Two situations:
1)
A: “I have a boy named Thomas”
B: “I have two children and one of them is a boy named Thomas”

In this case it is close to 50% that the other kid is a boy

2)
A: [something that has nothing to do with Thomas]
B: “I have two children and one of them is a boy named Thomas”

in this case B knows the names of the kids and the Thomas label does not change anything from “I have two children and one of them is a boy” situation

Dick Martin

Apparently you have lots of time. Of course I know what ambiguous means, please don’t try to teach me everything I know already.

You say: ““Event C” is a set of possible outcomes, not a description of one outcome.”. That’s exactly the question, Foshee’s question is one outcome, not a set.

Your coin game is ridiculous. Remember I win if the result is two tails.

One more time for JeffJo:

For me an ambiguous mathematic question is an invalid mathematic question.

I really hope you have nothing more to say to me.

Some say that some of their rudeness has been inherited, via the Viking invasions, by the British.

Anyone who does say that is an idiot.

Ambiguous means “can be interpreted in more than one way.” For example: I just placed two coins, face up, on the desk in front of me, making sure at least one is heads. What is the probability both are heads? Would you bet $1, to win $5, on this question? How often?

When the man says “at least one is a boy,” does that mean he choose a child and told us that child’s gender? Then the answer is 1/2. Or does it mean he choose a gender, and agreed that he had at least one matching child? Then the probability is 1/3. Setting reasonableness aside, either interpretation is possible, so the question is ambiguous.

Let’s try a slightly different coin-flipping game; Here’s a demonstration (and yes, I am actually doing this) of my game:

1) I flip two coins so you can’t see.

2) I tell you “At least one landed on heads.”

3) You can bet $1 on whether the two coins show the same side, or different.

If you win a bet that they are the same, I will pay you back $2.40. So you will win an average of $0.20 a game if the probability is 1/2, but I win an average of $0.20 if the probability is 1/3.

If you win a bet that they are different, I will pay you back $1.71. So you will win an average of $0.14 a game if the probability they are different is 2/3 (corresponding to “1/3” for “the same”), but I win an average of just over $0.14 if the probability is 1/2.

(By the way, “the same” won.)

I offer to repeat this game, according to a fixed procedure, as often as you want to play it. But only if you will as you bet the same way you did there each time. If you choose to bet on “different,” the procedure is that I tell you what came up each flip, and I win $0.14. If you bet on “the same,” the procedure is I reflip anytime I get two tails so that I can say “heads” as in the example, and I will win $0.20. In either case, did I contradict the example? I am not a croupier; but I might be someone who is looking for a heads, not looking for every result.

Now, I agree the first is not a good interpretation. All I’m saying is that it is a *possible* interpretation. And the only valid points of discussion here are why it is or isn’t, and why some people assume it is.

Vagn: “I think he contradicts himself. First he defines what I would call ‘the math puzzle genre,’ later he denies that it exists!”

The definition was for a conditional probability, and has nothing (directly) to do with puzzles. When you know an outcome was constrained, somehow, to event C, then the conditional probability (sometimes called posterior probability) of event E IS DEFINED TO BE the original probability of both events E and C happing at the same time, divided by the original probability of event C happening regardless of E.

So, if E is “family has two boys,” and C is “family has at least one boy,” then P(E&C)=1/4, P(C)=3/4, and P(E|C)=(1/4)/(3/4)=1/3. People get this answer, probably, by placing too much emphasis on the language used, and not recognizing that “Event C” is a set of possible outcomes, not a description of one outcome.

But if C is “Father chooses ‘at least one boy’ randomly from all of {‘at least one boy’, ‘at least one girl’} that apply,” which seems more likely to describe the set, then P(E&C)=1/4, P(C)=1/2, and P(E|C)=(1/4)/(2/4)=1/2.

But you have shown yourself to be someone who has little knowledge of the actual mathematics of probability, and little desire to discuss it. Your inability to grasp what others say goes beyond a language barrier; you seem to actively seek ways to misunderstand (or at least, you fail to try to understand when it is easier to misunderstand by reading only superficially). So unless that changes, I now have nothing more to say to you.

1. “A problem statement can be misleading if it is described as an example, because an example only describes posterior events. That is where we need to get the answer from, but their probabilities can never be calculated directly. They must be calculated from the prior probabilities. So we must extrapolate what the condition C was before it was applied, not after. Then, the posterior probability of any event E is the prior probability of E and C happening together, divided by the prior probability of C happening. This is a definition.”

and:

2. “There is no such thing as “the math puzzle genre,” that is a rationalization used to cover an assumption that is incorrect.”

I think he contradicts himself. First he defines what I would call “the math puzzle genre”, later he denies that it exists!

Let’s play the coin-flipping game:

a) We keep flipping two coins simultaneously.
b) If both coins are tails, I win, you give me $20.
c) Otherwise, I give you $15 if there is one head, and you give me $20 if there are two heads.

If the probability is 1/3, you will be making money. If it’s 1/2, I will.

Will you play?

Did you ever see a croupier cancel a game because he didn’t like the result? Your game exposes your lack of understanding.