Riddles

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.

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

BG
GB
GG

Nope.

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


G<-- girl you met ------> sibling B
G<-- girl you met ------> sibling G

Two possibilities.


 
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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.

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

BG
GB
GG

Nope.

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


G<-- girl you met ------> sibling B
G<-- girl you met ------> sibling G

Two possibilities.


No, that is incorrect. There are, in fact, four permutations of two children and two genders, not three. There are three combinations. But four permutations.

BB
BG
GB
GG

...all equally likely. As you can see, the probability of having one girl and one boy is 50%, if you have two children. But, given we have at least one girl, our sample space has been reduced to three permutations.

BG
GB
GG

...all equally likely. Now that we know at least one is a girl, the probability that there is one girl and one boy is 2/3.

1/3 is correct.
 
Same riddle, reiterated:

2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads".

What is the probability the other quarter will also show '"heads"?

1/3.
 
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.

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

BG
GB
GG

Nope.

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


G<-- girl you met ------> sibling B
G<-- girl you met ------> sibling G

Two possibilities.


No, that is incorrect. There are, in fact, four permutations of two children and two genders, not three. There are three combinations. But four permutations.

BB
BG
GB
GG

...all equally likely. As you can see, the probability of having one girl and one boy is 50%, if you have two children. But, given we have at least one girl, our sample space has been reduced to three permutations.

BG
GB
GG

...all equally likely. Now that we know at least one is a girl, the probability that there is one girl and one boy is 2/3.

1/3 is correct.

As you can see, the probability of having one girl and one boy is 50%,

Exactly. One girl......50% chance her sibling is a boy, 50% chance her sibling is a girl.

There are, in fact, four permutations of two children

We aren't looking for permutations of two children, because one was already determined.
 
Same riddle, reiterated:

2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads".

What is the probability the other quarter will also show '"heads"?

1/3.

2 quarters in a cup. You shake the cup and turn it over on a table. You reach under and slide out one quarter; it shows "heads".

What is the probability the other quarter will also show '"heads"?


50%.
 
Exactly. One girl......50% chance her sibling is a boy, 50% chance her sibling is a girl.
False.

We aren't looking for permutations of two children, because one was already determined.
Which eliminates only one of the four, equally likely permutations. You are left with three, equally likely permutations. Yes, the correct method is to consider the permutations.
 
Exactly. One girl......50% chance her sibling is a boy, 50% chance her sibling is a girl.
False.

We aren't looking for permutations of two children, because one was already determined.
Which eliminates only one of the four, equally likely permutations. You are left with three, equally likely permutations. Yes, the correct method is to consider the permutations.

False.

LOL!

Which eliminates only one of the four, equally likely permutations.

Nope. Girl eliminates half the permutations. Only two left.

Yes, the correct method is to consider the permutations.

Sure.

How many permutations are there for the second, unmet sibling?
 
Nope. Girl eliminates half the permutations. Only two left.
Obviously false. The following 3 permutations remain:

BG
GB
GG

You have, again, confused the concepts of "permutation" and "combination". And even then you are still wrong, as there are 3 possible combinations. Knowing one of the two children is a girl would only eliminate one of these 3 combinations, so not half.
 
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Nope. Girl eliminates half the permutations. Only two left.
Obviously false. The following 3 permutations remain:

BG
GB
GG

You have, again, confused the concepts of "permutation" and "combination".

Let's look at the possibilities, taking birth order into account.

We meet a girl, G.

She has either a younger or older brother (b) or a younger or older sister (s)

Gb
bG
Gs
sG

Any possibilities I missed?
 
Nope. Girl eliminates half the permutations. Only two left.
Obviously false. The following 3 permutations remain:

BG
GB
GG

You have, again, confused the concepts of "permutation" and "combination".

Let's look at the possibilities, taking birth order into account.

We meet a girl, G.

She has either a younger or older brother (b) or a younger or older sister (s)

Gb
bG
Gs
sG

Any possibilities I missed?
No need to complicate things and confuse yourself with odd annotation. We will just stick to the annotation used thus far:

BB
BG
GB
GG

These are, indeed, all four possible permutations. Keep in mind: this is your sample space before you know one of the children is a girl.
 
Nope. Girl eliminates half the permutations. Only two left.
Obviously false. The following 3 permutations remain:

BG
GB
GG

You have, again, confused the concepts of "permutation" and "combination".

Let's look at the possibilities, taking birth order into account.

We meet a girl, G.

She has either a younger or older brother (b) or a younger or older sister (s)

Gb
bG
Gs
sG

Any possibilities I missed?
No need to complicate things and confuse yourself with odd annotation. We will just stick to the annotation used thus far:

BB
BG
GB
GG

These are, indeed, all four possible permutations. Keep in mind: this is your sample space before you know one of the children is a girl.

No need to complicate things and confuse yourself with odd annotation.

I'm not confused at all by my clear annotation.

this is your sample space before you know one of the children is a girl.

BB
BG
GB
GG


This is your sample space after you know one of the children is a girl.

BG

GG
 
Another riddle, somewhat similar:

A dealer has an infinite deck of playing cards. He pulls one off the top and places it to his left. He pulls another card off the top and places it to his right. He flips the card to his left. It is red. What is the probability that the card to his right is also red?

50%

Now he pulls two cards from the top of the deck and shuffles the two cards. He places them to his left and right. He flips the one to his left, and it is red. What is the probability the other card is also red?

50%

Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?

33%
 
This is your sample space after you know one of the children is a girl.

BG

GG
False. Those are the possible combinations, not permutations. And they are not equally likely. This is the source of your errors.

You didn't like the real permutations.

Gb
bG
Gs
sG
That annotation gives no extra information and use 3 symbols instead of 2. So no, that is not correct. The third symbol is unnecessary.

And it obviously causes you to confuse yourself, as "Gs" and "sG" are not actually unique permutations. You have listed a permutation twice, which is an error.

If you insist on these being unique data points in the sample space, then you must modify their probability and reduce the probability of each by half. In which case, you will still arrive at the correct answer: 1/3.
 
This is your sample space after you know one of the children is a girl.

BG

GG
False. Those are the possible combinations, not permutations. And they are not equally likely. This is the source of your errors.

You didn't like the real permutations.

Gb
bG
Gs
sG
That annotatiin give no extra information and use 3 symbols instead of 2. So no, that is not correct. The third symbol is unnecessary.

And it obviously causes you to confuse yourself, as "Gs" and "sG" are not actually unique permutations. You have listed a permutation twice, which is an error.

That annotatiin give no extra information

It gave birth order info.

And it obviously causes you to confuse yourself, as "Gs" and "sG" are not actually unique permutations.

They are unique. In the first case, the girl we met (G) is the older sister, in the second case she is the younger.

You have listed a permutation twice, which is an error.

Like you did.....

BB
BG <here
GB <and here
GG
 
Another riddle, somewhat similar:

A dealer has an infinite deck of playing cards. He pulls one off the top and places it to his left. He pulls another card off the top and places it to his right. He flips the card to his left. It is red. What is the probability that the card to his right is also red?

50%

Now he pulls two cards from the top of the deck and shuffles the two cards. He places them to his left and right. He flips the one to his left, and it is red. What is the probability the other card is also red?

50%

Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?

33%

Now he pulls two cards from the deck and, without shuffling them, holds them behind his back. He shows you one of the cards, and it is red. What is the probability the other card is also red?

33%


50%
 

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