Riddles

[Fort Fun Indiana]

If you've already picked the first sibling, the girl you met, there are only combinations.

You did not "pick" anything in the riddle, though. That changes the riddle. And "picking" actually eliminates half of the BG combinations. So this reduction in the sample space is what leads to your second "pick", in your scenario, being, indeed, 50% likely to be a girl.

 

[Wuwei]

Wuwei

However, if you sent someone else into the house who opened a door (you don't know which door) and shouted out that he sees a girl, the probability, from your perspective, that he finds a girl in the other room as well is 1/3.

Why does the particular door matter? The problem seems to be symmetric.

 

[Toddsterpatriot]

If you've already picked the first sibling, the girl you met, there are only combinations.

You did not "pick" anything in the riddle, though. That changes the riddle. And "picking" actually eliminates half of the BG combinations. So this reduction in the sample space is what leads to your second "pick", in your scenario, being, indeed, 50% likely to be a girl.

You did not "pick" anything in the riddle, though

The first sibling was picked for you in post #1.


You meet a girl. She tells you she has one sibling.

 

[Fort Fun Indiana]

Why does the particular door matter?

Because, when you know which door has been opened, you have "ordered" your choice. Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the second element being behind the second door you open):

BB
BG
GB
GG

When you open the first door and see a girl, you immediately eliminate half the permutations from your sample space: BB and BG.

Now, you are left with a sample space of two equally likely permutations. The probability of finding a girl behind the second door is 1/2.

Now, if it is shouted outside the house to you that somene inside has opened a door and found a girl, but you don't know which door, this changes. You can only eliminate one permutation from this sample space -- BB -- and are left with THREE, equally likely permutations. The choice has not been "ordered". Now, to you, the probability that the other room contains a girl is only 1/3.

 

[cnm]

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

BG
GB
GG

No. We are told a girl has one sibling. The possibilities are

GG
GB

Age is not mentioned in the given information.

 

[cnm]

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.

You framed the question incorrectly for that solution to apply.

 

[luchitociencia]

There are two possibilities, heads or tails. The odds of either turning up are 50-50. Do you dispute this?

I do.

I dropped several times five coins until this happened

 

[Fort Fun Indiana]

So yes, if you want to scam someone and take some of their money, make a 4 card deck from two red and two black cards. Only allow bets when at least one card is red. Pull two cards. Keep them hidden from the mark, but peek at them. If at least one card is red, tell the mark this. Give him even money odds on betting that there are two red cards. Your expectation will be 0.33. If you start with $300, you will end up with $400. Your mark, who started with $300, will end up with $200.

 

[Fort Fun Indiana]

You framed the question incorrectly for that solution to apply.

I don't think so.

 

[Wuwei]

Because, when you know which door has been opened, you have "ordered" your choice. Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the secnd element being behind the second door you open):

BB
BG
GB
GG

When you open the first door and see a girl, you immediately eliminate half the permutations from your sample space: BB and BG.

Now, you are left with a sample space of two equally likely permutations.

Now, if it is shouted out to you that somene inside has opened a door and found a girl, but you don't know which door, this changes. You can only eliminate one permutation from this sample space -- BB -- and are left with THREE, equally likely permutations. The choice has not been "ordered". Now, to you, the probability that the other room contains a girl is only 1/3.

I think my objection is that I don't think it is an ordering problem.

 

[CrusaderFrank]

First one:

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

100%

 

[luchitociencia]

So yes, if you want to scam someone and take some of their money, make a 4 card deck from two red and two black cards. Only allow bets when one card is red. Pull two cards. Keep them hidden from the mark, but peek at them. If at least one card is red, tell the mark this. Give him even money odds on betting that there are two red cards. Your expectation will be 0.25. If you start with $300, you will end up with $375. Your mark, who started with $300, will end up with $225.

Your riddle was OK until you started with your nonsense. Now your riddle sucks.

 

[luchitociencia]

100%

That is correct, because being a girl she will say straight she has an older brother.

