Riddles

[Fort Fun Indiana]

In that case, BG and GB is also one permutation

False. You are confusing permutations and combinations again.

 

[MAGAman]

So it's really

BG
GG
GG
GB

Wrong. GG is one permutation, no matter whether you meet the younger or older sibling.

Then the chance of
BG
and
GB
are each 1/2 as likely as
GG

Either way, it's 50/50

 

[Toddsterpatriot]

In that case, BG and GB is also one permutation

False. You are confusing permutations and combinations again.

You are confusing permutations and combinations again.

You did, when you said, "GG is one permutation"

With no birth order, there are two combinations
1) she has a brother
2) she has a sister.

With birth order, there are four permutations,
1) she has an older brother
2) she has a younger brother
3) she has an older sister
4) she has an younger sister

Either way, there is a 50% chance she has a sister.

 

[Fort Fun Indiana]

Maybe this will help. Back to the card riddle.

Two cards are drawn from the infinite deck. They are laid in front of you. The dealer tells you that he knows at least one of the cards is red (he has xray vision). What are the odds both cards are red? 3:1. The probability is 1/3.

When you have two cards laid in front of you, you remove one third of the sample space by choosing one and flipping it to see it is red. You have now ordered your choice. If you do not know which card was flipped, the sample space is not reduced.

When reaching under the cup to grab a quarter or meeting the girl, you have not ordered your choice. When you dont know which card has been flipped, but only know at least one of the two cards is red, you have not ordered your choice.

 

[Fort Fun Indiana]

l"GG is one permutation"

Correct, it is. Describing the girl you have met to be the older or the younger describes the state of the girl. It does not describe unique permutations. It is just two ordered choices of whether the girl you met is the older or younger of the one GG permutation. And, to maintain the absolute fact that the sample space contains three equally likely permutations (BG, GB, GG) your descriptions of the girl's state each have probabilities that, taken together, sum to the probability of the permutation GG. There is no getting around that.

 

[Fort Fun Indiana]

Then the chance of
BG
and
GB
are each 1/2 as likely as
GG

Which is obviously absurd. You would then have to say that you are equally likely to have mixed sublings as you are to have two girls, should you have two children. Clearly you see that is false., assuming each birth has a 50/50 chance for boy or girl.

Surely you agree that, should you flip a coin twice, you are twice as likely to end up with one head and one tail (in any order) than you are two tails.

 

[Toddsterpatriot]

l"GG is one permutation"

Correct, it is. Describing the girl you have met to be the older or the younger describes the state of the girl. It does not describe unique permutations. It is just two ordered choices of whether the girl you met is the older or younger of the one GG permutation. And, to maintain the absolute fact that the sample space contains three equally likely permutations (BG, GB, GG, ) your description of the girl's state each have probabilities that, taken together, sum to the probability of the permutation GG. There is no getting around that.

It does not describe unique permutations.

She's older or she's younger. Unique permutations.

the sample space contains three equally likely permutations (BG, GB, GG, )

If you don't care about birth order for two girls, why do you use two permutations with a brother?

 

[Fort Fun Indiana]

She's older or she's younger. Unique permutations.

False. Only one permutation exists: GG.

Maybe it would help you if we added (unnecessary) subscripts: x1x2, x1 is the older sibling.

Whether the girl you meet is G1 or G2 does not have any bearing on the fact that G1G2 is a unique permutation. Same for B1G2 and G1B2. These are your three possible permutations. There are no others. There is no "G2G1" permutation. Your choice has not been ordered.

And, these three permutations, which are the entire sample space:

B1G2
G1B2
G1G2

...are equally likely. That is a fact that you cannot get around. Whether you meet the older or younger sibling (you dont know which, so the only information you have is "G"), this fact remains. Do you agree?

 

[Toddsterpatriot]

She's older or she's younger. Unique permutations.

False. Only one permutation exists: GG.

Maybe it would help you if we added (unnecessary) subscripts: x1x2, x1 is the older sibling.

Whether the girl you meet is G1 or G2 does not have any bearing on the fact that G1G2 is a unique permutation. Same for B1G2 and G1B2. These are your three possible permutations. There are no others. There is no "G2G1" permutation. Your choice has not been ordered.

And, these three permutations, which are the entire sample space:

B1G2
G1B2
G1G2

...are equally likely. That is a fact that you cannot get around. Whether you meet the older or younger sibling (you dont know which, so the only information you have is "G"), this fact remains. Do you agree?

Only one permutation exists: GG.

How can you have a permutation which disregards order?

And, these three permutations, which are the entire sample space:​

​

B1G2​

G1B2​

G1G2​

​

Nah.​

Every permutation has to include the girl(G) we met in post #1.

That's why this works.......

Gb
bG
Gs
sG


​

 

[Fort Fun Indiana]

Another reiteration:

So, let's say two siblngs are in a house with 2 rooms, a hallway, and one front door. This is all the information you have.

If you go inside and open one of the doors to find a girl, what is the probability that there is a girl in the other room?

50%.

Now, instead, you wait out front, one of the siblings is sent outside, and it is a girl. Now, what is the probability that the remaining sibling in the house is a girl?

1/3

 

[Fort Fun Indiana]

B1G2 G1B2 G1G2

Nah.
Every permutation has to include the girl(G) we met in post #1.​

Which all of those do. Who did you meet... G3? Doesn't matter if you met G1 or G100.... as far as you know, you only met G.

And no, that weird annotation doesn't "work" in that the four possibilities you listed are not equally likely. Do you get that?

