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        riddles >> easy >> Gold or Silver? 
        
(Message started by: THUDandBLUNDER on Jun 10th, 2003, 10:58am)
Title: Gold or Silver?
Post by THUDandBLUNDER on Jun 10th, 2003, 10:58am
[Not sure if this has been posted before.]  :-/

There are two boxes, each containing two coins.

One box contains a Gold coin and a Silver coin.

The other box contains two Silver coins.

A Silver coin is drawn from a box that is chosen at random.

What is the probability that the other coin in the box is also Silver?
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 11:27am
[hide]50%[/hide], because [hide]you chose the box, not a particular coin[/hide].

This is a little like the two-child family riddles in the medium section, isn't it?
Title: Re: Gold or Silver?
Post by towr on Jun 10th, 2003, 11:33am
Well, it depends, if the drawn silver coin is the result of the deliberate drawing of a silver coin, or the drawing of a random coin from the box. If the latter is the case than the answer may be different from 50%.
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 11:37am
It says


on 06/10/03 at 10:58:04, THUDandBLUNDER wrote:
A Silver coin is drawn from a box that is chosen at random.

To me, this means that the box is chosen at random. After that, it doesn't matter which silver coin you take if you chose the box with two silver coins.
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 10th, 2003, 11:40am

Quote:
This is a little like the two-child family riddles in the medium section, isn't it?  

Yes, it is a little like them.


Quote:
Well, it depends, if the drawn silver coin is the result of the deliberate drawing of a silver coin, or the drawing of a random coin from the box.

A silver coin is drawn at random.
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 11:42am

on 06/10/03 at 11:40:43, THUDandBLUNDER wrote:
A silver coin is drawn at random.

If you want to choose the coin (rather than the box) at random, why don't you say so in your original post? :)
Title: Re: Gold or Silver?
Post by MattyDK23 on Jun 10th, 2003, 11:51am
The way I read it...

You chose a box at random, and pick a coin.  You only look at cases where the first coin is silver.

In that case, the probability is [hide]66% (or 2/3)[/hide], isn't it?  I seem to remember that from my stats class two years ago.

I just sketched up a really crude C program that ran 10,000 iterations...it came up with roughly the same probability.
Title: Re: Gold or Silver?
Post by towr on Jun 10th, 2003, 11:52am
box1 silver
box1 gold
box2 silver
box2 silver

so if you drew a silver coin 2 to 3 it's box 2 with another silver.
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 11:57am
As (almost) always, it's a question of how you model the choosing and drawing process.

Matty, do you try to "choose" a silver coin from the box with the gold coin in it?

Towr, I completely agree with your last post. However, I choose a box first, so the second coin is silver in only 50% of the cases.
Title: Re: Gold or Silver?
Post by MattyDK23 on Jun 10th, 2003, 12:14pm

on 06/10/03 at 11:57:37, wowbagger wrote:
Matty, do you try to "choose" a silver coin from the box with the gold coin in it?

The selection of the first silver coin is an assumption in the 'riddle' / question, at least in the way I read it.  We're past the first stage, past the part where the first coin has already been chosen.

In my program, I have two 'stages'.  Everything has equal probability -- picking a random coin from a random box.  However, I only recorded the occurences of
1. The number of times the first coin was silver
2. The number of times both the first and second coin were silver

My output was:

Total Iterations: 10000
First Coin Was Silver: 7503 times
Both Were Silver: 5054 times
Conditional Probability: 5054/7503 = 0.6736

So, there was 10000 - 7503 = 2497 times that the first coin was gold.  And 7503 - 5054 = 2449 times that the second coin was gold when the first coin was silver.
Title: Re: Gold or Silver?
Post by towr on Jun 10th, 2003, 12:18pm

on 06/10/03 at 11:57:37, wowbagger wrote:
Towr, I completely agree with your last post. However, I choose a box first, so the second coin is silver in only 50% of the cases.

But you don't know which of the two boxes (identified by content) you choose.

Suppose you have two boxes, one with a million silver coins, and one with 999,999 gold coins and one silver coin. You pick a box, you randomly pull out a coin from 'your' box, and it happens to be silver. What are the chances you have the all silver box?

