Probability
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17-03-2013, 05:55 AM
RE: Probability
(17-03-2013 12:01 AM)Heywood Jahblome Wrote:  
(16-03-2013 05:49 PM)Chas Wrote:  You have no information on the distribution of colors, so you can't compute the probabilities. Maybe only white balls exist, maybe there are 100 colors. You will have to reframe your example.

You're grasping at straws Chas. You don't need to know how many other colors there are. Just lump every other potential color under the generic label of "non-white marbles". If you read my original example I said that black marbles stood for any colored marble that is not white.
The same thing applies.

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Science is not a subject, but a method.
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17-03-2013, 07:51 AM
RE: Probability
(17-03-2013 12:16 AM)Heywood Jahblome Wrote:  Another important assumption is that all possible combinations are equally likely.
Which is exactly what I criticized a few pages ago:

(16-03-2013 09:05 AM)Vosur Wrote:  I think I've already pointed out that you merely assumed the probabilities in your example, that you never calculated them to begin with because you do not know how many marbles of which color there are in total.

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17-03-2013, 11:16 AM
RE: Probability
(17-03-2013 07:51 AM)Vosur Wrote:  
(17-03-2013 12:16 AM)Heywood Jahblome Wrote:  Another important assumption is that all possible combinations are equally likely.
Which is exactly what I criticized a few pages ago:

(16-03-2013 09:05 AM)Vosur Wrote:  I think I've already pointed out that you merely assumed the probabilities in your example, that you never calculated them to begin with because you do not know how many marbles of which color there are in total.

And you this is an unreasonable assumption?
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17-03-2013, 01:26 PM
RE: Probability
(17-03-2013 11:16 AM)Heywood Jahblome Wrote:  And you this is an unreasonable assumption?
My personal opinion concerning your assumption is irrelevant; what matters is that it's unfounded because you have presented no evidence to indicate that it is true.

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18-03-2013, 12:47 AM
RE: Probability
(17-03-2013 01:26 PM)Vosur Wrote:  
(17-03-2013 11:16 AM)Heywood Jahblome Wrote:  And you this is an unreasonable assumption?
My personal opinion concerning your assumption is irrelevant; what matters is that it's unfounded because you have presented no evidence to indicate that it is true.

Vosur, if you accept the assumptions then the only way you can deny that the example makes my case is to find fault with the underlying logic.

Are the assumptions unreasonable or not?
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18-03-2013, 01:01 AM
RE: Probability
(16-03-2013 09:05 AM)Vosur Wrote:  
(16-03-2013 08:59 AM)Heywood Jahblome Wrote:  In the example I re-calculate the probability after each white marble is drawn and compare it to the previous.
I think I've already pointed out that you merely assumed the probabilities in your example, that you never calculated them to begin with because you do not know how many marbles of which color there are in total.

You should feel free to show me your math if you disagree.

Given the assumption that each possible combination is as likely as the other, and the fact that there are only 4 possible combinations, I can calculate that at the start of the experiment the probability of each possible combination is .25. I divide the one outcome by the 4 possible ones.

I should not have to show such simple maths to you.
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18-03-2013, 01:10 AM
RE: Probability
(15-03-2013 01:46 PM)Chas Wrote:  
(15-03-2013 01:38 PM)TheBeardedDude Wrote:  The odds of getting a single exact number (assuming sample size is infinity) is the same as getting a single exact number of any other.

So, the odds of getting one number in the powerball, is the same as the odds of getting any other. (once again, we are assuming that the balls are replaced once drawn).

That is because with each coin flip, the odds reset back to 50/50. They don't accumulate.

No. There is only 1 way, each, to get a trillion heads or tails.

There are a many, many ways to get half heads and half tails.

Chas, do you not agree that each time you draw a white marble, you eliminate a possible way that a black marble can exist?
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18-03-2013, 06:35 AM
RE: Probability
I got this from the wiki entry on probability

P(A|B) is the probability of A given B

furthermore

P(A|B) = P(AnB ) / P(B) (note P(B) in the denominator)

If P(B) = 0 then P(A|B) is undefined.
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18-03-2013, 06:57 AM
RE: Probability
(18-03-2013 12:47 AM)Heywood Jahblome Wrote:  Vosur, if you accept the assumptions then the only way you can deny that the example makes my case is to find fault with the underlying logic.

Are the assumptions unreasonable or not?
I already answered your question in my last post. I do not accept your assumptions because they are unsupported by evidence. They are unreasonable because you have presented no reason why they should be true.

(18-03-2013 01:01 AM)Heywood Jahblome Wrote:  Given the assumption that each possible combination is as likely as the other, and the fact that there are only 4 possible combinations, I can calculate that at the start of the experiment the probability of each possible combination is .25. I divide the one outcome by the 4 possible ones.

I should not have to show such simple maths to you.
Please read my posts with comprehension next time, thank you. I made a conditional request, saying that I want you to show me your math if you disagree with my claim that you merely assumed that the probabilities are this or that way. Since you do not disagree with said statement, the condition of my request is not fulfilled and there is no math that you could show me.

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18-03-2013, 07:33 AM
RE: Probability
(09-03-2013 01:50 PM)Heywood Jahblome Wrote:  [This is more than the mind creating some pattern. Observing only white marbles does increase the chance that all the marbles are white.

Let X = the probability that all the marbles are white.
Let An = n observations of white marbles without ever observing a non white marble.


Now from wikipedia entry on probability :-

P(A l B) is the probability of A given B

also

p(A l B) = P(A n B) / P(B) ( note the P(B) in the denominator )

if P(B) is zero then the expression is formally undefined.

With the given scenario, P(An) is either 1 if there are no non-white balls, or 0 if there are 1 or more non-white balls.

If there are no non-white balls then

P(X l An) = P(X n An) / P(An) = 1

This is a trivial example where if you know there are no non-white balls then the probability of all whites is 100%.

If there is at least 1 non-white ball, then P(An) = 0 (i.e. there are some values of n for which the nth ball is non-white)

Then P(X l An) = P(X n An) / 0

Since there is a zero in the denominator, the expression is formally undefined.

In laymans terms I think this means that the scenario you have described is not ammenable to mathematical analysis in the way you are attempting.

Regards

Grimesy
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