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Topic: Gold or Silver? (Read 2660 times) |
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towr
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Re: Gold or Silver?
« Reply #50 on: Jun 16th, 2003, 6:13am » |
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suppose in a game of coin flipping the sequence of one player is
11111011001001001111 (about 2/3 chance of 1)
If I knew that I'd choose
11111111111111111111 and score 13
or
11111101011101011111 (also 2/3s randomly chosen 1's) and score 15
but the opposite strategy is equally likely, unless it's given there is no validation to assume he has such a 2/3 sequence. And then I would score only 7 or maybe just 5
So not knowing which of the two strategies is used I choose randomly,
01000000111000001101
and score 10 out of 20 = 50% in both cases
(the random generator was nice in making this example)
Probability allways comes from imperfect knowledge in prediction. If I knew the future all my 'guesses' would be right.
If I knew a die was used and at rolling 1 the SG box was selected I'd use that information, P(SG) = P(SG | Strategy-1)= 1/6
But else I have to use P(SG) = Sum(P(SG | Strategy-i) * P(Strategy-i))
and since every strategy has an exact opposite which is a priori equally likely, I get P(SG) = 1/2
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| « Last Edit: Jun 16th, 2003, 6:22am by towr » |
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Sir Col
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Re: Gold or Silver?
« Reply #51 on: Jun 16th, 2003, 6:40am » |
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I'm sorry, Towr, but I am struggling to follow your logic. Let me summarise my case:
Firstly, the problem:
A box contains two coins, either two Silver or one Silver and one Gold. A coin is chosen at random.
It is Silver. What is the probability that the other coin is also Silver?
There exist two models that satisfy the givens:
(i) There is an equal chance of the box containing S1S2 or SG. Having taken a silver coin there are 3 ways this can happen, two of which leave a silver. So P(other coin is silver)=2/3.
(ii) The box contains two coins, each with equal chance of being G or S. There are four possible configurations: SS, SG, GS, GG (all with equal chance, 1/4), but as we are told that it contains SS or GS, we have three possible choices: SS, SG or GS. From this we can see that, having taken a silver, P(other coin is silver)=1/3.
I would argue that there are no reasonable grounds to assume that each model, based on the lack of further information, is equally likely. If you were given x+3=?, and it is not clear if the number on the right hand side is 6 or 8. You could, at least, assert that x=3 or x=5, but is it sensible to say that, given the lack of further information, x=4?
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towr
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Re: Gold or Silver?
« Reply #52 on: Jun 16th, 2003, 7:11am » |
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I disagree that model ii fits, it assumes things that simply aren't given. There is no reason to assume each coin in the box has 50-50 chance of being silver or gold.
you have SS or SG, that is given.
no information on how it chosen.
Actually, if both coins had 50% chance of being gold, the box should be able to contain GG as well. It is given that it can only contain SS or GS, so it is impossible that this method was used to fill the box.
Of course there are hundreds of other schemes to fill the box, put in one silver and throw a die, put in gold if it's one, or put in gold when it's not one. And for none of them is there any reason to assume they are used, since no information regarding it is given.
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| « Last Edit: Jun 16th, 2003, 7:30am by towr » |
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towr
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Re: Gold or Silver?
« Reply #53 on: Jun 16th, 2003, 7:24am » |
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on Jun 16th, 2003, 6:40am, Sir Col wrote:
If you were given x+3=?, and it is not clear if the number on the right hand side is 6 or 8. You could, at least, assert that x=3 or x=5, but is it sensible to say that, given the lack of further information, x=4? |
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If you role a die it never roles 3.5, yet that is still the expected value you get when you role.
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| « Last Edit: Jun 16th, 2003, 7:24am by towr » |
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towr
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Re: Gold or Silver?
« Reply #54 on: Jun 16th, 2003, 8:27am » |
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I'm really posting to many consequetive messages.. but anyway..
the problem
A box contains two coins, either two Silver or one Silver and one Gold. A coin is chosen at random.
It is Silver. What is the probability that the other coin is also Silver?
There exist an infinite number models that satisfy the givens. Let's examine a few.
