Originally posted by mtthwI'm telling you, he's the next Andy Kaufman. This thread will live in infamy as the breakthrough performance of the world's crappiest math comic. For posterity, I'd like to brand this type of humour as "macomicmatics", from "mathematics" and "comic" and the Latin meaning turd-gurgler. 😀
OK, I'm finished with you. You get caught in a statement so demonstrably false that even you can't deny you're wrong...so you ignore the refutation and fall back on the old "poorly worded question" defence. The question was rock solid. It's not just conditional probability you don't understand, it's the English language.
I think you're winding us up. I nd the universe it wrong.
Good luck everybody. See you in a different thread.
Originally posted by broblutothe whole point of whether or not it's a fair bet is to predict what would happen after many many tries, and then to make your decision based around that information. that is also how probability works - you keep saying that for a one time shot it's a 50/50 chance, but even your notation 50/50 implies a hundred trials, 50 of which are gold and 50 of which are silver... but you've said yourself that after a number of trials it becomes apparent that the odds are not equal.
I get that. Really I do. But if you apply that in application, it fails. 2/3 is saying that 2 out of 3 times I lose, right? Well, what happens on the first and only time? That's still a 50/50 chance.
Are you saying that if I lose the first time, and a silver side shows again, I have a 50/50 chance then?
so why do you insist that it's equally likely, when reality shows you that it's not by many different methods? just because you haven't REALIZED the odds as you put it, the predicted odds tell you that making this bet would be folly. and isn't that what gambling's all about? mathematically predicted odds?
Originally posted by AetheraelNicely put. 😏
the whole point of whether or not it's a fair bet is to [b] predict what would happen after many many tries, and then to make your decision based around that information. that is also how probability works - you keep saying that for a one time shot it's a 50/50 chance, but even your notation 50/50 implies a hundred trials, 50 of which are gold and 50 ...[text shortened]... olly. and isn't that what gambling's all about? mathematically predicted odds?[/b]
Originally posted by AetheraelI think he's snorkeling at the bottom of a bottle of Jack Daniels presently. I assume he'll be back once the sour mash wears off.
thank you i didn't even realize i was 4 pages behind in the thread though haha... i wish eldragonfly would try to bring me into the argument - he never answers my threads even when they talk specifically to him 🙂
Goodness gracious.. I missed a lot in this thread... Apologies for going AWOL by having to attend my job..
I will first make a personal note here to eldragonfly and his disciple.. Your arrogance and your ill-constructed arguments have been highly annoying after the amusement faded.. I have avoided personal comments and statement of impressions to focus on the content of the arguments, but while you refer to our arguments as old and tired, I find your convoluted and difficult to follow.
In addition, I have offered you full explanations with requests as to which particular step you disagreed with, in order to give you a chance to point out very specifically where you think I am wrong. I have seen no such response to these posts, wherein you explain to me HOW I have misapplied some concepts and WHY they don't apply.. Others have also tried to get this information from you, I believe.
Believe it or not, you are arguing against classic problems and basic principles which are almost universally accepted in the mathematics community, and mathematics does not often change.
So now, let me make the post I was going to make about 6-7-8 pages back, even though others have already made it repeatedly and it has been "refuted" by the gods of probability here, although not in a fashion I find in the least convincing.
What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?
Both questions are equivalent, in other words you can have a family with at least one boy, and that boy being the oldest.
Therefore both answers have to be the same.
The problem with this statement is as follows. If I say the oldest is a boy, I am not only telling you they have a boy, I am also telling you which specific child is a boy.
If I only tell you they have at least one boy, then I am not telling you which child it is. The 'eldest' part of the statement is additional information, since the children were born in a specific order.
Why does this matter? Well, let me ask you which of the two cases the following family fits under..
Let us say I know a specific family, the Smiths, who have two children. They have a 12-year-old daughter named Susan and a 6-year-old son named Peter.
1) Is the eldest a boy?
2) Do they have at least one boy?
3) Do they have at least one girl?
