together under the same category of "FAILURE TO ROLL X." So they are essentially the same.kojikun wrote:for the sake of arguing the fake theory:
there are multiple possible outcomes in the "not happen" category, not just one, so there are multiple ways it can not happen, and only one way it can.
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There might only be two possible outcomes, but 'not one' is five times more likely to occur than 'one'
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- Queeb Salaron
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True. But it is still only one of two possible outcomes. Therefore the ratio is 1:2.innerbrat wrote:There might only be two possible outcomes, but 'not one' is five times more likely to occur than 'one'
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- Ignorant twit
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You make the assumption that any given outcome is equally likely. I see no basis for this assumption.
In any event let's do this properly.
Given a number X on a six sided die with the integers 1-6 imprinted on various sides you assert that the odds the die will land X is 1/2, and the odds the die will land !X is 1/2
Assuming X != (1 or 6) you assume that there is a 1/3 chance the die lands X, a 1/3 chance the die lands > X, and a 1/3 chance the die lands < X.
Given that != is identical to > or < we can then sum the probability of the die landing > X and the die landing < X.
Thus you claim 1/3+1/3 = 1/2 or 2/3 = 1/2.
QED
In any event let's do this properly.
Given a number X on a six sided die with the integers 1-6 imprinted on various sides you assert that the odds the die will land X is 1/2, and the odds the die will land !X is 1/2
Assuming X != (1 or 6) you assume that there is a 1/3 chance the die lands X, a 1/3 chance the die lands > X, and a 1/3 chance the die lands < X.
Given that != is identical to > or < we can then sum the probability of the die landing > X and the die landing < X.
Thus you claim 1/3+1/3 = 1/2 or 2/3 = 1/2.
QED
- Queeb Salaron
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Before a di is rolled, we have no way of predicting what the "chance" of anything happening is. We can only postulate. And the best postulation that we can produce is that the expected action either WILL happen, or that it WILL NOT happen. We cannot say with any great certainty that the WILL NOT has a greater chance of occurring than the WILL without basing it on a similar postulation. We would have to be absolutely certain as to the nature of an ivory cube, its specific movement, and any other characteristics that might affect the rolling of the die. Because there is no way to be absolutely certain that rolling a three once out of every two rolls WILL NOT happen, (nor, admittedly, that it WILL happen) the only possible theory that we can produce is one based on results of which we are certain: That rolling a three either WILL happen, or it WILL NOT happen. Therefore, there is 1:2 probability of it occurring.Mad wrote:Yes, there are two outcomes, but they have an uneven chance of happening. One has a higher (often significantly higher) probability of occuring.
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At which point, you are no longer measuring the probability of something if you ignore the frequency with which it happens. It's a direct contradiction in terms. There is no valid ratio of possible outcomes in the hypothetical example you have provide, because the ratio directly involves quantity, and the quantity of "not one" is greater than the quantity of "one".Queeb Salaron wrote:True. But it is still only one of two possible outcomes. Therefore the ratio is 1:2.innerbrat wrote:There might only be two possible outcomes, but 'not one' is five times more likely to occur than 'one'
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::Claps:: Extremely well-done, Twit. Extremely well-done.Ignorant twit wrote:You make the assumption that any given outcome is equally likely. I see no basis for this assumption.
In any event let's do this properly.
Given a number X on a six sided die with the integers 1-6 imprinted on various sides you assert that the odds the die will land X is 1/2, and the odds the die will land !X is 1/2
Assuming X != (1 or 6) you assume that there is a 1/3 chance the die lands X, a 1/3 chance the die lands > X, and a 1/3 chance the die lands < X.
Given that != is identical to > or < we can then sum the probability of the die landing > X and the die landing < X.
Thus you claim 1/3+1/3 = 1/2 or 2/3 = 1/2.
QED
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The catch is this: We cannot simply assume the frequency with which anything will occur. Empircal data taken while rolling dice proves time and time again that these assumptions are not completely accurate. The only thing that is accurate, however, is the black-and-white fact that either the number X is rolled, or it is not rolled. Thus two categories, thus a 1:2 ratio.Hotfoot wrote:At which point, you are no longer measuring the probability of something if you ignore the frequency with which it happens. It's a direct contradiction in terms. There is no valid ratio of possible outcomes in the hypothetical example you have provide, because the ratio directly involves quantity, and the quantity of "not one" is greater than the quantity of "one".
