17 Apr 08
Originally posted by eldragonflyFor a person who seems to have a problem with the ad hominem, you sure engage in it a lot.
Wrong. Your ignorance is astounding, i must say.
I'll ask again, what is your goal in this thread?
Are you here to educate, help your fellow human beings develop into well-educated individuals?
Or are you here to insult?
Nemesio
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% 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.
Yes. There are two choices: the card could either be S/S or G/S. But those two choices are
not evenly distributed. If you buy a lotto ticked, there are two choices: winning or losing. But,
clearly, those choices are not evenly distributed.
The flaw in your reasoning is at #3. While it is true that the side to be revealed may be either
gold or silver, the likelihood that it will be silver is 2:1. That's what makes it a bad bet.
Nemesio
17 Apr 08
1 edit
Originally posted by geepamoogle
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).
Bingo.
Let's look at it this way.
A man has two cards in his hat. One is S/S the other is G/S. The likelihood that he will
pick either out of his hat is 50%. However, if you are privied to one side and are shown a
silver side, you have new information. Since one card has two silver sides, and the other one
has merely one, it's twice as likely to be the S/S card than the S/G because the former card
has twice as much chance of showing silver than the latter one.
Nemesio
17 Apr 08
Originally posted by NemesioI should have listed this one as an alternative to counterargument #1 as well (call it #1b).
Originally posted by geepamoogle
[b]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).
Bingo.
Let's look at it this way.
A ...[text shortened]... e the former card
has twice as much chance of showing silver than the latter one.
Nemesio[/b]
His response would likely be the same, to state that it was a card and not a side that was picked, and that therefore your argument is irrelevant.
He isn't interested in looking at sides, only cards when looking at odds.
17 Apr 08
2 edits
Originally posted by geepamoogleWrong. It is your reasoning that is faulty, not the other way around. In fact i have only restated here what i have stated countless times in this thread, you are being intellectually dishonest. You choose to remain ignorant, i can find no other reasonable explanation for your "sour grapes" approach to my rather sensible and direct statements. If you are functionally literate and/or math literate then you should have no problem understanding what i have stated here, you don't need to explain to me what i have written or what i have intended. 😉
Thank 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.
17 Apr 08
Originally posted by geepamoogleThis is nonsense. The better is looking at a "randomly" selected card, the dealer is showing the silver face then turns it over. Your solution is forced and artificial, the wording of the problem ambiguous.
I should have listed this one as an alternative to counterargument #1 as well (call it #1b).
His response would likely be the same, to state that it was a card and not a side that was picked, and that therefore your argument is irrelevant.
He isn't interested in looking at sides, only cards when looking at odds.
17 Apr 08
Originally posted by geepamoogleThat's akin to saying something like the following:
I should have listed this one as an alternative to counterargument #1 as well (call it #1b).
His response would likely be the same, to state that it was a card and not a side that was picked, and that therefore your argument is irrelevant.
He isn't interested in looking at sides, only cards when looking at odds.
I'm thinking of a number between 00 and 99. I'll give you 30:1 odds you can't guess it (a
bad bet).
Now, I'll tell you either the first or last digit is a 3. Now the odds are in your favor, because
you have new information. The likelihood that the number was 27 just went from 1 in a hundred
to zero. That probability has to be reassigned somewhere. So, there are only nineteen numbers
from which to choose (03, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, and 93),
which makes the bet a good one now.
He's not taking into account that he has new information. It's an intuitive mistake -- I'd be
lying if I said I understood the Monty Hall problem at first glance. Indeed, it took me days with
a statistical mathematician holding my hand to get me to understand it.
Nemesio
17 Apr 08
2 edits
Originally posted by NemesioWrong. There is no flaw in my reasoning, again the problem is poorly worded and ambiguous, your interpretations are careless.
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 ...[text shortened]... selected twice, this is rather arbitrary and is what makes the bayesian solution seem correct.
17 Apr 08
Originally posted by NemesioIrrelevant. Red herring = fallacy
For a person who seems to have a problem with the ad hominem, you sure engage in it a lot.
I'll ask again, [b]what is your goal in this thread?
Are you here to educate, help your fellow human beings develop into well-educated individuals?
Or are you here to insult?
Nemesio[/b]
17 Apr 08
2 edits
Originally posted by eldragonflyI have no problem following your reasoning as you have given it in that one post.
Wrong. It is your reasoning that is faulty, not the other way around. In fact i have only restated here what i have stated countless times in this thread, you are being intellectually dishonest. You choose to remain ignorant, i can find no other reasonable explanation for your "sour grapes" approach to my rather sensible and direct statements. If you are ...[text shortened]... ave stated here, you don't need to explain to me what i have written or what i have intended.
I believe it is erroneous however.
It is not intellectually dishonest solely to disagree with another's reasoning, even if you later find out you were wrong.
Even the best of us will sometimes be wrong, but we can be wrong and intellectually honest at the same time.
It's been said a myriad of times, including in the initial clarification of the problem when you voiced a complaint.. but not only is the card selected randomly, but the side which gets turned up is also selected randomly from the two sides of the card selected.
I believe firmly that you fail to take this into account. The SS card ALWAYS comes up silver, but the SG card only comes up silver half the time.
Your manner of calculating odds assumes the SG card always comes up silver, which is simply not supported in the problem. It is a false assumption.
Your statement about the 'phantom silver/silver card' is a misunderstanding of our position based on the fact you are looking at cards only and not sides. We are looking at sides.
For us, one side of the silver/silver card is distinct from the other side of the silver/silver side.
Either of the two has an equal chance of being the one which ends up visible on the table, and so to has the single silver face of the silver/gold card. So we are looking not at two cards, but at 3 silver faces.
