16 Apr 08
Originally posted by eldragonflyif you remove paragraph 4, there is not a single ad hominem argument in my post. everything applies directly counter or question things that you have said, and ideas you have presented, without bias or personal attack. i got away from myself with paragraph 4, and i apologize. i was very frustrated and i retract the comments that were inflammatory and perhaps mean.
Sorry, that ad hominem rich comment isn't even worthy of a response. 🙄
as an act of good faith, could you please respond to the rest of what i meant to be a probe into where the discrepancy is between our methods of analysis on these questions? and be specific - tell us what makes certain language ambiguous in these questions... perhaps two different ways a question could be interpreted and then the two differing results that come from it? or perhaps, why do the analogies between certain problem types we've presented (though clearly the literal definition of the problem is different) and the original problem inherently make the analogy a false one?
for example, besides the physicality of the action, how do a problem describing coin flips (equally likely heads/tails, and each flip being a separate individual probabalistic event), and a problem describing births of children (defined to be equally likely boy/girl, and each birth being a separate individual probabalistic event) differ? why does changing from births to coins automatically destroy the equivalence and the validity of the results?
16 Apr 08
Originally posted by eldragonflyIn that case, I will take my leave of you. Good day to you. Nothing I can and would tell you is of any value to you, and therefore of no value for me to say.
i already have, you choose to remain ignorant. i find the rest of your redundant and hollow commentary to be equally worthless and insubstantial.
16 Apr 08
eldragonfly, you know the only poster in this whole thread who commits ad hominem attacks on you is me, and I do it because you're a thick-headed troll who, despite being thrown at least 500 free pieces of "meat", still believes his waistline doesn't exceed a size 4.
Your objections amount to scavenging your grandfather's complaint letters to the Dumont Network for zingers. Your explanations are as thoughtful as a fortune cookie. Your diction is as tired as Rene Zellweger after a stair climb. But your repetition is insulting. If I wanted to be rebuffed by a Coke machine, I'd stay at your mom's place.
Put up your dukes, you munt!!
16 Apr 08
Originally posted by geepamoogleI can't argue with you. But I encourage you to stick around for the fireworks, I'm about to insult his dead uncles. 😵
In that case, I will take my leave of you. Good day to you. Nothing I can and would tell you is of any value to you, and therefore of no value for me to say.
17 Apr 08
Originally posted by eldragonflySeveral people have explained why your post on page five is incorrect. You have not addressed
nevermind
this issue, you've simply asserted that you're right and if we read your post enough times, we'll
see the light. But you refuse to address the responses to the post.
What is your goal here? Why do you return to this thread? I think most of the people in this thread
would want to be educated if they are wrong; I am. So, why don't you take the time to walk me
and the others through what you believe is the right answer? Several other people have shown you
the same courtesy.
Nemesio
17 Apr 08
The only logic I have seen and can recall from the 1/2 side is as follows.
There are two possible combinations that fit the extra information. Ergo the probability is 1/2.
I cannot recall any other specific reasoning being made to lead up to an answer of 1/2 (for either of the two main disputed problems).
eldragonfly is welcome to correct me if I have misstated his main argument throughout this entire thread, or if I missed any other main lines of reasoning. According to him, it would seem I have not understood what he has been saying, so you may not be able to rely upon my memory and understanding of it.
17 Apr 08
Originally posted by NemesioWrong. Your ignorance is astounding, i must say.
Several people have explained why your post on page five is incorrect. You have not addressed
this issue, you've simply asserted that you're right and if we read your post enough times, we'll
see the light. But you refuse to address the responses to the post.
What is your goal here? Why do you return to this thread? I think most of the people in thi ...[text shortened]... eve is the right answer? Several other people have shown you
the same courtesy.
Nemesio
17 Apr 08
4 edits
Originally posted by geepamoogleFrom the betters point of view.
The only logic I have seen and can recall from the 1/2 side is as follows.
