Simple gambling problem

Originally posted by brobluto
Your analogy is fundamentally flawed as the odds of rolling a G are more than rolling an S because one die has 19 G and only 1 S. The fact that you rolled an S has already defeated those odds and you're an idiot if you DON'T bet that you rolled the dice with all the S's.

Your analogy may make more sense if you switched it around to saying 1 all G die, 1 ...[text shortened]... pulling out silver or gold, whereas in your scenario, I have a better chance in rolling S.

i did not presume to state that the probability of my 20-sided "S" die and the probabilities of the original problem had anything to do with each other at all, and in fact my example could have used different colors, a different scenario altogether, etc. however... my basic point was to try to establish a "feel" for conditional probability for those who have a difficult time understanding the calculation behind it, and the scenarios in which it should apply.

the ultimately confusing aspect of the original problem (and seemingly the issue that you take with my analogous setup) is that it SEEMS like a 50/50 problem. two cards. two colors. 50/50. not so!

the idea behind my analogy was to show that one can determine the probability that the silver side being shown belongs to the gold/silver hybrid card (which we can call X), and THEN the probability it was the other card must in turn be (1-X) because there are no other cards with silver on them.

i tried to make this fact more obvious by heavily weighting my analogy in such a way that it became immediately apparent that the most likely card in the original problem was the silver/silver one. but perhaps the distinction between the actual numerical results of the original problem and my construction was not duly noted, and i apologize for any confusion.

cheers

Originally posted by mtthw
You know, you don't make someone wrong just by saying "fail". It just sounds silly. Especially when I've proved time and again in this thread that I understand it fine, whereas you've avoided several opportunities to show you do.

It's simple mtthw. Back up your rather strange "crystal ball" statement, or admit that you are wrong. That you and the others here find it necessary to keep regurgitating the same tired modified/redefined problem definition, and the same tired see-through explanations, does not change anything, and definitely does not make your solution correct. For example a gold/gold card and the gold/silver cards could be selected randomly, then of course the problem does not make any sense. This is a bayes theorem gee whiz problem not a conditional probability problem. Your solution sticks out like a sore thumb, hence the repeated and "forced" explanations and flat insights.

Originally posted by geepamoogle
1) Not only was the card chosen random, but so was the side.
2) Neither the man nor myself KNOW what the other side is, only the revealed side, not by sight or weight.
3) Gold COULD HAVE come up just as easily, but in this instance, it failed to.

If either 1 or 2 are false, then there is no odds, and this isn't a fair bet, nor is it an exercise in probability.

There you said it yourself. This problem is not "an exercise in probability." It is forced. If the silver/gold card came up with the gold side showing, or the gold/gold card was one of the two selected, then this problem does not make sense, hence it is ill-defined. This is what i have been saying all along.

Originally posted by eldragonfly
It's simple mtthw. Back up your rather strange "crystal ball" statement, or admit that you are wrong. That you and the others here find it necessary to keep regurgitating the same tired modified/redefined problem definition, and the same tired see-through explanations, does not change anything, and definitely does not make your solution correct. For exa ...[text shortened]... icks out like a sore thumb, hence the repeated and "forced" explanations and flat insights.

Let's make it clear, once more. The problem I have solved (correctly) is the one that was intended by the problem setter. So I'm not wrong - the only person who can confirm what was intended has confirmed that. Several people on this thread read that problem definition and interpreted the question as it was intended. This shows that the problem definition cannot have been that bad. I still think it is the only sensible interpretation, as any other interpretation does not give you enough information to solve the problem.

Basically, you fell into exactly the trap that the question is designed to set - you just won't accept it. Read the Wikipedia article - this is a classic problem.

Originally posted by kbaumen
That's the point. It's not a three card problem. It's a three side problem. You see one of the three possible silver sides. The man doesn't show you a card. He shows you side of it. You see only ONE side of the card. Now our task is to find out how big is the probability that the other side is silver. You can see one of the three possible silver sides where ...[text shortened]... even money that the other side of this card is silver too...whaddaya say, partner?"[/b]

Yet another unfortunate redefinition of the cards in the hat problem. First you say it's not a 3 card problem, then you say it is. Make up your mind kbaumen, your explanation is worthless.

Originally posted by eldragonfly
There you said it yourself. This problem is not "an exercise in probability." It is forced. If the silver/gold card came up with the gold side showing, or the gold/gold card was one of the two selected, then this problem does not make sense, hence it is ill-defined. This is what i have been saying all along.

"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..."

My emphasis. That phrasing eliminates your objection. If it had been gold he would have said "I'll bet you even money that the other side of this card is gold too..."