As for my stating my understanding of what you had argued, and how you have thought, I am trying my best to understand where you are coming from with this. The better I can understand your thought pattern on this, the better I can debate for or against it.
Have you attempted to understand why any of us argue as we do, or have you simply assumed that we have a different answer and that we must therefore be wrong?
17 Apr 08
1 edit
Originally posted by geepamoogleWrong. The problem of the silver/gold card always coming up with the silver side showing is a deficiency necessitated by the rather suspect problem definition. i have raised this very point numerous times, you choose to ignore it, now you choose to misrepresent what i have said, for reasons that can only be attributed to your own ignorance and/or the tedious reliance upon artificial and transparent solutions.
Your manner of calculating odds assumes the SG card always comes up silver, which is simply not supported in the problem. It is a false assumption.
Your statement about the 'phantom silver/silver card' is a misunderstanding of our position based on the fact you are looking at cards only and not sides.
17 Apr 08
Originally posted by eldragonflyThere is no deficiency in the problem as worded. The method of selection is very clearly defined, and it is abundantly clear that at every step of the way that the selections are random and fair, without the benefit of any preknowledge, and in addition to this that none of the involved parties have knowledge of the hidden side.
Wrong. The problem of the silver/gold card always coming up with the silver side showing is a deficiency necessitated by the rather suspect problem definition. i have raised this very point numerous times, you choose to ignore it, now you choose to misrepresent what i have said, for reasons that can only be attributed to your own ignorance and/or the tedious reliance upon artificial and transparent solutions.
What we do have is some knowledge which removes from consideration two of the potential possibilities for this one-time event.
The first eliminated potential was the potential to draw the gold/gold card, which would have resulted in a gold face showing 100% of the time. On this point we agree.
There is, however, a second eliminated potential.. That potential is the potential that the silver/gold card was selected, but placed gold facing up. Half the time the silver/gold card is selected, this happens.
This second one seems to be the sticking point, because what it means is that we have eliminated half the times the silver/gold card gets drawn.
Looking at the silver/silver card, we know that it will show a silver face all the time. No case where the silver/silver card gets drawn is eliminated.
We therefore know the following..
1) Before we knew which side was facing up, all 3 cards were equally likely to be drawn.
2) After seeing the silver face, we know the GG was not selected.
3) We also know the SG wasn't selected gold side up (which it does half the time). It is therefore only half as likely, because this card has a 50% of showing either face.
4) We know the silver/silver always shows silver, however.
Since the SS shows silver all the time, but SG only shows silver half the time, SS is twice as likely to have been the card picked in a random event, EVEN IF IT IS ONE TIME!
From your points made, I can see #3 is the one which causes you the most difficulty to understand. Think of it this way. If the silver/gold turned up on the gold side, it would invalidate the condition and therefore the problem. But rather than to simply ignore the fact it COULD turn up gold, you have to remove that chance from the calculation entirely.
I think it is very important to note here than before we found out anything about the results, that gold being turned up was perfectly likely. It helps clear your thinking about these things when you accept that there isn't anything particularly special about silver, save that it happened to be the side which turned up this time.
If this little experiment was repeated one more time, it could be gold which turns up.
17 Apr 08
2 edits
Originally posted by eldragonflyYour contention seems to be that for a one time event the odds are 50/50 yet after many trials the odds will gravitate to 2/3. Is this your position?
Wrong. The problem of the silver/gold card always coming up with the silver side showing is a deficiency necessitated by the rather suspect problem definition. i have raised this very point numerous times, you choose to ignore it, now you choose to misrepresent what i have said, for reasons that can only be attributed to your own ignorance and/or the tedious reliance upon artificial and transparent solutions.
Let's say that you are the last person in a 3000-strong queue each participating in one draw of the 'three card trick.' Aren't these just 3000 one time events? Those that win the bet stand on the left and those that lose stand on the right. Eventually it's your turn. As you step forward you see that the group on the right outnumbers the one on the left by twice as many (there are 1000 people on the left and 1999 on the right). Which group do you think you are more likely to join? Aren't the chances of you joining the group on the right 2/3?
Now imagine that there is a half hour lag between you and the person in front of you. In fact, you don't even know there is a queue and the participants don't stand in groups but rather leave immediately after playing. (The queue need not exist at all. Or you can imagine being the first person in the queue rather than the last). What are the chances of you losing the bet? Aren't they 2/3 even though this is a one-time event and not 50/50?
Please show me the error in my reasoning.
17 Apr 08
A man gives you a choice between two envelopes with a cheque in each one of them. The only thing you're told is that one of the cheques is for the double of the amount of the other one.
You pick one of those envelopes, open it up and see that the cheque is for €10 000. The man then asks you if you want to switch. Would you?
17 Apr 08
1 edit
Originally posted by PalynkaThis one does have an ambiguity in it. Depending on the man's method, we could argue stay with $10,000 or we could argue switch envelops.
A man gives you a choice between two envelopes with a cheque in each one of them. The only thing you're told is that one of the cheques is for the double of the amount of the other one.
You pick one of those envelopes, open it up and see that the cheque is for €10 000. The man then asks you if you want to switch. Would you?
Let me explain..
If the man would have offered to let us switch regardless of which envelop we picked (the greater or the lesser), then the other envelop is equally likely to contain $5,000 or $20,000, as far as we know.
In this case we get $12,500 on average if we switch, or $10,000 if we keep our envelop, and we want to switch.
If, however, the guy knows which envelop we picked, and would only make this offer to us if we chose the greater value, then the other envelop has to contain $5,000, and we had to have chosen the doubled amount. we wouldn't want to switch.
We don't, however, have any indication what the man's procedure is, and therefore have to make an assumption.
In my reading of this problem, I would read it as "the man always offers to switch envelops", as that seems most consistent with what I see of the setup.