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- Colonel Olrik
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All sets of interest in probability theory have elements taken from the largest set called a space and designated as S. Hence, all sets will be subsets of S. Supposing that the elements of a space consists of the six faces of a die, and that these faces are designated as 1,2,...,6, we'll have
S={1,2,3,4,5,6}
Subsets of S will be, for example, {1} or {1,4}, and can be identified with the events. Any event consisting of a single element is called an elementary event. To each event we assign a number, called the probability of the event. If the event is denoted A, the probability of that event will be named Pr(A).
If the elementary events are assumed to be equally probably, then each one will have a probability of 1/6.
Since the subset {3} is a elementary event, it has the probability of 1/6.
Since the subset {1,2,4,5,6} is not an elementary event, its probability is given by the sum of all the belonging elementary events, therefore being 1/6*5 = 5/6.
Therefore, you are wrong and God exists.
This thread is now dead.
S={1,2,3,4,5,6}
Subsets of S will be, for example, {1} or {1,4}, and can be identified with the events. Any event consisting of a single element is called an elementary event. To each event we assign a number, called the probability of the event. If the event is denoted A, the probability of that event will be named Pr(A).
If the elementary events are assumed to be equally probably, then each one will have a probability of 1/6.
Since the subset {3} is a elementary event, it has the probability of 1/6.
Since the subset {1,2,4,5,6} is not an elementary event, its probability is given by the sum of all the belonging elementary events, therefore being 1/6*5 = 5/6.
Therefore, you are wrong and God exists.
This thread is now dead.
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From a scientific point of view, it doesn't need to be perfectly accurate. In fact, there is no such thing as a truly random number generator in the world, so the mathematical formulas we use are little more than rough approximations of reality, or if you'd rather, how things should work in a perfect world. There is not a single die that has ever been created that has a perfect 1/6th chance of any one result. Yet the average experience tells us that 1/6th is a close enough approximation that we can safely use it without deviating terribly far in either direction.Queeb Salaron wrote:The catch is this: We cannot simply assume the frequency with which anything will occur. Empircal data taken while rolling dice proves time and time again that these assumptions are not completely accurate. The only thing that is accurate, however, is the black-and-white fact that either the number X is rolled, or it is not rolled. Thus two categories, thus a 1:2 ratio.
Your assertation that we simply throw experimentation out the window is curious, as it is a simple and effective error-checking device. Just because something is not completely accurate does not mean that it is completely inaccurate. Re: "But Scientists have been wrong before!" thread.
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I think the argument here is a "black and white" arguement, or something. It assumes only two states despite the fact that one state is actually 5 states not one. Its a fall assumption on the part of someone who makes this statement, and its also an incorrect statement to say that the chances of landing in the "not" state is the same as landing in the other state, because the not state contains 5 equally likely states within it.
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All your arguments are good and valid (and right, hehe) but I think Twit wins this one. Feel free to debate this some more, but I just wanted to see it disproven just to prove to myself that I hadn't stumbled across some great logical fallacy that had been widely accepted as truth.
Truth be told, even I couldn't disprove myself.
::Claps:: Good job everyone.
Truth be told, even I couldn't disprove myself.
::Claps:: Good job everyone.
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This argument is good if were were in a debate, but we're not debating it. We're trying to disprove it.kojikun wrote:I think the argument here is a "black and white" arguement, or something. It assumes only two states despite the fact that one state is actually 5 states not one. Its a fall assumption on the part of someone who makes this statement, and its also an incorrect statement to say that the chances of landing in the "not" state is the same as landing in the other state, because the not state contains 5 equally likely states within it.
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You didn't have probability classes at school? That stuff is idiotically simple to disprove with basic set theory.Queeb Salaron wrote:All your arguments are good and valid (and right, hehe) but I think Twit wins this one. Feel free to debate this some more, but I just wanted to see it disproven just to prove to myself that I hadn't stumbled across some great logical fallacy that had been widely accepted as truth.
