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Leejc is absolutely right. He did a better job expressing what I tried to say back in June, 2007. But while I agree with his final conclusion, it isn't quite the right argument to use, that ""we must assume the worst,"" to get it. So I thought I would try once again to explain how the different wordings of this problem, that seem to say the same thing, can lead to different solutions.

The ultimate cause of confusion is the math-specific definition of the term ""event."" When a random experiment produces a result, that result is not called an ""event,"" it is an ""occurrence."" ""Event"" is defined to be the subset of all possible occurrences that have some defining common property. So if I flip a coin a hundred times, there are only two possible events: ""heads"" and ""tails."" There will most likely be about fifty occurrences of each event; and each occurrence will fit into only one event.

If I draw a card from a standard deck of playing cards, there are lots of possible events and/or occurrences. Some of them overlap. They include ""black card,"" ""spade,"" ""honor card (TJQKA),"" ""Ace,"" and ""Ace of Spades."" The occurrence ""black card"" can be in the event ""spade"" or ""club"" as well. Note that an occurrence can fit into other events, but not vice versa. An occurrence of ""spade"" is also an occurrence of ""black card,"" but the event ""spade"" is most definitely not the event ""black card."" Since determining probabilities involves counting the number of elements in events, the difference can be crucial when less specific observations are applied.

There is a famous problem that isolates this difference between an occurrence and an event. It is called Bertrand's Box Paradox Suppose you have three indistinguishable boxes, each with two indistinguishable drawers. Inside each drawer is a coin. Box #1 has two gold coins, Box #2 has two bronze coins (I can't resist the temptation to use the initials B and G), and Box #3 has one of each. You pick a box at random, and open a drawer at random. If the coin is gold, what are the odds that the coin in the box's other drawer is also gold (i.e., that it is Box #1)?

The first issue here, common to story problems, is that this problem describes an occurrence, not an event. We must extrapolate what the possible outcomes of the actions are, to define the event (the set of possible occurrences) that defines the conditional probability.

The incorrect approach to Bertrand's Box Paradox is to say that you narrowed the identity of the box to two possibilities, Box #1 and Box #3. That the ""event"" matches the observation ""the box has at least one gold coin."" This is wrong because there are other ways to describe occurrences in that event. If you pick box #3, then no matter which drawer you open, the occurrence is in the event ""the box has at least one gold coin.""

The correct approach is to define the event as the set of all possible occurrences where you open a drawer with a gold coin. Every possible occurrence that is consistent with the problem description is in this event (remember, it's a set), and every possible element of this event is consistent with the problem description. Both of those conditions have to be true for the event set we extrapolate as the condition. This set has three elements, not two. In two of those three possible occurrences, the box's other coin is gold. The correct answer is 2/3.

I have seen a variation of Bertrand's Box Paradox used to represent the Boy or Girl Paradox. You are the organizer of a neighborhood swim meet. During the meet, you get a phone call from Mr. Smith, who is delayed and can't pick up his children after the meet. He says ""Tell my two kids to go home with a neighbor. Their names are Sarah and..."" and the phone connection goes dead. There are four families of two children in the meet, and they conveniently include the four possibilities BB, BG, GB, and GG. What are the odds that Mr. Smith has two daughters?

This occurrence can be described as ""a two-child family includes at least one girl."" But not the event you need to use as the condition. Mr. Smith could have mentioned a boy's name first; and if that boy had a sister, that occurrence would have been an element of the set ""a two-child family includes at least one girl."" So that event fails one of the two criteria. You get the answer 1/3 by erroneously using this event.

The correct event is just that Mr. Smith mentioned that he had a daughter, and compares to ""you opened a drawer with a gold coin"" in the Box Paradox. You can count four possible occurrences of that event. In two of those four, representing a single family, Sarah has a sister. The correct answer is 2/4, or 1/2.

If a conditional probability question describes an occurrence, and not an event, care needs to be taken whe