16 Apr 08
Originally posted by Nemesioeldragonfly, can I field this one?!?
I did read it. You have failed repeatedly to address the issue of the sides of the card. You simply
think it's a coin flip. It's not. I showed why it is not, elucidating that point with names of
boys and girls to illustrate that when you see a silver side, you are seeing one of three possible
silver sides just like you see one of three possible girls ...[text shortened]... ry little effort on your part. If you show that you are right, I will concede it.
Nemesio
Ad hominem = fallacy.
🙄
(Hehe...couldn't help myself)
16 Apr 08
Originally posted by NemesioNemesio, if you actually convince eldragonfly that you're right (which you are), I'll send you some money somehow, I swear it. 😵
I did read it. You have failed repeatedly to address the issue of the sides of the card. You simply
think it's a coin flip. It's not. I showed why it is not, elucidating that point with names of
boys and girls to illustrate that when you see a silver side, you are seeing one of three possible
silver sides just like you see one of three possible girls ...[text shortened]... ry little effort on your part. If you show that you are right, I will concede it.
Nemesio
16 Apr 08
1 edit
Originally posted by Nemesioi give the correct solution on page 5 of this thread. Many here, yourself included, have made many witless and incoherent assertions in an attempt to pretend that it just isn't there. And it appears that you keep switching from the cards in the hat problem to the boy-girl family hoax/paradox at random, please be consistent.
I did read it. You have failed repeatedly to address the issue of the sides of the card. You simply
think it's a coin flip. It's not. I showed why it is not, elucidating that point with names of
boys and girls to illustrate that when you see a silver side, you are seeing one of three possible
silver sides just like you see one of three possible girls ry little effort on your part. If you show that you are right, I will concede it.
Nemesio
16 Apr 08
Originally posted by eldragonflySwitching randomly or not, it still is an analogy. With the same medieval logic, both problems can be answered wrongly.
i give the correct solution on page 5 of this thread. Many here, yourself included, have made many witless and incoherent assertions in an attempt to pretend that it just isn't there. And it appears that you keep switching from the cards in the hat problem to the boy-girl family hoax/paradox at random, please be consistent.
Please explain why everything we (people who agree to 2/3 answer) say is incoherent, witless, etc. Otherwise, as I said before, there is no reason to take you seriously.
16 Apr 08
Originally posted by kbaumenthen of course you must realize that you are just modeling a simple two-coin toss permutation/combination maths problem. Like really.
Switching randomly or not, it still is an analogy. With the same medieval logic, both problems can be answered wrongly.
Please explain why everything we (people who agree to 2/3 answer) say is incoherent, witless, etc. Otherwise, as I said before, there is no reason to take you seriously.
16 Apr 08
Originally posted by eldragonflyBut these are probability exercises, not combinatorics. You have to compare the probability of two outcomes to find a solution.
then of course you must realize that you are just modeling a simple two-coin toss permutation/combination maths problem. Like really.
You haven't properly answered to the last Nemesio's post.
Did you look at the Monty Hall problem?
16 Apr 08
2 edits
Originally posted by PalynkaEvidently this post of mine was nearly impossible for the others to understand. Though my language was a bit strained, it would have done no harm to reword the problem correctly, point me to the wikipedia page where this problem was lifted from or admit that my solution was correct and/or reasonable, given the constraints of the original problem, which was improperly worded :
How did this turn into such a bus crash?
Originally posted by eldragonfly
Reword the problem then, this is nonsensical.
"A man with a hat shows you 3 cards, one completely gold, one completely silver, and one gold on one side and silver on the other. He puts them in his hat, and picks one at random. He then shows you one side of the card he picked, which happens to be silver. Now he says "I'll bet you even money that the other side of this card is silver too...whaddaya say, partner?"
If he held up a sliver card then there are only two choices, hence it is a 50/50 bet that the other side of a card showing a sliver side is silver.
He can't make this bet if he draws the gold/gold card. It is an even bet.
There is no other solution.
You don't make the case for holding up a gold sided card, so your "demonstration" is incorrect.
16 Apr 08
Originally posted by eldragonflyI'm going to play nice, so please play nice, too. No ad hominems between us from now on, okay?
iAnd it appears that you keep switching from the cards in the hat problem to the boy-girl family hoax/paradox at random, please be consistent.
Truce?
Let's look at it this way. There are three cards:
Card 1-Silver/Silver
Card 2-Silver/Gold
Card 3-Gold/Gold
In my analogue, there are three groups:
Group A-Sarah, Rebekah
Group B-Benjamin, Naomi
Group C-Isaac, David
As you can see, the genders of the three groups correspond directly to the colors on the card.
Now, I'm going to posit the original question simultaneously with my version:
Given that a silver side appears (given that a girl walks out), would you take an even money
bet against the claim that the other side is silver (that the other member of the group is a girl)?
You could be looking at the obverse of the S/S card, in which case the other side is silver (loser);
You could be looking at the reverse of the S/S card, in which case the other side is silver (loser);
or you could be looking at the silver side of the G/S card, in which case the other side is gold (winner).
Analogously:
The girl could be Sarah, which means Rebekah will walk out (loser).
The girl could be Rebekah, which means Sarah will walk out (loser).
The girl could be Naomi, which means Benjamin will walk out (winner).
Now, I have strived to show that these problems are identical, and that it is a bad decision
to take an even money bet on 2:1 odds.
So, since I have taken the time to give my summary of the problem as well as an analogue for
elucidation, I hope you will take the time to show how I have failed to understand the problem
(using a similar system, perhaps) or how you now understand why you've misunderstood the
math behind it.
Nemesio
16 Apr 08
Originally posted by kbaumenwrong : But these are probability exercises, not combinatorics. You have to compare the probability of two outcomes to find a solution.
What is wrong?
Answer to the questions. A question can't be wrong.
EDIT: and it seems you don't even know what Red Herring is, though that really is irrelevant.
red herring :You haven't properly answered to the last Nemesio's post.
red herring : Did you look at the Monty Hall problem?
a question can easily be wrong, reliant upon circular/superficial reasoning or be just plain misleading, surely you're joking my man. 😳
16 Apr 08
Originally posted by eldragonflyYes it's one of two possible cards. Yes, it's a silver side, one of three possible silver sides where to only one of them, the other side is gold. Three sides -> Three cases. In two cases the other side is silver.
Reword the problem then, this is nonsensical.
"A man with a hat shows you 3 cards, one completely gold, one completely silver, and one gold on one side and silver on the other. He puts them in his hat, and picks one at random. He then shows you one side of the card he picked, which happens to be silver. Now he says "I'll bet you even money that the o ...[text shortened]... the case for holding up a gold sided card, so your "demonstration" is incorrect.[/quote]
Which part of this do you disagree with?