Giveaway Stitches. WE HAVE A WINNER.
- Thread starter momayo
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Riiiiptide
Senior Member
1. Assuming a uniform distribution between boys and girls at birth i.e. P(boy) = P(girl) = 1/2, then the next child will be a girl is 1/2 since, as far as modern science goes, the events are independent.
2. D, E are definitely liars. E's statement is a contradiction and since E is a liar, and D is dependent on E since E is false, then D's statement must be false too.
Note that this means B's statement is true and C is automatically false and so A is true.
QED
3. There are two possible events, either you pick up the ace of spades before the 2 of spades or you don't. The "don't" i.e. the converse happens with equal probability, thus it is a 50% chance.
4. stuffin
5. For simplicity and without loss of generalization, define a new solar calender system measured on Jane's Birthday where New Year's Eve on which she was only 15 is 12/31/15.
In order for this to work, "last New Year's Eve" must mean the eve before the previous year i.e. 2 years ago.
Then on 12/31/16 she is 16 so on 12/31/17 she is 17 and turning 18 in 5 days say sometime in the early few days of january, for simplicity, say Jan 1st.
So her birthday must be on January 1st. Since no year was specified on the normal solar calender, there was indeed no loss of generalization in the linear year change. QED
2. D, E are definitely liars. E's statement is a contradiction and since E is a liar, and D is dependent on E since E is false, then D's statement must be false too.
Note that this means B's statement is true and C is automatically false and so A is true.
QED
3. There are two possible events, either you pick up the ace of spades before the 2 of spades or you don't. The "don't" i.e. the converse happens with equal probability, thus it is a 50% chance.
4. stuffin
5. For simplicity and without loss of generalization, define a new solar calender system measured on Jane's Birthday where New Year's Eve on which she was only 15 is 12/31/15.
In order for this to work, "last New Year's Eve" must mean the eve before the previous year i.e. 2 years ago.
Then on 12/31/16 she is 16 so on 12/31/17 she is 17 and turning 18 in 5 days say sometime in the early few days of january, for simplicity, say Jan 1st.
So her birthday must be on January 1st. Since no year was specified on the normal solar calender, there was indeed no loss of generalization in the linear year change. QED
Riiiiptide
Senior Member
1. Assuming a uniform distribution between boys and girls at birth i.e. P(boy) = P(girl) = 1/2, then the next child will be a girl is 1/2 since, as far as modern science goes, the events are independent.
2. D, E are definitely liars. E's statement is a contradiction and since E is a liar, and D is dependent on E since E is false, then D's statement must be false too.
Note that this means B's statement is true and C is automatically false and so A is true.
QED
3. There are two possible events, either you pick up the ace of spades before the 2 of spades or you don't. The "don't" i.e. the converse happens with equal probability, thus it is a 50% chance.
4. stuffin
5. For simplicity and without loss of generalization, define a new solar calender system measured on Jane's Birthday where New Year's Eve on which she was only 15 is 12/31/15.
In order for this to work, "last New Year's Eve" must mean the eve before the previous year i.e. 2 years ago.
Then on 12/31/16 she is 16 so on 12/31/17 she is 17 and turning 18 in 5 days say sometime in the early few days of january, for simplicity, say Jan 1st.
So her birthday must be on January 1st. Since no year was specified on the normal solar calender, there was indeed no loss of generalization in the linear year change. QED
2. D, E are definitely liars. E's statement is a contradiction and since E is a liar, and D is dependent on E since E is false, then D's statement must be false too.
Note that this means B's statement is true and C is automatically false and so A is true.
QED
3. There are two possible events, either you pick up the ace of spades before the 2 of spades or you don't. The "don't" i.e. the converse happens with equal probability, thus it is a 50% chance.
4. stuffin
5. For simplicity and without loss of generalization, define a new solar calender system measured on Jane's Birthday where New Year's Eve on which she was only 15 is 12/31/15.
In order for this to work, "last New Year's Eve" must mean the eve before the previous year i.e. 2 years ago.
Then on 12/31/16 she is 16 so on 12/31/17 she is 17 and turning 18 in 5 days say sometime in the early few days of january, for simplicity, say Jan 1st.
So her birthday must be on January 1st. Since no year was specified on the normal solar calender, there was indeed no loss of generalization in the linear year change. QED
- - - Post Merge - - -
1. Assuming a uniform distribution between boys and girls at birth i.e. P(boy) = P(girl) = 1/2, then the next child will be a girl is 1/2 since, as far as modern science goes, the events are independent.
2. D, E are definitely liars. E's statement is a contradiction and since E is a liar, and D is dependent on E since E is false, then D's statement must be false too.
Note that this means B's statement is true and C is automatically false and so A is true.