It's about knowledge. In a way you're right, but the extra knowledge would put the bayesian guess elsewhere.
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 10th, 2003, 12:20pm

Quote:
If you want to choose the coin (rather than the box) at random, why don't you say so in your original post?

OK, a Silver coin is chosen at random from a box that is chosen at random .
Title: Re: Gold or Silver?
Post by TenaliRaman on Jun 10th, 2003, 12:21pm
Does the knowledge that we have drawn one silver coin affect the probability in any way??

My guess it something like the Monty Hall Problem kinda thing!

So the question is (As i see it)
what is the probability that we choose a box containing 2 silver coins?

obviously i feel the answer is 50%
Title: Re: Gold or Silver?
Post by MattyDK23 on Jun 10th, 2003, 12:27pm
Yeah, sorry...I said my part kinda wrong. :-[  You still have to keep in mind the differences between the two boxes.  I kinda just overstated the part about the conditional / assuption part.

Just take what towr said, and multiply it by 2. ;)

Anyways, just look at the probabilities that the program spits out.

First coin silver: 75% of the time (0.75)
Both coins silver: 50% of the time (0.5)

Prob. of 2nd coin silver, given first coin is: 0.5 / 0.75 = 0.67
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 12:30pm

on 06/10/03 at 12:18:13, towr wrote:
But you don't know which of the two boxes (identified by content) you choose.

Suppose you have two boxes, one with a million silver coins, and one with 999,999 gold coins and one silver coin. You pick a box, you randomly pull out a coin from 'your' box, and it happens to be silver. What are the chances you have the all silver box?

Well, after picking a box (with equal probability, I assume), you have the all silver box with 50%.


Quote:
It's about knowledge. In a way you're right, but the extra knowledge would put the bayesian guess elsewhere.

Yes, knowledge can be important.  ;)
I don't know nothing about that bayesian guess though.
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 12:32pm

on 06/10/03 at 12:20:59, THUDandBLUNDER wrote:
OK, a Silver coin is chosen at random from a box that is chosen at random .

Hm, the box is chosen first, right? That's at least how I interpreted the original problem statement. So the probability of the second coin also being a silver coin has to be equal to the probability of choosing the all silver box in the first place.
Title: Re: Gold or Silver?
Post by towr on Jun 10th, 2003, 12:35pm
a different extreme case.

we have again two boxes, one with one silver coin, one with one gold coin. You select a box at random, and select a coin from that box at random, it is silver. What is the chance you had the box with only silver coins.

(to show it does matter, a lot, which coin is drawn, even if there is an apriori 50-50 chance of either box)
Title: Re: Gold or Silver?
Post by wowbagger on Jun 10th, 2003, 12:39pm
Of course you get a different result in your new extreme case, towr. Now the box with the gold coin isn't a valid result anymore (due to the extra knowledge, as you rightly point out), whereas in the original puzzle it is!
Title: Re: Gold or Silver?
Post by towr on Jun 10th, 2003, 12:47pm
actually, you get the exact same kind of information there as well, only less of it.

suppose you were to pick a random box, and then a random coin from the box.
you get
box1 silver
box1 gold
box2 silver1
box2 silver2

The only difference in the original puzzle is that the second occurence isn't a valid option any more.

In my extreme case you have
box1 gold
box2 silver
The parrallel is that one option from the 'pick a random coin' case is invalidated because you know part of the endresult. So instead of 1 out of 2 it goes to 1 out of 1 in my extrene case. And from 2 out of 4, to 2 out of 3 in the original puzzle.
Title: Re: Gold or Silver?
Post by towr on Jun 10th, 2003, 12:59pm
Here's the Bayes view on it.
P(A|B) = P(A & B) / P(B) = P(B|A) * P(A)/P(B)
P(S) = Sum over all i of P(S | Ei) * P(Ei))

P(box1) = 50% (the a priori chance of box1)
P(box2) = 50%
P(silver | box1) = 50%
P(silver | box2) = 100%
P(silver) =  100% * 50% + 50% * 50% = 75% (like we didn't know that beforehand, the a priori chance of picking silver randomly from a box)

P(box1 | silver) = P(silver | box1) * P(box1) / P(silver) = 50% * 50% / 75% = 33% (1/3)
P(box2 | silver) = P(silver | box2) * P(box1) / P(silver) = 100% * 50% / 75% = 67% (2/3)
Title: Re: Gold or Silver?
Post by Icarus on Jun 10th, 2003, 4:18pm
You are all wrong! The probability that the other coin is silver is 100%.
Why? Because the guy drawing the coin is extremely greedy, and would have drawn the gold coin if it had been available! 8)

This is the same argument as in the two kids puzzles, and in the Monty Hall puzzle and in some others.