 

[Toddsterpatriot]

B1G2 G1B2 G1G2

Nah.
Every permutation has to include the girl(G) we met in post #1.​

Which all of those do. Who did you meet... G3? Doesn't matter if you met G1 or G100.... as far as you know, you only met G.

And no, that weird annotation doesn't "work" in that the four possibilities you listed are not equally likely. Do you get that?

Which all of those do.

B1G2 G1B2 G1G2

G1 and G2 are different, they can't both exist with a brother in your "riddle".

And no, that weird annotation doesn't "work"

It works. Every permutation has the girl we met (G)
You can clearly see two permutations with a brother and two with a sister.

the four possibilities you listed are not equally likely.

Prove it.

 

[Fort Fun Indiana]

3rd riddle:

G1 and G2 are different, they can't both exist with a brother in your "riddle".

And, unfortunately for you who must solve the problem, you have no idea which one you have met. So you cannot eliminate anything from the sample space.You only know that at least one sibling is a girl. So, the fact remains that the siblings are twice as likely to be mixed as to be all girls.

 

[Toddsterpatriot]

3rd riddle:

G1 and G2 are different, they can't both exist with a brother in your "riddle".

And, unfortunately for you who must solve the problem, you have no idea which one you have met. So you cannot eliminate anything from the sample space.You only know that at least one sibling is a girl. So, the fact remains that the siblings are twice as likely to be mixed as to be all girls.

And, unfortunately for you who must solve the problem, you have no idea which one you have met.

You're lying.

You meet a girl. She tells you she has one sibling. What are the odds her sibling is also a girl?

 

[luchitociencia]

Permutations GED basic 1, Combinations PE QS 2600 2x 76R, 2x 77R QS Value: 2502 to find if she was a boy or a girl...

And I thought solving riddles was fun and like a poster says "with simple steps". Like Batman movies... sigh*

 

[OldLady]

OldLady

So, the answer is 1/3. (depotoo knew this already)

Why?

There are four equally likely possibilities (in this case, permutations) for two siblings, oldest listed first:

BB
BG
GB
GG

We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:

BG
GB
GG

As these are equally likely, the probability of each possible permutation is now 1/3. So, the probability that the other sibling is also a girl is 1/3.

Had you been given the information of whether the girl we met was the older or younger sibling, the answer would have been 1/2.

Interesting. I would have thought with each child the probability would start from a clean slate. Good question, Indiana. Not a riddle, though.

 

[Toddsterpatriot]

OldLady

So, the answer is 1/3. (depotoo knew this already)

Why?

There are four equally likely possibilities (in this case, permutations) for two siblings, oldest listed first:

BB
BG
GB
GG

We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:

BG
GB
GG

As these are equally likely, the probability of each possible permutation is now 1/3. So, the probability that the other sibling is also a girl is 1/3.

Had you been given the information of whether the girl we met was the older or younger sibling, the answer would have been 1/2.

Interesting. I would have thought with each child the probability would start from a clean slate. Good question, Indiana. Not a riddle, though.

I would have thought with each child the probability would start from a clean slate.

You would be correct.

 

[OldLady]

OldLady

So, the answer is 1/3. (depotoo knew this already)

Why?

There are four equally likely possibilities (in this case, permutations) for two siblings, oldest listed first:

BB
BG
GB
GG

We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:

BG
GB
GG

As these are equally likely, the probability of each possible permutation is now 1/3. So, the probability that the other sibling is also a girl is 1/3.

Had you been given the information of whether the girl we met was the older or younger sibling, the answer would have been 1/2.

Interesting. I would have thought with each child the probability would start from a clean slate. Good question, Indiana. Not a riddle, though.

I would have thought with each child the probability would start from a clean slate.

You would be correct.

You'll never win against these math minds, Todd. I Googled permutation, still have no idea what the diff is between that and the combinations, and you know what? Since I'm not a gambler in Vegas, I can get on with my life perfectly well NOT knowing.
I will remember the answer, 1/3, but this was a cheap trick to pull us into a statistics class.

 

[Toddsterpatriot]

OldLady

So, the answer is 1/3. (depotoo knew this already)

Why?

There are four equally likely possibilities (in this case, permutations) for two siblings, oldest listed first:

BB
BG
GB
GG

We have met one sibling, and she is a girl. So now, there exist three equally likely possibilities:

BG
GB
GG

As these are equally likely, the probability of each possible permutation is now 1/3. So, the probability that the other sibling is also a girl is 1/3.

Had you been given the information of whether the girl we met was the older or younger sibling, the answer would have been 1/2.

Interesting. I would have thought with each child the probability would start from a clean slate. Good question, Indiana. Not a riddle, though.

I would have thought with each child the probability would start from a clean slate.

You would be correct.

You'll never win against these math minds, Todd. I Googled permutation, still have no idea what the diff is between that and the combinations, and you know what? Since I'm not a gambler in Vegas, I can get on with my life perfectly well NOT knowing.
I will remember the answer, 1/3, but this was a cheap trick to pull us into a statistics class.

You'll never win against these math minds, Todd.

FF is far from a math mind.

I will remember the answer, 1/3,

Then you're remembering it wrong.

With no birth order, there are two combinations
1) she has a brother
2) she has a sister.

With birth order, there are four permutations,
1) she has an older brother
2) she has a younger brother
3) she has an older sister
4) she has an younger sister

Either way, there is a 50% chance she has a sister.

 

[K9Buck]

Every time you flip a coin, the odds of it coming up on one particular side is ALWAYS going to be 50-50, regardless of how it came up previously. This is simple logic, folks.