Truth be told, even I couldn't disprove myself.
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Not really... I was more of a letters-type student.Colonel Olrik wrote:You didn't have probability classes at school? That stuff is idiotically simple to disprove with basic set theory.Queeb Salaron wrote:All your arguments are good and valid (and right, hehe) but I think Twit wins this one. Feel free to debate this some more, but I just wanted to see it disproven just to prove to myself that I hadn't stumbled across some great logical fallacy that had been widely accepted as truth.
Truth be told, even I couldn't disprove myself.
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Actually, you have. It is widely accepted as truth, at least amongst people who lack the ability to rationally analyze things.Queeb Salaron wrote:All your arguments are good and valid (and right, hehe) but I think Twit wins this one. Feel free to debate this some more, but I just wanted to see it disproven just to prove to myself that I hadn't stumbled across some great logical fallacy that had been widely accepted as truth.
I want a cookie.Truth be told, even I couldn't disprove myself.
::Claps:: Good job everyone.
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The Realm of Confusion
"Every time you talk about Teal'c, I keep imagining Thor's ass. Thank you very much for that, you fucking fucker." -Marcao
SG-14: Because in some cases, "Recon" means "Blow up a fucking planet or die trying."
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- Slartibartfast
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Re: Probability Theory
Since there are 6 sides, the chances of ROLLING 1 and NOT ROLLING 1 aren't equal. No 50-50. There is 1/6 chance that you WILL roll a 1, and 5/6 chance that you WON'T roll a 1. Just like the probability that I'll be killed tomorrow and the probabilty that I WON'T aren't 50-50.Queeb Salaron wrote:But let's turn the situation around a bit: Let's say that you wanted to know what the odds would be for rolling a 1 with that same 6-sided die. I submit that there are only two possible outcomes: Either you WILL roll a 1, or you WON'T roll a 1. Because there are only two possible outcomes, the ratio is increased to 1:2.
Probability is based on calculating the likelihood of a given event, not in assuming that they are all equally likely.
There you go.I haven't found anyone who can disprove the theory without simply saying, "No, you're wrong."
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Interesting.Jeff My Roomate wrote:If you disregard percentages, you're still right. There is essentially a 1:2 probability that you will roll any given number on a 6-sided die, because you never claimed that those two options were of equal weight. Essentially, the "2" in "1:2" is lopsided.
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Your looking at a different result space from what people normally think of when rolling a dice. Normally you would think of the results breing the set {1, 2, 3, 4, 5, 6}. But what your looking at here is the set
{Dice rolls 1, Dice dosn't roll one}. So there are only two possible outcomes but the probabilities of each out come are not equal.
Another proof
[maths brain=on]
P(Dice=1) + P(dice!=1) = 1
Assuming a 1:2 ratio then P(dice=1)=P(dice!=1)=0.5
P(dice!=1)=P(Dice=2) + P(Dice=3) + P(Dice=4) + P(Dice=5) + P(Dice=6)
now IF it is a fair dice then the probabilities of rolling any number is equal
i.e.
P(dice=1)=P(dice=2)=P(dice=3)=...
so
P(dice!=1)=5*P(dice=1)
then
P(dice=1)+P(dice!=1)=0.5+5*.5=3>1
so assuming a fair die the probability of rolling 1 or not rolling 1 isn't 1:2.
[maths brain=off]
{Dice rolls 1, Dice dosn't roll one}. So there are only two possible outcomes but the probabilities of each out come are not equal.
Another proof
[maths brain=on]
P(Dice=1) + P(dice!=1) = 1
Assuming a 1:2 ratio then P(dice=1)=P(dice!=1)=0.5
P(dice!=1)=P(Dice=2) + P(Dice=3) + P(Dice=4) + P(Dice=5) + P(Dice=6)
now IF it is a fair dice then the probabilities of rolling any number is equal
i.e.
P(dice=1)=P(dice=2)=P(dice=3)=...
so
P(dice!=1)=5*P(dice=1)
then
P(dice=1)+P(dice!=1)=0.5+5*.5=3>1
so assuming a fair die the probability of rolling 1 or not rolling 1 isn't 1:2.
[maths brain=off]
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