It turns out that the actual puzzle is determining how to interpret the statements in the puzzle.

In this one T&B tells us clearly that the box was chosen at random. He tells us nothing about how the coin was chosen from the box - only that the result was a silver coin.

What probability you get depends on what assumptions you make about how the coin was chosen.
Title: Re: Gold or Silver?
Post by MattyDK23 on Jun 10th, 2003, 5:01pm
You've just expanded the question to include assumptions about the person's personality (greedy), the economy (value of gold vs. silver), etc...

One could also assume that the person was very modest...or that the value of gold had just plummeted...or that we're in an alternate universe where Gold is really Bronze, and vice versa...

We're working with what we're presented with...
Title: Re: Gold or Silver?
Post by Icarus on Jun 10th, 2003, 8:45pm
My point was that we were NOT presented enough to be able to calculate a probability.

However - in rereading the thread, I see that T&B has already addressed the problem by stating that the coin is selected at random from the box.

My apologies for scanning the thread too lightly, and reopening an already settled point.
Title: Re: Gold or Silver?
Post by wowbagger on Jun 11th, 2003, 2:25am

on 06/10/03 at 16:18:42, Icarus wrote:
It turns out that the actual puzzle is determining how to interpret the statements in the puzzle.

What probability you get depends on what assumptions you make about how the coin was chosen.

I'm glad you agree with me on this point.

Now for the finicky part:

on 06/10/03 at 11:40:43, THUDandBLUNDER wrote:
A silver coin is drawn at random.

Is this really clear-cut? I could argue in favour of an interpretation like: "Of all silver coins, one is drawn at random." Can anybody relate to this?

On the other hand, it can mean: "A coin is drawn at random, and it turns out to be silver."
In this case we have to deal with conditional probability and arrive at towr's result.
Title: Re: Gold or Silver?
Post by redPEPPER on Jun 11th, 2003, 2:45am

on 06/11/03 at 02:25:37, wowbagger wrote:
I could argue in favour of an interpretation like: "Of all silver coins, one is drawn at random." Can anybody relate to this?

Yep.  Although the other meaning is probably the intended meaning, the wording isn't very clear.

From T&B's update:


Quote:
OK, a Silver coin is chosen at random from a box that is chosen at random .

So, first, a box is chosen at random.  That's 50% to pick the box with gold.  In which case picking a silver coin at random will be easy, as there's only one siver coin.  The other coin in that case will be gold.  In the other 50% case, no matter what coin is picked, the other coin will be silver.  In conclusion, the probability would be 50% that the other coin is silver.

A more accurate wording for the other meaning would be:
"A coin is chosen at random from a box that is chosen at random. It turns out it's a silver coin."  And so on...
Title: Re: Gold or Silver?
Post by Sir Col on Jun 11th, 2003, 6:03am
By my interpretation of the problem I get 100%.

If a silver coin is chosen at random from a box that is chosen at random, that would imply you have chosen a box at random and are now faced with the choice of which silver coin to select. If you chose the box with silver and gold you would not be faced with a random choice, hence you must have chosen the box with the two silver coins.

I suspect that this was the clever twist T&B intended.
Title: Re: Gold or Silver?
Post by wowbagger on Jun 11th, 2003, 6:17am

on 06/11/03 at 06:03:28, Sir Col wrote:
I suspect that this was the clever twist T&B intended.

I don't think so, but who knows? Well, okay, T&B should know.

Anyway, you can choose one coin out of one. The number of possible outcomes is C(1,1) = 1. And please don't argue about whether this choice can be considered "random".  :)
Title: Re: Gold or Silver?
Post by otter on Jun 11th, 2003, 7:05am

on 06/11/03 at 02:25:37, wowbagger wrote:
On the other hand, it can mean: "A coin is drawn at random, and it turns out to be silver."
In this case we have to deal with conditional probability and arrive at towr's result.