Originally posted by mtthw
You know, you don't make someone wrong just by saying "fail". It just sounds silly. Especially when I've proved time and again in this thread that I understand it fine, whereas you've avoided several opportunities to show you do.

No mtthw back up this curious and childish statement.

Originally posted by mtthw
"He then shows you one side of the card he picked, [b]which happens to be silver. Now he says "I'll bet you even money that the other side of this card is silver too..."

My emphasis. That phrasing eliminates the problem. If it had been gold he would have said "I'll bet you even money that the other side of this card is gold too..."[/b]

Still don't need the bayes theorem here, the gold/gold card is automatically excluded from the sample space by definition, but i see your point.

Originally posted by mtthw
Let's make it clear, once more. The problem I have solved (correctly) is the one that was intended by the problem setter. So I'm not wrong - the only person who can confirm what was intended has confirmed that. Several people on this thread read that problem definition and interpreted the question as it was intended. This shows that the problem de ...[text shortened]... d to set - you just won't accept it. Read the Wikipedia article - this is a classic problem.

show me this wikipedia article you mention.

Originally posted by eldragonfly
No mtthw back up this curious and childish statement.

Nothing childish about it. Just an observation.

I've correctly solved several problems in this thread. Several times I have shown my full working to get to the solution. I have shown an understanding of conditional probability.

On the other hand, when challenged you have tended to avoid the issue. For instance, I printed my full calculation of that particular problem and asked you which bit you disagreed with. This was a genuine attempt to try and understand where the misunderstanding was. You avoided the issue. When other questions were set you avoided the issue. You haven't actually posted anything showing a full understanding of conditional probability. Hence my observation - I can only go on what you write - I can't guess what you're thinking.

And remember, my "childish statement" was in response to "Fail. If anyone needs to learn anything about conditional probability it is yourself.". You don't see how someone could find that irritating and not a useful contribution to the discussion?

Originally posted by eldragonfly
show me this wikipedia article you mention.

It was a couple of pages back: http://en.wikipedia.org/wiki/Three_cards_problem

Originally posted by eldragonfly
Still don't need the bayes theorem here, the gold/gold card is automatically excluded from the sample space by definition, but i see your point.

The Gold/Gold card is excluded, but so is one orientation of the Gold/Silver, leaving three possibilities.

Bayes' Theorem may not be needed, but it's always a valid approach. It's just another way of looking at the problem. My background is in mathematics, so my first instinct is to go looking for the mathematical solution. Not because I want to complicate things - just because I personally find it the most straightforward.

Anyway, got to go now!

Originally posted by mtthw
The Gold/Gold card is excluded, but so is one orientation of the Gold/Silver, leaving three possibilities.

Bayes' Theorem may not be needed, but it's always a valid approach. It's just another way of looking at the problem. My background is in mathematics, so my first instinct is to go looking for the mathematical solution. Not because I want to complica ...[text shortened]... hings - just because I personally find it the most straightforward.

Anyway, got to go now!

exactly.. no one ever said that Bayes' was the ONLY valid method to solving the problem. Bayes' theorem is, however, an interesting and elegant method for solving this problem, in addition to being a method that avoids the common probabalistic misconception that eldragonfly fell into his first go around on the problem.

and lastly, the "exclusion" of the gold/gold card as you (eldragonfly) put it has as little result on the clarity of the question, or on the accuracy of the calculation of the results, as does a "red herring" in a problem. the intent of the questioner was very clear (once it was stipulated that the card picker was doing so randomly and had no knowledge of the other side), and represented a single trial of this scenario. as originally stated, there is no need to even take into account the case where the picker chose a gold side. because he didn't in this trial. who cares? we are discussing the probability of a second silver side, given that the initial choice was a silver side. and whether you "exclude the gold/gold card" or not, any valid method of counting and sound leaps of logical reasoning will still predict the probability to be 2/3 that it's the double silver card. as even you have said you now come to realize, after further thought.

Originally posted by eldragonfly
show me this wikipedia article you mention.

Yeah, please take a look at the article. It mentions that this scenario has been named the "three-card swindle". Sorry, eldragonfly, but you are one of the many who would have been swindled.

Originally posted by mtthw
The Gold/Gold card is excluded, but so is one orientation of the Gold/Silver, leaving three possibilities.

Bayes' Theorem may not be needed, but it's always a valid approach. It's just another way of looking at the problem. My background is in mathematics, so my first instinct is to go looking for the mathematical solution. Not because I want to complica ...[text shortened]... hings - just because I personally find it the most straightforward.

Anyway, got to go now!

Thanks for a half-way coherent response. This is what i have been saying all along.