Riddles

[Fort Fun Indiana]

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

Knowing at least one child is a girl is the key to the riddle. If you think that, given only the information at least one child is a girl yields a 50% chance of the other child also being a girl, you are wrong. It is 33%.

 

[Fort Fun Indiana]

You meet a girl.

That is not you "picking" anything, any more than the dealer pulling one card from behind his back and showing it to you is "picking".

Now, if you had the two siblngs in two containers in front of you, then picked one and "met" a girl, that would be different.

 

[Fort Fun Indiana]

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 you roll a die and get 6, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider. There is only a sample space of 6 equally likely elements: 1-6. And yes, the odds of then rolling a total of 7 are one in 6.

Thats the riddle, really. While it must be accurately solved by counting permutations, this is unnecessary work, when you "open one of two doors and find a girl". At this point, you have ordered your two choices, and now all that remains is a sample space of two, equally likely elements: boy or girl.

Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?

 

[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.

Anyone? Anyone up for a game of cards? I'll deal. I'll give you 3:2, instead of even money.

 

[Fort Fun Indiana]

Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?

Worth noting:

In the sibling scenario, the odds got worse, when all the information you had was that at least one is a girl, compared to opening a door to find one (a dice roll).

In this scenario, you end up with better odds if you don't roll a die and instead are divined that at least one of two dice rolled is a 6. 11:2 are betters odds than 6:1.

Let there be six genders instead of two in the sibling riddle, and you will get the same results as the above dice riddle.

 

[Toddsterpatriot]

You meet a girl.

That is not you "picking" anything, any more than the dealer pulling one card from behind his back and showing it to you is "picking".

Now, if you had the two siblngs in two containers in front of you, then picked one and "met" a girl, that would be different.

That is not you "picking" anything,

She can't be a boy, the choice is made.

That eliminates permutations B1-B2 and B1-G2.
Only two left, G1-B2 and G1-G2.

 

[Toddsterpatriot]

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 you roll a die and get 6, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider. There is only a sample space of 6 equally likely elements: 1-6. And yes, the odds of then rolling a total of 7 are one in 6.

Thats the riddle, really. While it must be accurately solved by counting permutations, this is unnecessary work, when you "open one of two doors and find a girl". At this point, you have ordered your two choices, and now all that remains is a sample space of two, equally likely elements: boy or girl.

Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?

If you roll a die and get 6, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider. There is only a sample space of 6 equally likely elements: 1-6.

If you meet the first sibling and get a girl, you are right that the permutations are then "moot", because there no longer are any permutations or combinations to consider. There is only a sample space of 2 equally likely elements: boy or girl.

 

[Fort Fun Indiana]

If you meet the first sibling and get a girl, you are right that the permutations are then "moot"

False. Because you dont know which sibling you have met. Just as when the dealer holds the two cards behind his back, then hands you one red card. You never got to choose. You didn't order your choice. If the dealer draws two cards (from an infinite deck, for the hall monitors), holds them behind his back, then hands you one red card, the probability that the other card is also red is 1/3.

Now you have retreated to a semantic argument. So it appears that you DO understand that, if you are only given the information that at least one sibling is a girl, the probability of the other also being a girl is 1/3. Else you wouldn't be focused on calling this a "pick", or a card flip, using the word "meet"..

So, good on you, you do understand. However, "meeting" the girl doesn't mean you have ordered your choice. As I will show in the next post.

 

[Fort Fun Indiana]

Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to meet her and shake her hand and exchange pleasantries. They don't just meet, they go on to get married and have 11 kids.

What is the probability that there remains a girl behind the door?

Correct answer: 1/3.

 

[Fort Fun Indiana]

Now, Toddsterpatriot walks into a house with two rooms and a hallway. Two siblings are in the house, one in each room. He opens one door. He meets a girl and she spurns him; she plays for the wrong team, or something. What is the probability a girl is behind the other door?

1/2

 

[Fort Fun Indiana]

Now consider:

If I roll two dice in a cup and look at them and tell you at least one is a 6 (thus "giving" you the information that one die shows 6), what are the odds the dice total 7? You have to count permutations.

The odds are 11:2. Do you know why?

 

[Toddsterpatriot]

If you meet the first sibling and get a girl, you are right that the permutations are then "moot"

False. Because you dont know which sibling you have met. Just as when the dealer holds the two cards behind his back, then hands you one red card. You never got to choose. You didn't order your choice. If the dealer draws two cards (from an infinite deck, for the hall monitors), holds them behind his back, then hands you one red card, the probability that the other card is also red is 1/3.

Now you have retreated to a semantic argument. So it appears that you DO understand that, if you are only given the information that at least one sibling is a girl, the probability of the other also being a girl is 1/3. Else you wouldn't be focused on calling this a "pick", or a card flip, using the word "meet"..

So, good on you, you do understand. However, "meeting" the girl doesn't mean you have ordered your choice. As I will show in the next post.

