Best...Puzzle...Ever....

[MerakSpielman]

If you say one is Tom, you get a different riddle. Let me give them a different naming schema: Mom decided her firstborn son would be Harry, her second son Tom; her first daughter would be Mary, her second Jane. Then, given the first bit of information (two kids), you know that they consist of one of the following four pairs, though you don't know which pair:

Mary Harry
Harry Tom
Harry Mary
Mary Jane

Note that pairs 1 and 3 look similar, but in pair 1, Mary's the older and Harry's the younger; pair 3 is reversed. There are no other possible combinations, given the first bit of information we have.

Now you find out that one of the kids is a boy. What possible pairs are you left with?

If this is too confusing, we can come up with a more elaborate naming schema, in which mother chooses kids names based on both their order and their gender, but all that'll do is further distinguish 1 from 3. Lemme know if you need me to do that.

Daniel

Spoiler

OK, so you have 2 children, one of whom is older (O) and one who is younger (Y)

Your possibilities are, again, 4-fold, and you layed them out rather nicely.

O=Mary, Y=Harry
O=Harry, Y=Tom
O=Harry, Y=Mary
O=Mary, Y=Jane

Now, it's been revealed that one of the children is a boy. Fine.

You have 2 possibilities for this boy, then:

A) the older child is the boy (in which case he would be named Harry)
B) the younger child is the boy (in which case he is Harry if the first child was a girl, or Tom if the first child was a boy).

If the situation is A), then we have 2 possibities:

A1) O=Harry, Y=Tom
A2) O=Harry, Y=Mary

If the situation is B), then we still have 2 possibities:

B1) O=Mary, Y=Harry
B2) O=Harry, Y=Tom

With either A) or B), there is still a 50-50 chance of the remaining child being a girl or a boy.

 

[Pielorinho]

Merak, you keep changing the riddle. The riddle isn't, "I have two children, and the first is a boy"; the riddle is, "I have two children, and one is a boy."

Let's look further at your analysis.

Either child A or Child B is a boy, and there are four possibilities:
1) Ab Bb,
2) Ab Bg,
3) Ag Bb,
4) Ag Bg.

With me so far?

One of them is a boy. If it's A, then we end up looking at two of the four possibilities:
1) Ab Bb
2) Ab Bg

If it's B, then we end up looking at two of the four possibilities:
1) Ab Bb
3) Ag Bb

The problem is that we're looking at one of the four possibilities (1) twice. That doesn't mean that possibility is twice as likely, though; it's still as likely as all the others. If you weigh it equally each time, you're actually suggesting that it's likelier than the other possibilities.

This really isn't a matter of debate. If you really don't believe me, try some experiments, as follows:

Take those four combinations and give them to a friend.

Have the friend roll 1d4 secretly to determine which of the combinations to choose.

If "4" comes up on the die, discard it as not relevant to this experiment (we're told one of the two is a girl, and in 4, that's not true).

Then have the friend tell you that one of the two children is a boy, and you guess whether the other one is a girl.

I'll bet you five bucks that if you always guess girl, you'll be right 2/3 of the time .

Daniel

 

[Pielorinho]

If the situation is A), then we have 2 possibities:

A1) O=Harry, Y=Tom
A2) O=Harry, Y=Mary

If the situation is B), then we still have 2 possibities:

B1) O=Mary, Y=Harry
B2) O=Harry, Y=Tom

With either A) or B), there is still a 50-50 chance of the remaining child being a girl or a boy.

B2 and A1 are the same possibility. You're counting it twice instead of once.

Daniel

 

[Pyske]

It doesn't matter. A single child has been revealed as a boy. This boy either has to be the first or the second child. Either way, it works out. See my most recent post on the topic above, in response to Pielorinho.

Giving up on the spoilering, eh? Forgive if I keep at it a bit longer.

Spoiler

You are assuming, I think, that you know which child it is (first or second) after I tell you one is a boy. I haven't told you that, however. For the purposes of this excercise, shall we assume the order is birth order (first is older)? In that case, if I told you "my oldest child is a boy" the probability would be 50% as to the gender of the second child. In fact, the probability for the second child is always 50% once we fix the order. The problem here is that the order was not fixed when I told you that one is a boy. I can give you either child (older or younger) as the boy in the question provided. Like the Monty example, my choice of which child to reveal is not random.

. . . . . . . -- Eric

 

[PennStud77]

The problem is that the riddle doesn't ask this question; when you add it in, you get a very different riddle. We don't need to know which child is the boy; we don't get to know that. By adding in an extra step, you end up getting an incorrect answer.

Daniel

I am done with this argument. Merak put it much better than I, and actually has actually proven it. (Good job, Merak). Those of you still adhering to "Your answer is incorrect" without doing anything to back up that idea will a) never have any good evidence, and b) never admit that you could be wrong..... so there's no point in arguing further.

Have a Great One

 

[MerakSpielman]

Merak, you keep changing the riddle. The riddle isn't, "I have two children, and the first is a boy"; the riddle is, "I have two children, and one is a boy."

Let's look further at your analysis.

Either child A or Child B is a boy, and there are four possibilities:
1) Ab Bb,
2) Ab Bg,
3) Ag Bb,
4) Ag Bg.

With me so far?

One of them is a boy. If it's A, then we end up looking at two of the four possibilities:
1) Ab Bb
2) Ab Bg

If it's B, then we end up looking at two of the four possibilities:
1) Ab Bb
3) Ag Bb

The problem is that we're looking at one of the four possibilities (1) twice. That doesn't mean that possibility is twice as likely, though; it's still as likely as all the others. If you weigh it equally each time, you're actually suggesting that it's likelier than the other possibilities.

This really isn't a matter of debate. If you really don't believe me, try some experiments, as follows:

Take those four combinations and give them to a friend.

Have the friend roll 1d4 secretly to determine which of the combinations to choose.

If "4" comes up on the die, discard it as not relevant to this experiment (we're told one of the two is a girl, and in 4, that's not true).

Then have the friend tell you that one of the two children is a boy, and you guess whether the other one is a girl.

I'll bet you five bucks that if you always guess girl, you'll be right 2/3 of the time .

Daniel

No, what you're saying is that if you mark two sides of a coin "boy" and "girl," then flip it and get "boy," then the next flip is more likely to be a "girl." This is incorrect.

And you know what? If you flip the coin 3 times, and it comes up boy each time, the next flip is still 50-50 for being a girl.

If you don't believe me, try it. Flip a coin. Wait for a run of 3 heads or tails in a row. Write down the result of the next flip.

This flip will be heads (boys) half the time and tails (girls) half the time.

You have no idea how well I know this subject, Pie.

 

[Pielorinho]

Pennstud, I was serious about my bet. Choose someone impartial on this board to be the diceroller, do fifty trials, and my bet (I'll increase it to fifty bucks to sweeten the pot) is that your end result will be closer to 2/3 correct if you always choose girl.

Daniel

 

[PennStud77]

My apologies to any I offended. This thread moves so fast, by the time it went through the other side DID start to back it up. Let's see where it goes, though.

 

[MerakSpielman]

B2 and A1 are the same possibility. You're counting it twice instead of once.

Daniel

Yup. Can you think of a reason why I shouldn't?

 

[Pielorinho]

Merak, I got a pretty good idea of how well you know this subject . Your alternate bet is nothing like the riddle in question, and I agree that in that case, you've got a 50-50 chance.

Question: do you agree that with the bet I set up, only a sucker would be against me?

Here's a page that explains the odds better.

Daniel