There are two possible combinations that fit the extra information. Ergo the probability is 1/2.
I cannot recall any other specific reasoning being made to lead up to an answer of 1/2 (for either of the two main disputed problems).
eldragonfly is welcome to correct me if ...[text shortened]... what he has been saying, so you may not be able to rely upon my memory and understanding of it.
1) The dealer shows one side of one card, the silver side of either a SG or SS card.
2) The dealer then offers even money the other side of that card is silver.
3) the dealer then reveals the other side of the card, which can only be either gold or silver.
4) As a one shot one draw deal, it is an even bet, 50-50% odds.
5) Your simulations assume the existence of a "ghost" silver/silver card, which can be selected twice, this is rather arbitrary and is what makes the bayesian solution seem correct.
From the dealers POV :
5) the silver/silver card can be selected twice, therefore the bayesian solution appears to be correct, but only for repeated trials.
conclusion : the word problem is ambiguous and poorly worded, these repeated faux analogies are a bit labored.
17 Apr 08
Originally posted by eldragonfly😴
From the betters point of view.
1) The dealer shows one side of one card, the silver side of either a SG or SS card.
2) The dealer then offers even money the other side of that card is silver.
3) the dealer then reveals the other side of the card, which can only be either gold or silver.
4) As a one shot one draw deal, it is an even bet, 50-50% o ...[text shortened]... word problem is ambiguous and poorly worded, these repeated faux analogies are a bit labored.
17 Apr 08
New problem!
Assume that no person has more than 200,000 hairs on their head.
(a) Prove that at least 2 people in New York City have the same number of hairs on their heads.
(b) Can you prove that this must be the case in Vatican City as well?
(c) Why do people with hair have more hair than people with hairs? 😉
17 Apr 08
Originally posted by eldragonflyThank you finally for a succinct and comprehensive statement of the full process of your reasoning. This was one of the things I had tried to get you to give me, rather than merely calling opposing reasoning a variety of rhetorical names and descriptions without pointing out the specific disagreement.
From the betters point of view.
1) The dealer shows one side of one card, the silver side of either a SG or SS card.
2) The dealer then offers even money the other side of that card is silver.
3) the dealer then reveals the other side of the card, which can only be either gold or silver.
4) As a one shot one draw deal, it is an even bet, 50-50% o ...[text shortened]... word problem is ambiguous and poorly worded, these repeated faux analogies are a bit labored.
Rhetoric was not useful; this is.
For the benefit I will try to list of the arguments against this line of reasoning, and to the best of my ability your response.
1) It has been argued that you are not selecting whole cards without reference to sides, but that you are actually selecting the side which faces up.
One of the related arguments based on this is that half the times the SG card would be selected, it would be selected gold-side-up. The same is not true of the SS card.
The argument continues that since half the time the SG is picked, it wouldn't meet the condition set forth in the problem, but that the SS would always meet the condition, that the SG card is therefore half as likely (and that the SS is therefore twice as likely).
You response, as I best recall, was to say that it was the card selected, but not the side, and therefore that which side comes up is smokescreen and irrelevant.
2) In an alternative to #1, but running parallel to it, people have listed all the sides, including marking each side of the SS card as distinct and separate. The then crossed out the gold sides and left 3 silver sides, two of which were on the SS card and therefore opposite a silver side. (The "phantom" SS card in your mind).
Your response here was largely similar to #1, but you also seemed to indicate that distinguishing between the silver sides on the SS card was an error in your judgment, presumably because you pick the card, not the side.
3) People have also argued (as an alternative method of answering the problem) that two of the three cards have matching sides on the back, and given the symmetrical nature of gold and silver, that the odds are therefore 2/3 that the back of the card matches the side you see.
Your response was that gold WASN'T selected and therefore this reasoning isn't relevant.
That's all of them I can think of for the moment, but there has been some debate on some other points touching on it as well.