Riddles

It gave birth order info.
Which is unnecessary, and causes you to make yet another error. As the probability of these two permutations is 1/2 the probability of "GG", you still arrive at 1/3 as the correct answer and have not actually added any information. You have just forced yourself to multiply by 2 then multiply by 1/2. No net change.
 
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%
Wrong. For all the same reasons. That's why the riddle is called a "paradox"... Even though it actually isn't. When you don't fully understand the problem -- as you clearly do not -- it appears as a paradox.
 
It gave birth order info.
Which is unnecessary, and causes you to make yet another error. As the probability of these two permutations is 1/2 the probability of "GG", you still arrive at 1/3 as the correct answer and have not actually added any information. You have just forced yourself to multiply by 2 then multiply by 1/2. No net change.

Which is unnecessary

You didn't want permutations?
 
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%
Wrong. For all the same reasons. That's why the riddle is called a "paradox"... Even though it actually isn't. When you don't fully understand the problem -- as you clearly do not -- it appears as a paradox.

Your confusion doesn't make these a paradox.
 
Theres 3 doors. One has a prize.

You get to choose a door.

Before it's opened, the host (who knows where the prize is), opens one of the doors that has no prize.

He asks you if youd like to switch to the other door left, before the reveal.

Should you switch?


Yepp, odds on your door = 33%
Odds on winning if you switch = 66%


Most initially intuit that its 50/50. Its not.
 
I came up with approximately 33.6%. There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same. This also may change slightly across different populations.
 
I came up with approximately 33.6%. There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same. This also may change slightly across different populations.
Ha! I will make it clear next time that they are not identical twins.
 
Theres 3 doors. One has a prize.

You get to choose a door.

Before it's opened, the host (who knows where the prize is), opens one of the doors that has no prize.

He asks you if youd like to switch to the other door left, before the reveal.

Should you switch?


Yepp, odds on your door = 33%
Odds on winning if you switch = 66%


Most initially intuit that its 50/50. Its not.
Love it.
 
I came up with approximately 33.6%. There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same. This also may change slightly across different populations.
Ha! I will make it clear next time that they are not identical twins.

I'm not certain, except it might be more clear to also specify whether she is the older or younger sibling, anything that clearly designates which one she is, as opposed to simply a sibling.
 
This is a riddle that needs Android 101 fonts. No other way to find the answer. You can download it from

Android 101 Font | dafont.com

This is a riddle from the 60s when Texas Instrument watches were in the market like crazy. If you are old like me, you might know the answer already, but please give it using the Android 101 fonts.

IMG_1509[1].JPG
 
I came up with approximately 33.6%. There is a .8% chance that her unidentified sibling is an identical twin and this slightly increases the likelihood of the sex being the same. This also may change slightly across different populations.
Ha! I will make it clear next time that they are not identical twins.

I'm not certain, except it might be more clear to also specify whether she is the older or younger sibling, anything that clearly designates which one she is, as opposed to simply a sibling.

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.
 
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.

First, those are not four distinct permutations. Sorry. You were wrong the first time you said it, and you are still wrong.

Furthermore, those four states of the girl you met are not equally likely. This is easily shown by proof by contradiction.

1 and 2 each have a probability of 33.33%, while 3 and 4 each have a probability of 16.67%.
 
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.
No.

If she's older its 50/50

If she's younger its 50/50

BG
GG
 
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.

First, those are not four distinct permutations. Sorry. You were wrong the first time you said it, and you are still wrong.

Furthermore, those four states of the girl you met are not equally likely. This is easily shown by proof by contradiction.

1 and 2 each have a probability of 33.33%, while 3 and 4 each have a probability of 16.67%.

First, those are not four distinct permutations

Why not?

This is easily shown by proof by contradiction.

Okay, prove it.

1 and 2 each have a probability of 33.33%

Your confusion still doesn't prove your claim.
 

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