GS and SS are equally likely, you can get this in many ways, the easiest:
- take one silver coin, then randomly with equal chance add either a gold or silver coin
The probability of GS is twice as high as SS, again there are many ways to get this.
- For instance twice choose uniformly random a gold or silver coin, if not both are gold you're done, else start anew.
- Or the easier way, take one silver coin, and use a die to choose either a silver (sides 3 or 6), or gold coin (sides 1,2,4 or 5)
The probability of GS is half that of SS
- Take one silver coin, and use a die to choose either a silver (sides 1,2,3 or 4), or gold coin (sides 5 or 6)
- Or twice choose uniformly random a gold or silver coin, if not both are silver replace gold with silver and vice versa and you're done, else start anew.
there are ways to get any probability ratio of GS to SS, and you can always simply replace GS with SS or vice versa.
If all strategies are equally likely (and there is no information supporting the opposite), it evens out to both being equal.
And to minimize risk we have to assume all strategies are equally likely, because we don't know which would be more likely and thus can't use it to improve our odds.
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| « Last Edit: Jun 16th, 2003, 8:31am by towr » |
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Sir Col
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Re: Gold or Silver?
« Reply #55 on: Jun 16th, 2003, 8:27am » |
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Sorry, but I believe that the two models I propose are valid. It all stems from the first sentence:
A box contains two coins, either two Silver or one Silver and one Gold.
Ignoring the given statement for the moment, if a box contains two coins (silver and/or gold), there are four equally like configurations: SS, SG, GS and GG.
Now we apply the given (either two silver or one silver and one gold) and we can interpret that two ways:
(i) it is equally likely to be SS or GS (order unimportant).
(ii) it is equally likely to be SS, SG or GS (eliminating the GG possibility).
The first model leads to P(other coin silver)=2/3 and the second leads to P(other coin silver)=1/3.
I would also be bold enough to say that there is no other model for this problem, unless...
We argue that it cannot be assumed that a single coin being silver/gold is equally likely. I accept this possibility, but it still reinforces the point that the problem, as it stands, lacks a definite solution. I believe we're singing from the same hymn sheet on this issue? 
With regards to your expected outcome analogy, expected outcomes are quite different to probability distributions. Expected outcomes can only be applied to quantitative data (numerical types), and whereas probability can be applied to both types, the silver/gold situation is entirely qualitative.
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| « Last Edit: Jun 16th, 2003, 8:32am by Sir Col » |
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towr
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Re: Gold or Silver?
« Reply #56 on: Jun 16th, 2003, 8:34am » |
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I sure wouldn't mind having you as a customer in my casino (if I had one) 
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| « Last Edit: Jun 16th, 2003, 8:35am by towr » |
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redPEPPER
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Re: Gold or Silver?
« Reply #57 on: Jun 16th, 2003, 9:25am » |
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towr, I now understand a little better. Here's where we differ:
You are examining the probability of something happening given what you know. I'm examining the probability given that there's something I don't know.
Let's define probability as the portion of time something occurs if the experiment is reproduced a sufficient number of time. This is probably not an accurate definition but it'll do. Example: spin a balanced coin, it will come as heads about 500 times out of 1000. The probability is about 1/2.
Now let's imagine the coin might not be balanced, or might be tricked. You say it doesn't matter because it can be tricked in symmetrical ways that even out. If we reproduce the experiment 1000 time while keeping a constant only the things you know (you flip a coin which will result in either heads or tail) then the probability remains 1/2. One time you'll flip a coin that has two heads, another time the coin will have two tails, one time it'll be heavier on one side, the other time heavier on the other side... It evens out and you still get about 500 heads out of 1000... Assuming the coins are chosen at random among all the possible coins!
But if we reproduce the experiment 1000 times while keeping a constant the things you know AND the things you don't know (for example the coin has two tails) then the probability changes drastically.
In other words, the probability that a random coin, tricked or not, comes as head is 1/2. But the probability that this specific coin here comes as heads might not be 1/2. If you think it is, then I wouldn't mind having you at my casino, betting on it coming as heads
But I think you're right: of all the systems that will put two coins in a box, one being silver and the other one silver or gold, the probability that they're both silver will be 1/2. While I was focused on this specific system, which had a unique way (unknown to me) to determine whether to put silver or gold. I was focused on the probability of one system, you calculated the probability of any system that met the requirement, which is probably a better answer.