The answer to #1 and #2 are different in this case, and so it is ludicrous to claim that they are equivalent conditions, and yet this is required for your argument to be valid. In order for the questions to be identical, the conditions given have to be identical as well. If you can point to a case which fits one condition but not the other, the conditions and hence the problem is not identical.
In fact Bayes theorem does not apply, conditional probabilities does not apply, in fact gives the wrong answer. And once again your coin analogy is bogus and erroneous, it is a combinations problem, not a permutations problem.
Bayes formula can rightly be applied to any probability problem where we are given information on the results which eliminate some of the possible answers that the set-up alone does not eliminate. Even if the additional information does not actually eliminate any possibilities or alter any odds, it will STILL result in a correct answer given accurate information.
In this case, both questions eliminate the possibility the family has two girls, although we could not eliminate that possibility before being told something of the gender make-up of the family. Since we have reduced our sample space, Bayes formula is valid for recalculating odds in the smaller sample space.
The first question also eliminates those families with mixed gender children where the daughter is older than the son. The second does not do this. The result is that the sample space is different for both questions, which means the questions AREN'T identical, and can have different answers.
The binomial formula is valid for calculating combinations of things as well, by the way. As a matter of fact, that is its primary purpose whenever you expand the terms of equations like (x+1)^n.
(a+b)^2 = a^2 + 2ab + b^2.
If a is the chance the child is a boy, and b is the chance the child is a girl, and a=b, then the combination of boy and girl is twice as likely in a family with two children as a family with 2 boys.
We also see that in a family with two children, if we don't have any information on their gender, they have a 75% chance that at least one is a girl.
Also a note on probability for one-time events. Your odds will match the long term odds or repeated runs of the same identical event. When I roll a fair six-sided die once, I have a 1/6 chance to get 6, not a 1/2 chance.
When people talk about 50/50, that means more than "it could either way". It means "It's equally likely to go either way."
For some events, this may be true, for a great many, it is not. A lottery ticket could be win or lose, but it's not 50/50 because it's far more likely to be the latter.
FINAL COMMENTS
I'm about ready to start or enter another brainteaser thread with a different theme, this one is getting old. Again, I give out my thanks to mtthw, PBE6, and everyone who participated in the fistfi.. err, I mean debate.
Originally posted by kbaumenWrong.
No one here denies that it's two cards. What most of us are implying, is that [b]the probability of it being SS is twice as big as it being GS, hence the probability of losing the bet is twice as big. Of course there are only two possible outcomes, but one of them is more probable than the other.[/b]
Originally posted by Green PaladinPoint taken, but your "randomness" of only choosing between the SS and the SG cards makes this a moot point.
You say it's a one time event, well, imagine standing at the back of a line of people all participating in the card game. Each person has one bash at the game. As the queue grows shorter you start to notice that 2/3 of the people are handing over money while only 1/3 are collecting. Aren't these all one time events? I don't know about you but I would step out of the line at that point.
Originally posted by mtthwWrong.
No, I'm saying that it's not a 50/50 chance, it's a 2/1 chance. Yes, you'll either win or lose. But it's completely misleading to say it's 50/50. Because to me, 50/50 means "one or the other with equal likelihood". Tossing a coin is 50/50. If by 50/50 you just mean "one or the other", then fair enough. Personally I don't think that's a useful concept.
...[text shortened]... e're not arguing about the calculation any more, though, so I'll leave it at that.
Originally posted by PBE6
One more new problem! This one is called the Mr. Smith 3-Card Monty problem.
I have 3 cards: the Ace of Hearts, the Ace of Clubs, and a Joker. I pick 2 of these cards at random (each card is equally likely to be chosen, chance of any card being chosen is 1/3), and look at them. I then tell you that I hold at least one ace...
(a) What is the chance the other card I hold is also an ace?
(b) I now tell you that the ace I am speaking of is actually the Ace of Hearts. What is the chance the other card I hold is an ace?
(a) 1/3
(b) 1/2