I believe this was the intent of the original post.  One of two boxes is randomly chosen and one coin is randomly removed from the box, which turns out to be silver.  As wowbagger said, we then arrive at towr's answer.   :D
Title: Re: Gold or Silver?
Post by Icarus on Jun 11th, 2003, 4:01pm

on 06/11/03 at 06:03:28, Sir Col wrote:
By my interpretation of the problem I get 100%.

If a silver coin is chosen at random from a box that is chosen at random, that would imply you have chosen a box at random and are now faced with the choice of which silver coin to select. If you chose the box with silver and gold you would not be faced with a random choice, hence you must have chosen the box with the two silver coins.

I suspect that this was the clever twist T&B intended.


If this is what T&B intended, then he duffed the statement very badly indeed.

How about it, T&B? Any comment?
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 13th, 2003, 12:18pm
Sorry, guys, I think I screwed up. I was trying to remember a puzzle from way back, and posed a different one.
Anyway, you seem have covered every possibility, and then some.

I think this was the one I had in mind:

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?
Title: Re: Gold or Silver?
Post by Sir Col on Jun 13th, 2003, 4:42pm
In which case, as you said, it's been covered.

Assume box 1 contains: silver, silver and box 2 contains: silver, gold.

There are three possible ways, with equal chance, of having taken a silver coin:

Selected box 1 and took 1st silver coin; 2nd silver coin remains.
Selected box 1 and took 2nd silver coin; 1st silver coin remains.
Selected box 2 and took silver coin; gold coin remains.

Hence P(remaining coin is silver)=2/3.
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 13th, 2003, 5:18pm
But here there is only ONE box.
Title: Re: Gold or Silver?
Post by Chronos on Jun 13th, 2003, 11:25pm
We need to know the probability on your either/or.  Is it equally likely, a priori, to be a GS box or an SS box?  Because if that's the case, then just say that the box in question is the box we randomly chose in the previous version of the problem.  Really, would you expect the answer to change just because there's some other box floating around somewhere in the Universe?
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 14th, 2003, 3:19am

Quote:
Is it equally likely, a priori, to be a GS box or an SS box?

No a priori is given.  :-X
Title: Re: Gold or Silver?
Post by Sir Col on Jun 14th, 2003, 4:06am

on 06/13/03 at 17:18:09, THUDandBLUNDER wrote:
But here there is only ONE box.

:-[ I missed that!

However, It may still hold. As Chronos said, we're dealing with the concept of a box that contains silver/silver (box 1) or silver/gold (box 2).

Having said that, and as we have no more information, I believe there are two ways to interpret the problem:

(i) Equal chance of the box being SS or SG. So P(other coin is silver)=2/3 [as outlined above].

(ii) The contents of the box can be generated from a principle of there being an equal chance of each of the two coins being silver/gold. That is, P(SS)=P(SG)=P(GS)=P(GG). Using the first given (it contains two silver or silver and gold), we reduce to three possible configurations: SS,SG or GS. We then find that the second given (the first coin taken is silver), is redundant, as all three boxes satisfy this. Hence we deduce that P(other coin is silver)=1/3.

I would suggest that we are unable to solve this problem from the information given, unless I've missed something. What is the probability of that? :)
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 14th, 2003, 2:10pm

Quote:
I would suggest that we are unable to solve this problem from the information given...

I agree.
Title: Re: Gold or Silver?
Post by Sir Col on Jun 14th, 2003, 4:13pm
If we're right, isn't it strangely satisfying that an apparently simple and familiar problem cannot be solved – rather, the solution is that it has no exact solution. For all intents and purposes it appeared to be a version of the three door problem or the boy/girl problem. Thanks for sharing it, T&B.
Title: Re: Gold or Silver?
Post by towr on Jun 15th, 2003, 8:10am

on 06/14/03 at 04:06:13, Sir Col wrote:
I would suggest that we are unable to solve this problem from the information given, unless I've missed something. What is the probability of that? :)
Without any information the educated guess is equal chance, 50-50 in this case (since there are two options).