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towr
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Re: Gold or Silver?
« Reply #58 on: Jun 16th, 2003, 9:47am » |
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on Jun 16th, 2003, 8:27am, Sir Col wrote:Sorry, but I believe that the two models I propose are valid. It all stems from the first sentence:
A box contains two coins, either two Silver or one Silver and one Gold. |
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Note that there is no mention at all how the box is filled.
Quote:| Ignoring the given statement for the moment, if a box contains two coins (silver and/or gold), there are four equally like configurations: SS, SG, GS and GG. |
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note that the box cannot contain GG, and this thus cannot be a valid way to fill the box.
Quote:Now we apply the given (either two silver or one silver and one gold) and we can interpret that two ways:
(i) it is equally likely to be SS or GS (order unimportant).
(ii) it is equally likely to be SS, SG or GS (eliminating the GG possibility). |
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note that you make an assumption about the distribution, without there being any information to justify it.
Quote:| I would also be bold enough to say that there is no other model for this problem, unless... |
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I allready gave a way to get an infinite amount of other models that fit (but are equally unsupported by the information)
Quote:| We argue that it cannot be assumed that a single coin being silver/gold is equally likely. I accept this possibility, but it still reinforces the point that the problem, as it stands, lacks a definite solution. I believe we're singing from the same hymn sheet on this issue? |
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Probability theory can _allways_ give an answer. Lack of information means you have an even guess, that is the best way to limit risk/loss = maximize gain.
Quote:| With regards to your expected outcome analogy, expected outcomes are quite different to probability distributions. Expected outcomes can only be applied to quantitative data (numerical types), and whereas probability can be applied to both types, the silver/gold situation is entirely qualitative. |
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The expected outcome is the average of the probability distribution, it'd be odd if they were identical. But I never even implied they were.
Expected values can be gotten from qualitive data, the side of the die the lands up is qualitive data. Form the qualitive data you can get quantitive data (the value), and from that an expected value.
In the silver/gold case you can do the same with expected winnings.
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towr
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Re: Gold or Silver?
« Reply #59 on: Jun 16th, 2003, 10:24am » |
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on Jun 16th, 2003, 9:25am, redPEPPER wrote:
But if we reproduce the experiment 1000 times while keeping a constant the things you know AND the things you don't know (for example the coin has two tails) then the probability changes drastically. |
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Of course. But there's still no way to know what you don't know, so you can't use that the real probability is drastically biased. You don't know which of the coins you have, even if you might know that it is a biased/trick coin.
So the best way to maximize your gain/minimize loss is bet randomly, that will make it fair.
And if you can change your bets along the line you can switch to whatever turns up most after let's say ten flips.
(Might be interesting to look up the computer roshamboo competition, it's computer programs playing a sort of rock-paper-scissor, trying to beat each other with strategies, in part based on recognizing the strategy of the other player. And once you're sufficiently ahead you can stay ahead by playing randomly.)
Quote:| In other words, the probability that a random coin, tricked or not, comes as head is 1/2. But the probability that this specific coin here comes as heads might not be 1/2. If you think it is, then I wouldn't mind having you at my casino, betting on it coming as heads |
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I wouldn't play heads all the time, but randomly switch :p
Quote:| But I think you're right: of all the systems that will put two coins in a box, one being silver and the other one silver or gold, the probability that they're both silver will be 1/2. While I was focused on this specific system, which had a unique way (unknown to me) to determine whether to put silver or gold. I was focused on the probability of one system, you calculated the probability of any system that met the requirement, which is probably a better answer. |
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I'm glad someone understands me
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Icarus
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Re: Gold or Silver?
« Reply #60 on: Jun 16th, 2003, 5:30pm » |
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Here is my probable 2 cents worth:
(1) T&B - Infinities I've studied well. Certain other branches of mathematics highly unlikely to come up in this forum, I can pontificate on endlessly (Okay - I do that on anything - but in these I can with authority). Probability is not among them, however, so don't expect me to sort this out!
Having said that...