QED
3. There are two possible events, either you pick up the ace of spades before the 2 of spades or you don't. The "don't" i.e. the inverse happens with equal probability, thus it is a 50% chance.
4. stuffin
5. For simplicity and without loss of generalization, define a new solar calender system measured on Jane's Birthday where New Year's Eve on which she was only 15 is 12/31/15.
In order for this to work, "last New Year's Eve" must mean the eve before the previous year i.e. 2 years ago.
Then on 12/31/16 she is 16 so on 12/31/17 she is 17 and turning 18 in 5 days say sometime in the early few days of january, for simplicity, say Jan 1st.
So her birthday must be on January 1st. Since no year was specified on the normal solar calender, there was indeed no loss of generalization in the linear year change. QED
2. D, E are definitely liars. E's statement is a contradiction and since E is a liar, and D is dependent on E since E is false, then D's statement must be false too.
Note that this means B's statement is true and C is automatically false and so A is true.
QED
3. There are two possible events, either you pick up the ace of spades before the 2 of spades or you don't. The "don't" i.e. the inverse happens with equal probability, thus it is a 50% chance.
4. stuffin
5. For simplicity and without loss of generalization, define a new solar calender system measured on Jane's Birthday where New Year's Eve on which she was only 15 is 12/31/15.
In order for this to work, "last New Year's Eve" must mean the eve before the previous year i.e. 2 years ago.
Then on 12/31/16 she is 16 so on 12/31/17 she is 17 and turning 18 in 5 days say sometime in the early few days of january, for simplicity, say Jan 1st.
So her birthday must be on January 1st. Since no year was specified on the normal solar calender, there was indeed no loss of generalization in the linear year change. QED
invertedpolkadots
Senior Member
Thank you so much!
I was devastated after I edited it and realized what I did..
Let's try again!
Hopefully this is right this time. Who knows! ><
I was devastated after I edited it and realized what I did..
Let's try again!
1. Sticking with 50%
2. Gonna stick to my guns with A and B telling the truth and the rest lying.
3. If I actually apply probability it would be 50% I think..
4. What is Stitches' default catchphrase? stuffin
5. I can't believe I didn't realize this one when I'm born SO close to that date. It's the leap year day! Soooo February 29th!
2. Gonna stick to my guns with A and B telling the truth and the rest lying.
3. If I actually apply probability it would be 50% I think..
4. What is Stitches' default catchphrase? stuffin
5. I can't believe I didn't realize this one when I'm born SO close to that date. It's the leap year day! Soooo February 29th!
Hopefully this is right this time. Who knows! ><
invertedpolkadots
Senior Member
1. Sticking with 50%
2. Gonna stick to my guns with A and B telling the truth and the rest lying.
3. If I actually apply probability it would be 50% I think..
4. What is Stitches' default catchphrase? stuffin
5. Ugh I'm gonna change it to January 1st again :/ the more I think about it the leap year thing only works as a clever riddle, not actually on paper...
2. Gonna stick to my guns with A and B telling the truth and the rest lying.
3. If I actually apply probability it would be 50% I think..
4. What is Stitches' default catchphrase? stuffin
5. Ugh I'm gonna change it to January 1st again :/ the more I think about it the leap year thing only works as a clever riddle, not actually on paper...
My last attempt I guess :/ started second guessing myself like crazy. Sorry Stitches lol!
Geebusjas
Senior Member
1. You have two children. Knowing that one of them is a girl, what is the probability that the other child is also a girl?
50%
2. You meet five people (let's call them A, B, C, D, and E) on the road. You know for sure that at least one of them is a liar.
A says: "C and D are definitely lying."
B mutters: "I'm not sure...but I think A or D are big fat liars."
C scoffs: "A is so full of it! He's the only liar here."
D sighs: "At least B and E are both telling the truth."
E smiles: "Everybody's a liar."
Who are the liars, and who are telling the truth?
Well D is a liar because he says both B and E are telling the truth, which means that everybody would not be lying.
C is a liar because he said A is the only liar
E is a liar because if every person was a liar then he would be a liar.
That leaves A with the only true statement, and so since A is telling the truth B is lying.
3. 52 cards in a deck have been randomly thrown, face-down, on a flat surface. You are trying to pick up all cards one by one. What is the probability that you will pick up an ace of spades before the two of spades? 1/52
4. What is Stitches' default catchphrase? stuffin
5. Jane will turn 18 years old in five days. On New Year's Eve last year, however, she was only 15. When is her birthday?
Jan 5th...?
50%
2. You meet five people (let's call them A, B, C, D, and E) on the road. You know for sure that at least one of them is a liar.
A says: "C and D are definitely lying."