But there is information, and that gives the answer you gave earlier, 2/3. Because without further information the either/or gives a 50-50 chance of it being either box, which is equal in every relevant way to first choosing a box, and than randomly pulling out a coin that is revealed to be silver.
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 15th, 2003, 9:01am

Quote:
But there is information...

Where? '50/50' is merely a subjective preference.
Title: Re: Gold or Silver?
Post by Sir Col on Jun 15th, 2003, 12:07pm
LOL  ;D

When I said, "What is the probability of that?" I meant, what is the probability of me missing somehting.
Title: Re: Gold or Silver?
Post by towr on Jun 15th, 2003, 2:31pm

on 06/15/03 at 09:01:21, THUDandBLUNDER wrote:
Where? '50/50' is merely a subjective preference.

The coin is silver. Which is usefull information (just apply Bayes)

50-50 is not a subjective preference. When there are two choices, the 'real' probability is either 100% or 0% because it will allways be one or the other, not (part of) both.
The educated guess without any knowledge is therefore 50% (in the middle), giving no opportunity for a 'profitable bet'.
Any information brings it closer to the 'real' probability, and perfect knowledge allows you to predict/classify perfectly (allowing you to know the future past and present).
Title: Re: Gold or Silver?
Post by Sir Col on Jun 15th, 2003, 3:42pm
Philosophically speaking, can we be certain of anything? I know that Pascal said, "It is not certain that everything is uncertain." Which is doubly ironic, as it was Pascal who invented the system of probability to settle a dispute when a gambling game was interrupted. He was asked to determine who was more likely to win, if it had been allowed to be completed.

Anyway, I digress. The reason I mention certainties is to challenge whether or not we may permit talk of certainties in probability at all. Aside from the existential philosophical issues, we can also ask, if probability is definied as a measure of the likelihood of an event happening, is it reasonable to talk of an event for which there is no measure of uncertainty?

Back to the current question. As we've established that the probability of the other coin being silver is 2/3 or 1/3, depending on the model, can we reasonably interpolate by a linear method? Is it more likely to be at one extreme or the other, or is it sensible to suppose that both are equally likely? I would dare to suggest that we have no reasonable grounds to suppose on thing or the other without further information. It would be as foolish as arguing that as I can win £1 million in a lottery or win nothing, my expected winnings is £0.5 million. Unless you are provided with the system that determines one outcome against the other, you simply cannot assume. Now what's the probability of that?  ;)
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 15th, 2003, 7:01pm

Quote:
Without any information the educated guess is equal chance, 50-50 in this case (since there are two options).

But there is information, and that gives the answer you gave earlier, 2/3.

I throw a die.
If it comes up a 1, I put SS in the box;
Otherwise, I put SG in the box.
(Or I could do it the other way round. Or use another method to get whatever probabilities I choose.)

Without any a priori information on how the coins were put in the box, any 'answers' will be based on
unjustified assumptions and/or blind guesswork.
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 15th, 2003, 9:59pm

Quote:
Anyway, I digress. The reason I mention certainties is to challenge whether or not we may permit talk of certainties in probability at all. Aside from the existential philosophical issues, we can also ask, if probability is definied as a measure of the likelihood of an event happening, is it reasonable to talk of an event for which there is no measure of uncertainty?

MY 2 CENTS:

In higher maths (and I'm no expert), probability is considered in terms of measure theory.
Probability is basically a special kind of measure. Considering 'impossibility', certainty that an event cannot occur:

(i) Zero probability events have a probability of occurring of zero, but are not impossible.
(ii) Impossible events also have a probabilty of occurring of zero, and are impossible.

eg, the probability of choosing a particular real in the interval [0,1] = zero, but we know that it is not impossible.
Conversely, the probability of not choosing the number is 1, but it is not certain that it will not be chosen.

Also, the Law of Large Numbers doesn't say that the ratio is guaranteed to approach the probability, p,
of the event one is observing.  It merely says that the event approaches p 'with probability 1'.
Again, there's a difference. The outcome is not certain, but it is almost certain.