(2) Probability can be expressed as measure theory with the restriction that the measure of the whole space is 1. However this really doesn't add much to your knowledge of probability. The only advantage to be gained by bringing measure theory in is to increase the scope of events to which a probability can be defined -- but NOT in a useful fashion! (Measure theory allows you to provide probabilities to sets which you can show must exist, but cannot actually define.)
(3) It seems to me that what towr is describing could be described as "metaprobability" or "2nd Order probability". In ordinary probability, you have a probability density function (pdf) described in some way, and you use it to calculate the probability of various events. For instance, if f is the pdf of a standard die, f(1)=f(2)=f(3)=f(4)=f(5)=f(6) = 1/6. From which you can calculate, for instance, that the probability of rolling a number divisible by 3 is f(3)+f(6) = 1/3.
But in this situation, the pdf itself is unknown. Towr's approach is to back up one step, asking first "what is the probability of each particular pdf, and given that pdf, what is the probability of a second silver coin?". This is a valid approach, provided one can answer the first part of the question.
Here is where I have to disagree: Towr assumes that, in the lack of all other information, all pdfs are equally probable. As he puts it, "even chance". I don't agree that this is a good assumption, even lacking further information. In particular, in this case you are never totally lacking in further information. For instance I would definitely assume that the pdf giving equal chances to each outcome is far more likely to be the case than the pdf making gold exactly 123653242093543781439231 times more likely than silver.
A second objection is deeper: Towr says "even chance". I say "even with respect to what?". The concept of uniform probability is only definable in terms of a "naturally occuring" distribution. For events with a finite number of possible outcomes this "naturally occuring" distribution is easily recognized: the probability of all events are the same. For events with an infinite number of possible outcomes, things are trickier. Sometimes a choice is available because of some aspect of the outcomes. For instance, if the outcomes correspond to all real numbers between 0 and 1, then you can define the pdf which gives the probability that X < y to be y, as being uniform.
In other cases, no good choice of uniform probability exists. If the outcomes are all Natural numbers, there is no non-arbitrary way of defining uniform probability. As far as I see, pdfs themselves are another example. Can you define a pdf of all possible pdfs in this problem that reasonably can be considered uniform?
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Re: Gold or Silver?
« Reply #61 on: Jun 16th, 2003, 9:20pm » |
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Icarus, what do you think of Sir Col's idea that maybe nothing is certain?
If at the microscopic level of the physical world there is always uncertainty, and 'an event happens with probability 1' does not mean it is certain to happen, then perhaps the only true certainty we have is logical certainty.
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towr
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Re: Gold or Silver?
« Reply #62 on: Jun 16th, 2003, 11:37pm » |
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on Jun 16th, 2003, 5:30pm, Icarus wrote:
(3) It seems to me that what towr is describing could be described as "metaprobability" or "2nd Order probability".
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It's only one way to in a sense justify that without information probability is 50-50. The real statistician probably has a better proof, but it is nonetheless the standard. No information -> even chance.
Quote:| For instance I would definitely assume that the pdf giving equal chances to each outcome is far more likely to be the case than the pdf making gold exactly 123653242093543781439231 times more likely than silver. |
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gold being exactly 123653242093543781439231 times more likely than silver
is balanced by
silver being exactly 123653242093543781439231 times more likely than gold
And it doesn't even matter how much more or less likely they are than the case of gold being equally likely as silver.
And also, you are using supposed information about the world that isn't given, nor justified.
If nothing else I can justify myself by using Occams razor. Even chance is the simplest model you can get.
Quote:| A second objection is deeper: Towr says "even chance". I say "even with respect to what?". |
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Even with repect to each other. Like each side of a die having even chance.
Quote:Can you define a pdf of all possible pdfs in this problem that reasonably can be considered uniform?
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It isn't even needed, any symmetrical pdf will do, to show the same result. But yes, I think there is. Every possible pdf for the problem is defined by one (positive) real, and the opposite by its reciprocal. The reciprocal is allways between 0 and 1, and you just showed a uniform distribution for it.
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redPEPPER
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Re: Gold or Silver?