 

[Toddsterpatriot]

Why does the particular door matter?

Because, when you know which door has been opened, you have "ordered" your choice. Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the second element being behind the second door you open):

BB
BG
GB
GG

When you open the first door and see a girl, you immediately eliminate half the permutations from your sample space: BB and BG.

Now, you are left with a sample space of two equally likely permutations. The probability of finding a girl behind the second door is 1/2.

Now, if it is shouted outside the house to you that somene inside has opened a door and found a girl, but you don't know which door, this changes. You can only eliminate one permutation from this sample space -- BB -- and are left with THREE, equally likely permutations. The choice has not been "ordered". Now, to you, the probability that the other room contains a girl is only 1/3.

Before opening the door, you were faced with a sample space of 4, equally likely permutations (you can look at them as the first element being behind the first door you open, with the second element being behind the second door you open):

BB
BG
GB
GG

There are two choices for each room.

Room 1, boy (B1) or
Room 1, girl (G1)

Room 2, boy (B2) or
Room 2, girl (G2)

Doesn't matter if your friend opens door one and finds G1 or door two and finds G2.
In the first case, the odds of a girl behind door two is 50%.
In the second case, the odds of a girl behind door one is 50%.​

 

[cnm]

You did not "pick" anything in the riddle, though

The first sibling was picked for you in post #1.


You meet a girl. She tells you she has one sibling.

Yes.

The chances of throwing 7 with two dice are 6 in 36

If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 6. The permutations are rendered moot.

 

[Toddsterpatriot]

You did not "pick" anything in the riddle, though

The first sibling was picked for you in post #1.


You meet a girl. She tells you she has one sibling.

Yes.

The chances of throwing 7 with two dice are 6 in 36

If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 6. The permutations are rendered moot.

If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 36.

No. If the first die shows 6, there is a one-in-six chance the second die will show a 1.

 

[cnm]

You framed the question incorrectly for that solution to apply.

I don't think so.

Stating the gender of one of the children killed it stone dead.

 

[cnm]

You did not "pick" anything in the riddle, though

The first sibling was picked for you in post #1.


You meet a girl. She tells you she has one sibling.

Yes.

The chances of throwing 7 with two dice are 6 in 36

If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 6. The permutations are rendered moot.

If it is given one die shows 6, for example, the chance of totalling 7 with the second die is one in 36.

No. If the first die shows 6, there is a one-in-six chance the second die will show a 1.

Already corrected.

 

[Wuwei]

BB
BG
GB
GG

When you open the first door and see a girl, you immediately eliminate half the permutations from your sample space: BB and BG.

Now, you are left with a sample space of two equally likely permutations. The probability of finding a girl behind the second door is 1/2.

This is the way I look at it. There are just two kids each having one of two properties: B or G.
These are all the possibilities of properties:
BB, BG, GG
BB and GG each have probabilities of 1/4
BG has a probability of 2/4
(Note: BG and GB are two different ways of saying the same thing: "one of each" That assumes ordering is not involved. I will just use the first label, BG).

If either kid is identified as B, the BB and BG possibilities remain.
Probabilities BG = 1/2, BB = 1/4.
If either kid is identified as G, the GG and BG possibilities remain.
Probabilities BG = 1/2, GG = 1/4.

If you normalize the two cases after one kid is identified, you get 2/3 and 1/3.

That is the analysis if ordering is not an issue.
(Maybe.)
.

 

[Fort Fun Indiana]

These are all the possibilities of properties:
BB, BG, GG

Yes, the three combinations. And they have the following probabilities:

BB: 25%
BG: 50%
GG: 25%

Half of the BG combinations are the permutation BG, and half are the permutation GB.

When you "order" your choice by opening a door, you eliminate all the BB combinations and half the BG combinations (specifically, all the BG permutations).

Not knowing which door was opened keeps your choice unordered, and you can only eliminate the BB combinations from your sample space. This leaves you with a 66% chance of a BG combination and a 33% chance of a GG combination.