(To sum up, I would say that when we are dealing with infinite sets (and possibly finite sets, too),
there is no certainty. I guess Icarus will have a few things to say about your philosophical musings,
besides being well booked-up on the above.)  :)

PS Don't think about this stuff while you are driving or you are certain to have an accident!  ;)
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 1:49am

on 06/15/03 at 19:01:20, THUDandBLUNDER wrote:
I throw a die.
If it comes up a 1, I put SS in the box;
Otherwise, I put SG in the box.
(Or I could do it the other way round. Or use another method to get whatever probabilities I choose.)

If I know you used a die it gives extra information.
simple as that.


Quote:
Without any a priori information on how the coins were put in the box, any 'answers' will be based on  
unjustified assumptions and/or blind guesswork.

Precisely, justified blind guesswork. Because without information you are blind.
That's the whole point. Even (blind) chance, because any other assumption is unjustified.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 1:53am

on 06/15/03 at 15:42:51, Sir Col wrote:
Back to the current question. As we've established that the probability of the other coin being silver is 2/3 or 1/3, depending on the model, can we reasonably interpolate by a linear method?
Firstly the other model is wrong, since it isn't supported by any information, the other is supported by the lack of the information supporting the other model.
Lastly when there is insufficient information the uniform distribution gives the best educated guess, since it doesn't favour any outcome (and no outcome should be favoured as there is no information).


Quote:
It would be as foolish as arguing that as I can win £1 million in a lottery or win nothing, my expected winnings is £0.5 million.
That is _only_ the case because you _know_ the lottery isn't an even-bet game. you know they want to make a profit, you know many people enter it and only one can win.
If you didn't know anything about it, nor about human nature, nor had any intuition (which is on avarage a surprinsingly good predictor) etc, it would be foolish to assume the expected winnings was not 0.5 million.

Because without any information, you only know that either you win, or you don't win, and you do not know which is more likely, so you can only assume they are equally likely. Favouring neither outcome over the other as there is no information to justify doing so.
(But your instincts will probably tell you there is no free lunch, and an average winning of 0.5 million is impossible, so you'll never be in this situation.)
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 2:46am
suppose we play a game.

rules:
We can't see each other
We both flip a coin.
We write down the side that faces up on a piece of paper and hand it to an independant and fair arbitrer.
If both pieces of paper say the same thing you win, else I win.

What is the chance you win?

What is the chance you win if only I cheat and:
1. always write heads
2. write heads 1/3 of the time and tail 2/3 of the time.
3. use any other strategy

What is the chance you win if only you cheat (using whatever strategy you want)

What is the chance you win if we both cheat (and have to determine our strategy beforehand)

What is the chance you win if you know my strategy

"knowledge is power"
"when in doubt, scream and shout"
Title: Re: Gold or Silver?
Post by redPEPPER on Jun 16th, 2003, 3:52am
I'm no expert in probabilities, but isn't there a difference between a probability based on complete and incomplete information?

In towr's example, the probability to win if at least one player doesn't cheat is 1/2.  If you reproduce the experiment a number of times, each will win about half of the time.

But what about the probability to win if you don't know if the players cheat?  You would give a probability of 1/2 for lack of information?  But if you reproduce the experiment a number of times, you might not verify that probability.  Say, if one player uses strategy 1 and the other strategy 2, one of them will win 2/3 of the time.

So is it better to say the probability is 1/2, or to say you can't decide for lack of information?  The "real" probability is not 1/2, it's 2/3.  You just have no way to say it's 2/3 but you know enough to see it might not be 1/2.  So what's the value of such an uninformed "guess"?
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 4:24am
For every strategy there is an exact opposite strategy.
what you win with one you loose in the other. And you don't know which strategy is being used. So it is a fair game.

The average gain would be the weighed sum over all strategies for both players. And since every strategy has an exact opposite you end up at even chance.

In the end you can always make a guess, that's why it's called a guess. If you don't have any information, in other words the educated guess is 50%, you know you can't make a profitable bet, so you might as well not bet.
(Aside from an average gain there is also something to be said for the smallest variance. The smallest variance for no gain is attained by not betting.
Betting 1 dollar against 100  on a 1/100 chance 100 times is worse than 1 against 2 dollars on a 1/2 chance 100 times )
Title: Re: Gold or Silver?
Post by redPEPPER on Jun 16th, 2003, 4:55am
Okay, I understand how, if the players independently choose how to play at each iteration, the probability is 1/2.  But what about a case where they use the same strategy each time, but we don't know that strategy?