« Reply #63 on: Jun 17th, 2003, 3:53am » |
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on Jun 16th, 2003, 11:37pm, towr wrote:
gold being exactly 123653242093543781439231 times more likely than silver
is balanced by
silver being exactly 123653242093543781439231 times more likely than gold
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But how can you be sure of that assumption? There's less gold on earth than silver, so wouldn't the second be more likely? I mean, if you measured that and created a significant number of systems, you're claiming that you'd make about as many systems where gold is n times more likely as the opposite systems where gold is n times less likely. But maybe that assumption is invalid?
I understood your meta modelization as such: if you don't know an information, calculate the probability for all systems, that comply with the informations you have, regardless of that information you don't have. But I'm not sure you can assume all these systems are evenly balanced. Not knowing is not the same as knowing they're equally probable.
I'll try to illustrate with an example:
What is the probability to roll a die and make 7? There probably exists 6-faced dice with a 7 on one of the sides, and there certainly exists dice with more than 6 faces. But these aren't anywhere as likely as a regular die, with faces numbered from 1 to 6.
You don't know how much more likely these are though. And you don't know what die will be rolled. So how do you process that? Do you ignore the information? Is 7 as probable as, say 3? Or as 2645?
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towr
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Re: Gold or Silver?
« Reply #64 on: Jun 17th, 2003, 4:39am » |
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on Jun 17th, 2003, 3:53am, redPEPPER wrote:
But how can you be sure of that assumption? There's less gold on earth than silver, so wouldn't the second be more likely? |
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There is no certainty. Aside from that the box can only contain one configuration at a time.
But if I don't have any information I also don't know if there is less gold than silver on earth. Besides, there's just two coins, not mountains of the stuff.
There is a lot more real world information you could use, but none of it is substantiated by the puzzle.
Quote:| I understood your meta modelization as such: if you don't know an information, calculate the probability for all systems, that comply with the informations you have, regardless of that information you don't have. But I'm not sure you can assume all these systems are evenly balanced. Not knowing is not the same as knowing they're equally probable. |
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As long as you don't know otherwise I claim they are. It's the only way to limit your risk and make a safe guess, which will have the same expected gain as allways choosing one option, or the other, when you repeat it any number of times without changing the underlying process.
Quote:I'll try to illustrate with an example:
What is the probability to roll a die and make 7? There probably exists 6-faced dice with a 7 on one of the sides, and there certainly exists dice with more than 6 faces. But these aren't anywhere as likely as a regular die, with faces numbered from 1 to 6. |
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But only because you have a lot of knowledge about dice.
Quote:You don't know how much more likely these are though. And you don't know what die will be rolled. So how do you process that? Do you ignore the information? Is 7 as probable as, say 3? Or as 2645?
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If you don't know anything about dice, yes.
If you don't know anything about the distribution, the safe bet is even chance. Else you have to guess at what distribution it is, and you're just as likely to get everything right, as everything wrong.
We allready know that the box contains either SG or SS with 100% and the other with 0%, because they're allready in the box. (Unless you want to allow a superpositionlike state like with schrodingers cat, but let's not..)
What is your best guess at this point? That's what its all about.
I say you can't know which is more likely, because there is no mention about the process that led to how the box was filled, no mention of gold-shortage, no mention about the wealth of the game-master or any other relevant factor.
So 50-50 gives you the best result; smallest variance, same mean. Regardless of the distribution you break even, it's a baseline.
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| « Last Edit: Jun 17th, 2003, 4:47am by towr » |
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Sir Col
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Re: Gold or Silver?
« Reply #65 on: Jun 17th, 2003, 5:02am » |
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Sorry to sound like a stuck record, and I know records are practically obsolete now before anyone says anything! 
Towr, I'm curious about your reproach of my post. You take quotes, object to a point, then add the next quote in which I, myself, qualify my statements?
However, I believe that we are discussing something entirely different. My argument was to demonstrate that, despite the two possible interpretations using the simplified model (equal chance of a single coin being silver or gold), we have no rational grounds to determine which interpretation is more or less likely. Conclusion: there is insufficient information to solve the problem.
I then admitted a naivety in my base assumption (using the simplified model), and accepted that the unknown random probability distribution for a silver/gold coin creates infinitely many possible scenarios; for which we have no rational grounds to determine which model is more or less likely. Conclusion: there is insufficient information to solve the problem.
