Originally posted by kbaumenYes, your analogy is now making 2 beans equal to 1 card, and 1 bean equal to another card. There are two cards, so show me the example using 2 beans.
Did you read the explanation here - http://en.wikipedia.org/wiki/three_cards_problem. It should make everything clear. 50% would be correct if all four sides were silver. Then there is an even number of cases.
Now here the cards are different and one of the cards has silver side on both sides but the other only on one side. If we randomly choose a silver ...[text shortened]... e red one, hence 66% that we see a silver side from the completely silver side.
Still wrong?
Further, more. What will the result be after the card is revealed only performing it one time?
It's either SS or SG, is it not? There's no 1/3 SG or 66% SS or anything else. In REALITY, after one try, between SS or SG, it is either SS or SG. 50/50. I concede that multiple tries being logged will yield a lopsided result in the favor of the man with the hat. But, it does not exist in a one shot deal.
Both sides of the equation have to equal up. When the question is between SS and SG, the answer can NOT be 1/3 SG and 2/3 SS, it is either SS or SG.
2 edits
Originally posted by broblutoIn REALITY you either win the lottery or you don't. You still think that's 50/50? Surely you can see that's not how probability works.
It's either SS or SG, is it not? There's no 1/3 SG or 66% SS or anything else. In REALITY, after one try, between SS or SG, it is either SS or SG. 50/50.
Besides, if you insist on looking at it that way, the side isn't 50% gold 50% silver either. If we accepted your argument, you've just proved it can't be 50-50 either. It must be 100% one way or the other. Yet that doesn't mean one of them is impossible.
Originally posted by geepamoogleIrrelevant. geepamoogle once again your explanation is worthless. You can jump through all the hoops that you want, your answers and shallow justifications are still in error.
[b]PART 1
Let me get straight to the point. With the children, each birth is a distinct and independent event. I made this abundantly clear by letting you know they weren't twins (and hence that the gender of the first birth does not affect the odds of the second birth at all.)
Both births HAVE occurred in the problem, although we are ...[text shortened]... will either win or I will lose. That does not mean I have a 50% chance at the jackpot.[/b]
read these two questions again.
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.
Yet you, PBE6 and the others give different answers!
😕
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.
Originally posted by eldragonflyWhat? You don't see the difference?
What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?[/quote]
Both questions are equivalent, in other words you can have a family with at least one boy, and that boy being the oldest.
In the 2nd question, in that family can be only one boy, who in fact is the youngest of two kids.
1 edit
Originally posted by kbaumenWrong. Your foaming at the mouth explanations are worthless, you can pretend all you want, but this is a question of semantics.
What? You don't see the difference?
In the 2nd question, in that family can be only one boy, who in fact is the youngest of two kids.
From wikipedia:
An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"
Originally posted by eldragonflyTrue. But it was funny. 😀
Irrelevant personal attack PBE6. Both questions are the same, you can't really be that stoopid, but obviously you are.
As for the two questions about the probability of having at least one girl being the same, they are not. That's the whole point of asking them together!! It's all about context, baby.
Originally posted by eldragonflySurely you can see that the following scenario:What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?
"The youngest child is a boy, the oldest child is a girl"
Satisfies (2) but not (1).
Therefore, they are not equivalent questions. Therefore they can have a different answer.
Originally posted by eldragonflyHAHAHAHAH!!!!! 😵
read these two questions again.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.
Originally posted by eldragonflyWe have three possible combinations then
Wrong. Your foaming at the mouth explanations are worthless, you can pretend all you want, but this is a question of semantics.
From wikipedia:An equivalent and perhaps clearer way of stating the problem is "Excluding the case of two girls, what is the probability that two random children are of different gender?"
1)older brother, younger sister
2)older sister, younger brother
3)two brothers
so 66% that they're different gender.
And, please, don't say it's a worthless solution. Instead, explain why so, if you think it's worthless.
New problems! Exquisitely simple, but hopefully illuminating.
1) What is the probability of a standard roulette wheel (36 numbers, plus 0 and 00 for a total of 38 possibilities):
(a) landing on 7 on the first spin?
(b) landing on 7 on the second spin, given that it already landed on 7 on the first spin?
2) What is the probability of:
(a) being dealt a natural blackjack (an Ace plus any face card) on a draw of 2 cards from a single deck, if you are playing alone and you can't see the dealer's cards?
(b) being dealt a natural blackjack on a draw of 2 cards from a single deck, if you are playing alone but the dealer has revealed one of his cards to be an Ace?
1 edit
Oh, and...
In fact Bayes theorem does not apply, conditional probabilities does not apply, in fact gives the wrong answer.
Bayes' theorem is a statement of equivalence. It is simply true. It might not always be useful, but it never gives the wrong answer. If it does, you're applying it wrong.
Originally posted by mtthwDo you deny that the man with the hat can either be holding the SS card or the SG card?
In REALITY you either win the lottery or you don't. You still think that's 50/50? Surely you can see that's not how probability works.
Besides, if you insist on looking at it that way, the side isn't 50% gold 50% silver either. If we accepted your argument, you've just proved it can't be 50-50 either. It must be 100% one way or the other. Yet that doesn't mean one of them is impossible.
Do you deny, that in order for an equation to be correct, it must be equal on both sides?
If you said No to both of these questions, then the answer is that the odds are 50/50 because in a ONE TIME shot between SS and SG, the answer will either be SS OR SG, nothing more, nothing less. One try equals EITHER SS or SG. Multiple tries will yield a ratio of 2SS to 1SG.
With the lottery, you can have multiple different combinations be the answer, not analogous.
He is holding 1 of 2 possible cards in his hand. Once it is revealed which one it is, the result of the equation will be either SS or SG.
If we want to talk sides. Then the GG bar is G1 and G2, SS is S1 and S2, and SG is S3 and G3. We know that a side is S, but we don't know if it's S1, S2, or S3. Knowing if it was S1, S2, or S3 would mean that we would know what bar it is immediately. Not knowing, means that I have a 1 in 3 chance of guessing what side it is. But, to win the bet, for ONE time, I don't NEED to know what side it is. The side is already chosen, so the option of S1 or S2 go out the door. It's only a decision between SG and SS, and even if it is SS then I still don't know which side is S1 or S2 because they're both silver, nor do I care because I wouldn't be playing again because it's a ONE TIME event and playing more than once would put the odds against me.
BTW, in order for something to be compared, at least the results have to be similar. The lottery and jelly beans don't work because even after 60 tries, I would still probably lose each of those times. In the card example I would've won about 20 of those 60 times.
1 edit
Originally posted by broblutoNo one here denies that it's two cards. What most of us are implying, is that 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.
Do you deny that the man with the hat can either be holding the SS card or the SG card?
Do you deny, that in order for an equation to be correct, it must be equal on both sides?
If you said No to both of these questions, then the answer is that the odds are 50/50 because in a ONE TIME shot between SS and SG, the answer will either be SS OR SG, nothin se each of those times. In the card example I would've won about 20 of those 60 times.