I'm referring to your earlier posts, about the 1/2 probability to win the lottery if you don't have more infos, or the die to decide whether to put gold or silver in the box.

What difference do you make between a 1/2 probability because you don't know a die was thrown to decide between SG and SS, and the 1/6 probability you would have if you knew that?  The situation is the same, so the "real" probability is the same, but your guess changes.

Or to look at it from the other way: what difference do you make between a 1/2 probability because you don't know how the choice between SG and SS was made, and a 1/2 probability because you know the choice was made at random, with equal chance for each?  I view these as very different: the latter is an accurate probability (if you repeated the action several times you'd win 1/2 of the time) while the former is only a best guess, in which case I'd personally view "I can't know the probability" as a better answer.  But you don't seem to be making that distinction, and seem to view 1/2 as a very valid answer to the first situation.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 6:13am
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
Title: Re: Gold or Silver?
Post by Sir Col on Jun 16th, 2003, 6:40am
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?
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 7:11am
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.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 7:24am

on 06/16/03 at 06:40:35, 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?
If you role a die it never roles 3.5, yet that is still the expected value you get when you role.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 8:27am
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.
Title: Re: Gold or Silver?
Post by Sir Col on Jun 16th, 2003, 8:27am
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.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 8:34am
I sure wouldn't mind having you as a customer in my casino (if I had one) ;)
Title: Re: Gold or Silver?
Post by redPEPPER on Jun 16th, 2003, 9:25am
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.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 9:47am

on 06/16/03 at 08:27:30, 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.

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.

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).

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...

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? :)

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.

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.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 10:24am

on 06/16/03 at 09:25:39, 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.

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 ;)

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.

I'm glad someone understands me

(I guess it's a good thing I'm not a teacher)
Title: Re: Gold or Silver?
Post by Icarus on Jun 16th, 2003, 5:30pm
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?
Title: Re: Gold or Silver?
Post by THUDandBLUNDER on Jun 16th, 2003, 9:20pm
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.
Title: Re: Gold or Silver?
Post by towr on Jun 16th, 2003, 11:37pm

on 06/16/03 at 17:30:45, Icarus wrote:
(3) It seems to me that what towr is describing could be described as "metaprobability" or "2nd Order probability".
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.

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?".
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?
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.
Title: Re: Gold or Silver?
Post by redPEPPER on Jun 17th, 2003, 3:53am

on 06/16/03 at 23:37:31, towr wrote:
gold being exactly 123653242093543781439231 times more likely than silver
is balanced by
silver being exactly 123653242093543781439231 times more likely than gold

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?
Title: Re: Gold or Silver?
Post by towr on Jun 17th, 2003, 4:39am

on 06/17/03 at 03:53:03, 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?

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.
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.
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?
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.
Title: Re: Gold or Silver?
Post by Sir Col on Jun 17th, 2003, 5:02am
Sorry to sound like a stuck record, and I know records are practically obsolete now before anyone says anything!  :P

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?
Title: Re: Gold or Silver?
Post by towr on Jun 17th, 2003, 7:14am
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.
Title: Re: Gold or Silver?
Post by Icarus on Jun 17th, 2003, 5:39pm
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.

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.

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.

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.
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.

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".
Title: Re: Gold or Silver?
Post by towr on Jun 18th, 2003, 1:51am

on 06/17/03 at 17:39:35, 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.
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.
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.
There is no observed situation yet. When there is it gives plenty of information to change the model using bayes.
Title: Re: Gold or Silver?
Post by Icarus on Jun 18th, 2003, 3:38pm

on 06/18/03 at 01:51:05, 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.

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.


Quote:
Yes, but given that models have the same explanatory/predictive power (which is the case here), the simplest is closest to correct.

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.


But some possible "observable situations" are more likely than others. This opens the door to refining the numbers before observation begins.
Title: Re: Gold or Silver?
Post by towr on Jun 18th, 2003, 11:10pm

on 06/18/03 at 15:38:55, 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.

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.
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.
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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