In both cases, I still believe that we are not permitted to apply symmetries to probability models if we lack information. If I ask you to determine the probability of drawing a red disc out of a bag of coloured discs, you simply cannot solve the problem without more information. This, I believe, is analogous to the original problem.
I would still like to know what people think about certainties in the realm of probabilities. I appreciate that the total sum of all probable events must equal one, but can we apply probabilities to a single probable event and define it as being certain? Is it reasonable to say that the probability of rolling a 1-6 on an ordinary die is 1, or do probabilities simply not apply?
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towr
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Re: Gold or Silver?
« Reply #66 on: Jun 17th, 2003, 7:14am » |
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When it comes to statistical problems there is never to little data to solve the problem. You simple make the best guess you can with the information that is there.
And since rolling a die is a very physical problem once you have all information on how it is thrown and where you can simply apply mechanics and calculate which side ends up. So there it will 100% be 1 or 0%. It's just very hard to get the information, even though it exists.
Due to chaotic effects and lack of knowledge about the starting conditions the best guess is usually 1/6th.
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| « Last Edit: Jun 17th, 2003, 7:28am by towr » |
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Icarus
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Re: Gold or Silver?
« Reply #67 on: Jun 17th, 2003, 5:39pm » |
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If the only acceptable solution to the problem is to produce a number, then towr's approach is clearly the best way to go. But it would be better to call that number a "metaprobability" than "the probability". It is a probability over a broader event space than that directly encompassing the problem.
Occasions where it is necessary to produce a statistical result without sufficient information are numerous (I've had to do so several times in the course of my job), but I don't hold that this problem is one of them. It is enough here to point out that the information given is insufficient to calculate the "actual" or "first-order" probability.
On the other hand, it is informative to go ahead and give the metaprobability solution. But when giving the metaprobability, it needs to be clearly stated what assumptions are being made about the corresponding "metadistribution".
To me, either answer "solves" the problem as stated (you could state it differently to force one or the other answer).
Quote:| And also, you are using supposed information about the world that isn't given, nor justified. |
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I disagree. I find no cause at all to reject this as "unjustified". You give me any situation in any society on any planet or any other "any" that you wish, which has not been purposely chosen to go the other way, and I say that my "information" is almost certainly true for that situation. To throw out this information is unjustified in my opinion.
Quote:If nothing else I can justify myself by using Occams razor. Even chance is the simplest model you can get.
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Occam's razor is a tool of science, not of mathematics at all, and even in science it is not a very good one. I can give you a vastly simpler explanation of the universe than that currently in vogue: Everything is the result of random chance. This model (fleshed out a bit) completely explains all observations past and future. It is extremely simple and by Occam's razor should be viewed as best. It is also completely useless. Therefore by the razor of science, "The explanation that best predicts future observations is the closest to correct", it is discarded for the very complex and convoluted explanations of modern particle theories.
Similarly here, your "even chance" produces a number. But perhaps a different distribution of pdfs would produce a probability more likely to match observed situations.
Quote:| Even with repect to each other. Like each side of a die having even chance. |
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This definition only works when there are a finite number of possible events. When there are an infinite number of events, it is insufficient to define what "even chance" is: It will tell you that the probability of any individual event is 0, but there are infinitely many distributions that satisfy this requirement.
Quote:| [A pdf of pdfs] isn't even needed, any symmetrical pdf will do, to show the same result. |
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True, but the result should be stated as contigent on this assumption.
Quote:| But yes, I think there is. Every possible pdf for the problem is defined by one (positive) real, and the opposite by its reciprocal. The reciprocal is allways between 0 and 1, and you just showed a uniform distribution for it. |
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True as well - my objection was more of a general one. For this particular problem, the pdfs boil down to a probability that two silver coins were placed in the box, so you can assume the uniform distribution on the interval [0,1]. But I would wager that a bell-shaped distribution (I don't recall what the standard one is for a bounded interval) would give you a number more likely to occur in any "real-world" situation.
For problems with more complexity in the possible pdfs, it may well be that there is no natural way of defining "even chance".
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towr
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Re: Gold or Silver?