B mutters: "I'm not sure...but I think A or D are big fat liars."
C scoffs: "A is so full of it! He's the only liar here."
D sighs: "At least B and E are both telling the truth."
E smiles: "Everybody's a liar."
Who are the liars, and who are telling the truth?
Well D is a liar because he says both B and E are telling the truth, which means that everybody would not be lying.
C is a liar because he said A is the only liar
E is a liar because if every person was a liar then he would be a liar.
That leaves A with the only true statement, and so since A is telling the truth B is lying.
3. 52 cards in a deck have been randomly thrown, face-down, on a flat surface. You are trying to pick up all cards one by one. What is the probability that you will pick up an ace of spades before the two of spades? 1/52
4. What is Stitches' default catchphrase? stuffin
5. Jane will turn 18 years old in five days. On New Year's Eve last year, however, she was only 15. When is her birthday?
Jan 5th...?
- - - Post Merge - - -
1. You have two children. Knowing that one of them is a girl, what is the probability that the other child is also a girl?
50%
2. You meet five people (let's call them A, B, C, D, and E) on the road. You know for sure that at least one of them is a liar.
A says: "C and D are definitely lying."
B mutters: "I'm not sure...but I think A or D are big fat liars."
C scoffs: "A is so full of it! He's the only liar here."
D sighs: "At least B and E are both telling the truth."
E smiles: "Everybody's a liar."
Who are the liars, and who are telling the truth?
Well D is a liar because he says both B and E are telling the truth, which means that everybody would not be lying.
C is a liar because he said A is the only liar
E is a liar because if every person was a liar then he would be a liar.
That leaves A and B with the only true statements.
(B isnt a liar because opinion is not fact)
3. 52 cards in a deck have been randomly thrown, face-down, on a flat surface. You are trying to pick up all cards one by one. What is the probability that you will pick up an ace of spades before the two of spades? 1/52
4. What is Stitches' default catchphrase? stuffin
5. Jane will turn 18 years old in five days. On New Year's Eve last year, however, she was only 15. When is her birthday?
Jan 5th...?
50%
2. You meet five people (let's call them A, B, C, D, and E) on the road. You know for sure that at least one of them is a liar.
A says: "C and D are definitely lying."
B mutters: "I'm not sure...but I think A or D are big fat liars."
C scoffs: "A is so full of it! He's the only liar here."
D sighs: "At least B and E are both telling the truth."
E smiles: "Everybody's a liar."
Who are the liars, and who are telling the truth?
Well D is a liar because he says both B and E are telling the truth, which means that everybody would not be lying.
C is a liar because he said A is the only liar
E is a liar because if every person was a liar then he would be a liar.
That leaves A and B with the only true statements.
(B isnt a liar because opinion is not fact)
3. 52 cards in a deck have been randomly thrown, face-down, on a flat surface. You are trying to pick up all cards one by one. What is the probability that you will pick up an ace of spades before the two of spades? 1/52
4. What is Stitches' default catchphrase? stuffin
5. Jane will turn 18 years old in five days. On New Year's Eve last year, however, she was only 15. When is her birthday?
Jan 5th...?
ayeeprill
The Hash Slinging Slasher
oooh i love brainteasers and would love Stitches! My kinda giveaway!
1. The probability would always be 50%. The gender of your previous child is a nonfactor.
2. E has to be lying. If everybody is lying, then that means E would be telling the truth and therefore not everyone would be lying. If E is lying, then D is lying too. From there, you can work out that A and B are telling the truth, but C, D, and E are lying.
3. This seems like it could go one of two ways. But I'm going to say 50% as in-disregard the other fifty cards, you either draw the ace first or you draw the two first.
4. stuffin'
5. January 2?
2. E has to be lying. If everybody is lying, then that means E would be telling the truth and therefore not everyone would be lying. If E is lying, then D is lying too. From there, you can work out that A and B are telling the truth, but C, D, and E are lying.
3. This seems like it could go one of two ways. But I'm going to say 50% as in-disregard the other fifty cards, you either draw the ace first or you draw the two first.
4. stuffin'
5. January 2?
momayo
*sheep noises*
Thanks for everyone who's responded so far! Most of you actually got most of the questions correct already 
I'm moving the deadline a few hours earlier since I just realized most of you guys will be asleep at that time ^^; Time zone shenanigans and all. Answers will be revealed 10 pm, Thursday CST time.
Here's a bit of a hint. There's only one question so far that nobody has gotten right yet. Which question? Why, it just so happens to be the one question that looks the most straightforward...
I'm moving the deadline a few hours earlier since I just realized most of you guys will be asleep at that time ^^; Time zone shenanigans and all. Answers will be revealed 10 pm, Thursday CST time.
