I have a simple riddle

Very nice trolling.

1/3 is the only answer being discussed that can't be justified.

Before you have any children the probably is 25% for each of these BB, BG, GB, GG. But after you have one you've eliminated 2 out of the four combinations. Once you know one is a girl, it is already determined that she is either the first OR the second in the combination. She CANNOT be both, so if she is first then options are GG or GB. If she is second then the options are GG or BG.

Also, those of you saying 2/3 are attempting to answer two questions...what is the sex of the other child and what is the order of the children. The OP is only asking for the sex of the other child. There are only two possible answers.

This marble example is a little bit incorrect. Each marble should be labeled with an 'L' or 'R'. Then your blue/red combos have 25 blue with an L and 25 blue with an R. Now if Bob holds a red marble which has an L . In order to be accurate he can then only choose from those combos that do not have L. This eliminates 50% of the options.

Hgrsfh wrote:

Before you have any children the probably is 25% for each of these BB, BG, GB, GG. But after you have one you've eliminated 2 out of the four combinations. Once you know one is a girl, it is already determined that she is either the first OR the second in the combination. She CANNOT be both, so if she is first then options are GG or GB. If she is second then the options are GG or BG.

Yes. But given the first case, GG is less likely than GB, and given the second, GG is less likely than BG. The reason is that in the GG case, there was only half the chance that she was the "chosen girl," since there was another girl she was "competing" with to be "the one."

Hgrsfh wrote:

Also, those of you saying 2/3 are attempting to answer two questions...what is the sex of the other child and what is the order of the children. The OP is only asking for the sex of the other child. There are only two possible answers.

I have two kids, one is a girl. What is the probability that the other kid is a boy?

The question is pretty straight forward and makes no mention of whether or not the girl was first or second. So the answer is 2/3 as others have shown with detailed explanations. You have to make unfounded assumptions about the question to come to any other conclusion.

Correct answer is 1/2 or 50%. Only valid options are GG or GB. Flip GB to BG if you want it doesn't matter because we're only worried about whether one boy exists, not the order of boys and girls. Man you guys are stupid.

Youz all dummies wrote:

Correct answer is 1/2 or 50%. Only valid options are GG or GB. Flip GB to BG if you want it doesn't matter because we're only worried about whether one boy exists, not the order of boys and girls. Man you guys are stupid.

Think of it this way.

Ask someone on the street how many kids they have. For those who say 2, ask them if one of them is a girl. For those who say yes the probability that the other kid is a boy is 2/3 because you have already filtered out the BB possibility.

GG

BG

GB

That's the logic and it's very similar to the Monty Hall Paradox where information is inserted into the equation which changes the probability of the outcome for the remaining unknowns.

It certainly isn't obvious.

DiscoGary wrote:

Youz all dummies wrote:

Correct answer is 1/2 or 50%. Only valid options are GG or GB. Flip GB to BG if you want it doesn't matter because we're only worried about whether one boy exists, not the order of boys and girls. Man you guys are stupid.

Think of it this way.

Ask someone on the street how many kids they have. For those who say 2, ask them if one of them is a girl. For those who say yes the probability that the other kid is a boy is 2/3 because you have already filtered out the BB possibility.

GG

BG

GB

GG isn't an option because the question states that ONE is a girl.

2 is not 1 wrote:

DiscoGary wrote:

Think of it this way.

Ask someone on the street how many kids they have. For those who say 2, ask them if one of them is a girl. For those who say yes the probability that the other kid is a boy is 2/3 because you have already filtered out the BB possibility.

GG

BG

GB

GG isn't an option because the question states that ONE is a girl.

If you interpret the question to say "Only" one of them is a girl, then there would be a 100% chance that the other was a boy.... but that's not what the question says.

G =known g=unknown b=unknown

The possible outcomes are

Gg

Gb

gG

bG

Or GG=2 (50%) BG=(25%) GB=(25%)

DiscoGary wrote:

2 is not 1 wrote:

GG isn't an option because the question states that ONE is a girl.

If you interpret the question to say "Only" one of them is a girl, then there would be a 100% chance that the other was a boy.... but that's not what the question says.

Yes, that is what the question says. One. Not "at least one." One.

DiscoGary wrote:

Youz all dummies wrote:

Correct answer is 1/2 or 50%. Only valid options are GG or GB. Flip GB to BG if you want it doesn't matter because we're only worried about whether one boy exists, not the order of boys and girls. Man you guys are stupid.

Think of it this way.

Ask someone on the street how many kids they have. For those who say 2, ask them if one of them is a girl. For those who say yes the probability that the other kid is a boy is 2/3 because you have already filtered out the BB possibility.

GG

BG

GB

That's the logic and it's very similar to the Monty Hall Paradox where information is inserted into the equation which changes the probability of the outcome for the remaining unknowns.

It certainly isn't obvious.

I feel you but don't view this as the Monty Hall problem. OP asked, KNOWING THAT ONE CHILD IS A GIRL, what is the probability the other child is a boy. Only two options are G and G or G and B.

DiscoGary wrote:

Youz all dummies wrote:

Correct answer is 1/2 or 50%. Only valid options are GG or GB. Flip GB to BG if you want it doesn't matter because we're only worried about whether one boy exists, not the order of boys and girls. Man you guys are stupid.

Think of it this way.

Ask someone on the street how many kids they have. For those who say 2, ask them if one of them is a girl. For those who say yes the probability that the other kid is a boy is 2/3 because you have already filtered out the BB possibility.

GG

BG

GB

That's the logic and it's very similar to the Monty Hall Paradox where information is inserted into the equation which changes the probability of the outcome for the remaining unknowns.

It certainly isn't obvious.

How would twins affect this?

the probability is high that these boys and girls are the same posters that had difficulty proving the earth is round.

Also a high probability, say 2/3 that this thread will continue for another several pages.

Be Patient.... wrote:

Also a high probability, say 2/3 that this thread will continue for another several pages.

Did we ever figure out if the airplane will take off?

Hgrsfh wrote:

Before you have any children the probably is 25% for each of these BB, BG, GB, GG. But after you have one you've eliminated 2 out of the four combinations. Once you know one is a girl, it is already determined that she is either the first OR the second in the combination. She CANNOT be both, so if she is first then options are GG or GB. If she is second then the options are GG or BG.

This guy gets it.

^ correct, pretty much what I mean. We don't know whether Bob is treating the marbles as ordered pairs or not. He might be proceeding by looking at the marble on his left rather than considering both marbles, in which case the result has no bearing at all on the other marble's color.

That's why the MH problem should include a step where the contestant eliminates one element of a non-ordered set of possibilities. I first heard it as a card trick where three cards are dealt face down, two red, one black. The contestant points to a card, then the dealer flips over one of the remaining cards (one or both of which are red) leading to a 2/3 probability that the remaining card is black. With 3 cards it's easy to test and see how this works.

But if it's the dealer selecting the first card, that selection can't be considered random as it is with the contestant. For example, suppose the dealer selects by first making a 50/50 choice between red and black, and if red, another 50/50 choice between high red and low red. And when revealing one of two red cards in the second step, selects 50/50 there too. These independent outcomes (selected card, revealed card, other card) can happen:

B-RH-RL (p = .5 x .5 = .25)

B-RL-RH (p = .5 x .5 = .25)

RL - RH - B (p = .25 x 1 = .25)

RH - RL - B (p = .25)

so that, at the second stage, the probability of the 3rd card being black is .25 + .25 = .50. Even if the contestant is unaware of this, it affects the outcome over many iterations. That can be tested with three cards and a coin, or writing some code to save time.