« Reply #68 on: Jun 18th, 2003, 1:51am » |
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on Jun 17th, 2003, 5:39pm, Icarus wrote:
I disagree. I find no cause at all to reject this as "unjustified". You give me any situation in any society on any planet or any other "any" that you wish, which has not been purposely chosen to go the other way, and I say that my "information" is almost certainly true for that situation. To throw out this information is unjustified in my opinion. |
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There are only two coins, the amount of gold on the planet doesn't matter.
And if you can use this information, why not use other information. Anyone setting up such a game is likely to want to trick you. Anyone using a silver coin for such a game probably has a gold one as well. People wouldn't generally say either silver and silver or gold and silver if they didn't have a gold coin to make the latter possible. etc etc.
Why not call the psychic hotline and ask them which box it is.
Quote:| Occam's razor is a tool of science, not of mathematics at all, and even in science it is not a very good one. I can give you a vastly simpler explanation of the universe than that currently in vogue: Everything is the result of random chance. This model (fleshed out a bit) completely explains all observations past and future. It is extremely simple and by Occam's razor should be viewed as best. It is also completely useless. Therefore by the razor of science, "The explanation that best predicts future observations is the closest to correct", it is discarded for the very complex and convoluted explanations of modern particle theories. |
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Yes, but given that models have the same explanatory/predictive power (which is the case here), the simplest is closest to correct.
Quote:| Similarly here, your "even chance" produces a number. But perhaps a different distribution of pdfs would produce a probability more likely to match observed situations. |
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There is no observed situation yet. When there is it gives plenty of information to change the model using bayes.
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Re: Gold or Silver?
« Reply #69 on: Jun 18th, 2003, 3:38pm » |
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on Jun 18th, 2003, 1:51am, towr wrote:
There are only two coins, the amount of gold on the planet doesn't matter.
And if you can use this information, why not use other information. Anyone setting up such a game is likely to want to trick you. Anyone using a silver coin for such a game probably has a gold one as well. People wouldn't generally say either silver and silver or gold and silver if they didn't have a gold coin to make the latter possible. etc etc.
Why not call the psychic hotline and ask them which box it is. |
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My comments were not based on any assumption of how common gold or silver is. They were based only on the idea that some probabilities are in and of themselves more likely to come up than others. I find it very likely that someone may choose to put the coins in with equal chance for gold or silver, or to put them in with no chance for gold or with no chance of silver. I find it extremely unlikely that they would choose in a fashion that gives the gold coin (or the silver coin) an ~ 10-25 chance of being in the box.
Figuring this out does not need "psychic ability" - just common sense.
My point is that there is more information to be used. You might as well make the best use of it that you can.
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Yes, but given that models have the same explanatory/predictive power (which is the case here), the simplest is closest to correct.
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I disagree that this "is the case here". By claiming this, you are asserting that no other choice for pdf of pdfs will predict the overall outcomes more accurately than the uniform one. You can't demonstrate this.
(And that the simplest is not always closest to correct - Occam's Razor only asserts that it is more likely to be correct, not that it definitely is, and the "science razor" only makes assertions for predictive power, not simplicity.)
Quote:| There is no observed situation yet. When there is it gives plenty of information to change the model using bayes. |
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But some possible "observable situations" are more likely than others. This opens the door to refining the numbers before observation begins.
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Re: Gold or Silver?
« Reply #70 on: Jun 18th, 2003, 11:10pm » |
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on Jun 18th, 2003, 3:38pm, Icarus wrote:| I find it extremely unlikely that they would choose in a fashion that gives the gold coin (or the silver coin) an ~ 10-25 chance of being in the box. |
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If you want to introduce your personal bias into a probabilistic problem, well, that's your choice..
I wouldn't call it a mathematicle approach though..
Quote:| I disagree that this "is the case here". By claiming this, you are asserting that no other choice for pdf of pdfs will predict the overall outcomes more accurately than the uniform one. You can't demonstrate this. |
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No, I claim that you cannot know which pdf will predict the overall outcome more accurately, there is no way to make a better guess.
Quote:| But some possible "observable situations" are more likely than others. This opens the door to refining the numbers before observation begins. |
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Only when you allready have a model/theory of the universe you're in.
If you don't think this is a math problem, just say so..
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