17 Apr 08
Originally posted by geepamoogleIt's agreed beforehand that he will ask you to switch. Sorry if that wasn't clear.
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.
If you know it's ALWAYS better to switch, why isn't taking the other envelope first and refusing to switch optimal?
17 Apr 08
Originally posted by PalynkaI thought as much, but it's no biggie. At the very least, the ambiguity is cleared up, and the question is a good one in my opinion.
It's agreed beforehand that he will ask you to switch. Sorry if that wasn't clear.
If you know it's ALWAYS better to switch, why isn't taking the other envelope first and refusing to switch optimal?
As for your follow-up question, I do not have an answer for you besides the fact that that is simply the way the numbers work out. It is very paradoxical, no?
17 Apr 08
Originally posted by eldragonflyYou assertions to the contrary, there is a flaw in your reasoning. The fact that you cannot
Wrong. There is no flaw in my reasoning, again the problem is poorly worded and ambiguous, your interpretations are careless.
respond to the actual content of people's objections is testimony to that.
Nemesio
17 Apr 08
Originally posted by geepamooglegeepamoogle you are backpedalling here, fact is the problem was very poorly stated. You hanging your explanation on this idea of extra information is quite silly and deficient.
There 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.
What we do have is som ...[text shortened]... ime.
If this little experiment was repeated one more time, it could be gold which turns up.
17 Apr 08
1 edit
Originally posted by Green Paladini have already conceded this point, however for your string of single betters it is a 50-50% shot per better. And this string of one time betters is not the same as a one-shot one draw one selection experiment. You also fail to explain how the sample space is always reduced to the two card experiment, excluding the gold/gold card. This is hardly the random experiment you pretend it to be. No, realistically your dealer picks at random from one of two cards showing a silver side, the SS and the SG cards, and makes his even money bet. Since he has only a SS and a SG card to choose from it is clearly a 50-50% bet.
Your contention seems to be that for a [b]one time event the odds are 50/50 yet after many trials the odds will gravitate to 2/3. Is this your position?
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 sta is a one-time event and not 50/50?
Please show me the error in my reasoning.[/b]
17 Apr 08
2 edits
Originally posted by eldragonflyThis is where your reasoning goes wrong.
Since he has only a SS and a SG card to choose from it is clearly a 50-50% bet.
The problem states that he show us a side which happens to be silver. Which silver side is it more likely to be? The one from SG or one of the two SS silver sides? The fact that there are only two cards from which to choose, doesn't mean that these two cards are equally probable to be chosen, which we evaluate from the additional information (that we one side is silver).
What is NOT SO IMPORTANT:
We see two cards
What is VERY IMPORTANT:
We see one side of a card.
Analogy:
3 beans, 2 red (2 silver sides from SS), 1 white (1 silver side from SG). You choose one randomly. What are the odds that you get a red bean? Same thing in the original question: what are the odds that you see a silver side from the SS?
17 Apr 08
1 edit
Originally posted by eldragonflyNevermind
i have already conceded this point, however for your string of single betters it is a 50-50% shot per better. And this string of one time betters is not the same as a one-shot one draw one selection experiment. You also fail to explain how the sample space is always reduced to the two card experiment, excluding the gold/gold card. This is hardly the rand ...[text shortened]... even money bet. Since he has only a SS and a SG card to choose from it is clearly a 50-50% bet.
17 Apr 08
Originally posted by kbaumenThat false analogy has already been debunked and is totally inaccurate.
This is where your reasoning goes wrong.
The problem states that he show us a side which happens to be silver. Which silver side is it more likely to be? The one from SG or one of the two SS silver sides? The fact that there are only two cards from which to choose, doesn't mean that these two cards are equally probable to be chosen, which we evaluate from ...[text shortened]... Same thing in the original question: what are the odds that you see a silver side from the SS?
17 Apr 08
1 edit
Originally posted by eldragonflySeems like I was wrong about you debating as a reasonable person.
And once again you are backpedalling, my explanations have always been consistent. Let's move on...
What's wrong with my explanation? With which part exactly do you disagree or don't understand?
EDIT: Why is the analogy inaccurate? In what what way does it differ from the original problem?
17 Apr 08
Originally posted by kbaumenWrong. Then you couldn't perform the card trick as stated in the original, i.e. not the wikipedia page, word problem. You start of with three cards, then the gold/gold card is *magically* eliminated, leaving only two cards to choose from, the SS and the SG, by definition. Indeed this is the foundation of this paradox.
What is [b]NOT SO IMPORTANT:
We see two cards[/b]
Analogy:[/b]