in terms of a "fool-proof" explanation, let's try an analog to the explanation that most often works for the monty hall problem (but let's not diverge further into chaos by saying ANYTHING more about that problem haha 🙂)
imagine for a moment, a situation where instead of looking at two-sided cards, we have 20-sided dice (thank you dungeons and dragons). one of the dice has a G on every face, one of them has an S on only one face and a G on every other, and the last has an S on every face.
at random, you choose one of the dice (blindfolded) and roll it. a friend records the result, notes in his own mind which dice was the one you rolled, and puts them back in the bag. you remove the blindfold.
now, your friend reveals to you that you rolled an S. he asks you to guess which dice you rolled, and offers you even money on your decision. i think anyone with any worldly intuition at all would have an overwhelming suspicion that they chose the dice covered with S's. and only an idiot would guess the dice that had no S's at all. now, this does not mean that you didn't actually get the one with mostly G's and a single S! in fact, it's entirely possible that you did... but don't you at least have a gut feeling that it's more likely you chose the one with all S's? this should give you an inkling about conditional probability: that because the odds of an S are so loaded in the favor of the 3rd dice, that a result of S means an increased likelihood that the one you picked was the one covered in S's.
now let's go back to our unhappy, bickering reality:
so, in PBE6's original problem there are really two odds to consider. you know that S was the result. first, given the result of S, what are the odds that the guy with the cards chose the gold/silver card out of the hat? i think everyone agrees it's 1/3. some agree for different reasons. i don't care to explain it again - the "real" reason has been explained about a hundred times and still evades absorption for some reason... whatever. we all agree the odds of the gold/silver were 1/3.
now the silver/silver card (since it is the only other card with silver things on it) MUST be the one he chose if he did not chose the gold/silver card. i repeat. if he didn't chose the gold/silver card, but his result was silver, he MUST have chosen the silver/silver card. so then the probability it was that card is 1-P(gold/silver) = 1 - (1/3) = 2/3.
note: don't try to apply this to my problem's setup until you absorb the reasoning that the odds of the gold/silver one being 1/3 is because of the fact it has one out of the three possible silver SIDES and not because it was one out of the three possible cards to chose.
that's my $0.02
Originally posted by eldragonflyYou 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.
Fail. If anyone needs to learn anything about conditional probability it is yourself.
Originally posted by PBE6I fail to see how this is "forced". Is it a bit cheesy? Yes. (What do you expect from a guy with a dancing banana peel as an avatar?)
This one comes from the Old West, apparently.
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?"
Is this bet a fair one?
But behind the cheesiness is a fairly simple logic for the answer. There aren't any huge complications or hidden tricks. The problem is straightforward. I had heard an equivalent problem before and recognized it right away.
Of course, I am assuming the following things.
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. The problem therefore would have no answer, especially not the 1/2 some are touting.
The third point seems to be the sticking point.
I submit that the problem some like yourself are answering is the following.
"Three cards are placed in a hat, a silver, a gold, and a silver/gold. The man then draws one of the three randomly and informs you that the card has a silver side. He then offers even money that BOTH sides are silver."
In this case, the event is the selection of a card and not a side..
Originally posted by AetheraelYour 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.
in terms of a "fool-proof" explanation, let's try an analog to the explanation that most often works for the monty hall problem (but let's not diverge further into chaos by saying ANYTHING more about that problem haha 🙂)
imagine for a moment, a situation where instead of looking at two-sided cards, we have 20-sided dice (thank you dungeons and dragons) e it was one out of the three possible cards to chose.
that's my $0.02
Your analogy may make more sense if you switched it around to saying 1 all G die, 1 all S die, 1 19 S and 1 G side die, and you roll an S. Now, which one do you bet on rolling, the all S die, or the 19S side die? Your "gut" now tells you it's closer to 50/50.
However, this is still fundamentally flawed as an analogy, because in the original problem, the man had an equal chance of pulling out silver or gold, whereas in your scenario, I have a better chance in rolling S.
Originally posted by eldragonflyThat'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 to only one of them, the other side is golden. And the problem isn't poorly worded.
Fail.
It's not a three card problem, the gold card is "automatically" excluded by definition.
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?"
Originally posted by mtthwI'm expecting the answer that "there is a lot of crap in Wikipedia".
It should. I'll believe it when I see it though!
EDIT: Btw, For those, to whom this Wikipedia article seems false, search for the Monty Hall problem. It's similar. Also counter-intuitive.
EDIT2: There is a link to the MH problem in the page I gave.
Originally posted by kbaumenPretty close to 100%, if I applied Bayes formula right..
😀 Can you figure out the probability of this answer being written?
Anyways, (moving on,) here's another pair of probability problems.
Suppose a family has two children. The children are not twins, and each child is either a boy or a girl, and equally likely to be either.
What are the chances the family has a girl if...
1) the oldest is a boy?
2) they have at least one boy?