False. Because you dont know which sibling you have met.

I met a girl sibling.

You meet a girl.

What are the other possible siblings?

However, "meeting" the girl doesn't mean you have ordered your choice.

I don't have to order my choice, because I didn't make one.

 

[Fort Fun Indiana]

Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to meet her and shake her hand and exchange pleasantries. They don't just meet, they go on to get married and have 11 kids.

What is the probability that there remains a girl behind the door?

Correct answer: 1/3.

Now, Toddsterpatriot walks into a house with two rooms and a hallway. Two siblings are in the house, one in each room. He opens one door. He meets a girl and she spurns him; she plays for the wrong team, or something. What is the probability a girl is behind the other door?

1/2

 

[Toddsterpatriot]

Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to meet her and shake her hand and exchange pleasantries. They don't just meet, they go on to get married and have 11 kids.

What is the probability that there remains a girl behind the door?

Correct answer: 1/3.

Two siblings are in a room behind a door,

Excellent!
Let's look at the possibilities.
Sibling one is either B1 (boy) or G1 (girl). Sibling two is either B2 (boy) or G2 (girl)

One is sent out, and it is a girl.

Excellent!
Let's look at the possibilities.
She's either G1 or G2, the only two possible girl siblings.

If she's G1, the remaining sibling is either B2 (boy) or G2 (girl).

If she's G2, the remaining sibling is either B1 (boy) or G1 (girl).
​

Correct answer: 1/3.​

There aren't 3 siblings remaining behind the door, there aren't 3 choices. There aren't 3 possibilities.

One sibling left, 2 possibilities. Still 1/2.​

 

[Fort Fun Indiana]

If she's G1, the remaining sibling is either B2 (boy) or G2 (girl).

If she's G2, the remaining sibling is either B1 (boy) or G1 (girl).

Yep, three equally possible permutations.

BG
GB
GG

You're just not getting this.

 

[Wuwei]

Two siblings are in a room behind a door, with Toddsterpatriot waiting outside . One is sent out, and it is a girl. Toddsterpatriot gets to meet her and shake her hand and exchange pleasantries. They don't just meet, they go on to get married and have 11 kids.

What is the probability that there remains a girl behind the door?

Correct answer: 1/3.

Probability = 0. It is neither a boy or girl. It's either a really ticked off woman or a man.... Waiting in a room for a couple of decades is going to do that to you while Todd is out having fun overpopulating the world with your sister who never writes to you.

Now, Toddsterpatriot walks into a house with two rooms and a hallway. Two siblings are in the house, one in each room. He opens one door. He meets a girl and she spurns him; she plays for the wrong team, or something. What is the probability a girl is behind the other door?

1/2

But seriously folks, I think we are overthinking it. I don't think ordering plays a role.

At the start of this puzzle, the premise is that the genders of the two kids are statistically independent, namely 50:50 for each. Suppose you went through life meeting thousands of parents with their two kids visible or known. You will compile these statistics.
(Notation: OE logically means BG ∪ GB or One of Each. The order doesn't matter)

BB 25%
OE 50%
GG 25%

So, if for the first time you encounter one kid, a G, and you don't know the gender of the other kid, There can be many circumstances (The first kid is discovered behind a door, walking in the street..., the second kid is locked in a room, or in jail...) You will eliminate the BB possibility and resort to your statistics. You will see that the OE possibility is twice the GG possibility. That means the chances of a second G is half that for OE. So the chance is 1/3 that the second is a G and 2/3 that its an OE (in this case, a boy).

I think.
.

 

[Fort Fun Indiana]

But seriously folks, I think we are overthinking it. I don't think ordering plays a role.

And you are wrong. By rolling a die or opening a door, you have ordered your choice of the two elements which constitute the permutation. It's a simple mathematical fact that you cannot get around.

I think tackling the dice riddle i just posted might give you some clarity on this. Note the difference in probability when you roll a die and get 6, versus the probability when someone tells you that at least one of two rolled dice shows a 6.

 

[Dogmaphobe]

Damn, and here I thought Fort Fun was a right, proper lefty. Is he actually implying there are just two genders?

Shouldn't we, like, try to get him fired from his job or something?

 

[Fort Fun Indiana]

Damn, and here I thought Fort Fun was a right, proper lefty. Is he actually implying there are just two genders?

Shouldn't we, like, try to get him fired from his job or something?

For the sake of the math, there are two, equally likely genders. If you feel left out, I can add a third, just to include you. Androgynous, or hermaphrodite? You tell me.

 

[Dogmaphobe]

Damn, and here I thought Fort Fun was a right, proper lefty. Is he actually implying there are just two genders?

Shouldn't we, like, try to get him fired from his job or something?

For the sake of the math, there are two, equally likely genders. If you feel left out, I can add a third, just to include you. Androgynous, or hermaphrodite? You tell me.

I'm selecting hermaphrodite.

It makes things a lot more fun when you tell me to go fuck myself.