Here's a bit of a hint. There's only one question so far that nobody has gotten right yet. Which question? Why, it just so happens to be the one question that looks the most straightforward...
Kikiyama
Senior Member
Trying again!
1. 3:1
It can't be 50/50 because females are dominant in probability. So because the chances it will be Female are higher, 50% is inaccurate. 3:1 (in the female favor)
Would argue it's unanswerable because there are physical factors involved in the probability of a newborn being male or female, but that would be relevant to location and other things which aren't mentioned in the question, so the answer lies in the basic probability regarding sexes as the question is stated.
2. A and B are telling the truth, and C, D, and E are liars
3. 1/51
4. stuffin
5. February 29
It can't be 50/50 because females are dominant in probability. So because the chances it will be Female are higher, 50% is inaccurate. 3:1 (in the female favor)
Would argue it's unanswerable because there are physical factors involved in the probability of a newborn being male or female, but that would be relevant to location and other things which aren't mentioned in the question, so the answer lies in the basic probability regarding sexes as the question is stated.
2. A and B are telling the truth, and C, D, and E are liars
3. 1/51
4. stuffin
5. February 29
invertedpolkadots
Senior Member
Hmm, the most straight forward?
ONE MORE TIME it's worth it for Stitches!
ONE MORE TIME it's worth it for Stitches!
1. There are four possibilities with two children. boy/boy girl/boy boy/girl girl/girl
But the question doesn't state if it's the first or second child that female.. It just says other.
My bio teacher used to get us with this all the time and the answer would be 1/3 because you eliminate the boy/boy option,
And the chance would be 1/3 because you have to factor in the two types of boy and girl set up
But this question doesn't call for first or second, just other so I'm kinda confused.
It might still be 50%.. But this is the only one tha seemed 'straightforward' to me.
2. Gonna stick to my guns with A and B telling the truth and the rest lying.
3. If I actually apply probability it would be 50% I think..
4. stuffin
5. Ugh I'm gonna change it to January 1st again :/ the more I think about it the leap year thing only works as a clever riddle, not actually on paper...
But the question doesn't state if it's the first or second child that female.. It just says other.
My bio teacher used to get us with this all the time and the answer would be 1/3 because you eliminate the boy/boy option,
And the chance would be 1/3 because you have to factor in the two types of boy and girl set up
But this question doesn't call for first or second, just other so I'm kinda confused.
It might still be 50%.. But this is the only one tha seemed 'straightforward' to me.
2. Gonna stick to my guns with A and B telling the truth and the rest lying.
3. If I actually apply probability it would be 50% I think..
4. stuffin
5. Ugh I'm gonna change it to January 1st again :/ the more I think about it the leap year thing only works as a clever riddle, not actually on paper...
Kanapachi
Senior Member
1. Knowing that there are only two children, and we know the gender of one of them already, the probability of the other being a girl would be 1/2.
2. A and B are telling the truth, the rest are liars.
3. 50% 4. stuffin'
5. Possible options include January 1st, 2nd, 3rd, 4th and 5th.
2. A and B are telling the truth, the rest are liars.
3. 50% 4. stuffin'
5. Possible options include January 1st, 2nd, 3rd, 4th and 5th.
Last edited:
momayo
*sheep noises*
........
........
Congratulations!
invertedpolkadots got it right!!!!
If you're curious about the answers, here are the solution:
1. The first question is the trickiest one. But invertedpolkadots has the right solution.
There are only four possible combinations for any two kids: BB, GB, BG, and GG.
The only possible combinations that have a girl in it are GB, BG, and GG.
The only combination that has two girls in it is GG. Which is 1 combination out of the possible 3.
Now, why is this question misleading? Because the question never mentioned which child we're talking about. The only thing we know is that one of them is a girl. If the question was phrased, "If the first child was a girl, what is the probability that the second child is also a girl?" then the answer would indeed be 50%. But since you have to factor in the possibility the girl could be the first or the second child, it gets a bit complicated.
Another way to look at it is, "What is the probability that both children are girls given that there is zero possibility that both of them are boys?"
2. A and B are telling the truth. C, D and E are lying. The only tricky part here is picking up their 'and/or'-type of statements.
3. 50%. It doesn't matter how many other cards there are! The probability is the same whether or not there are 52, 100, or 1000 cards in the table, as long as you are only concerned with those two cards.
4. Stuffin'! Sorry if this was a sneaky question, I couldn't resist it ^^;
5. Her birthday lies on any of the dates between January 1 to January 5. The only possible birthday she can have is up to 5 days after New Year's Eve. I kinda messed this question up a bit so I'm allowing a range of answers for this one.
There are only four possible combinations for any two kids: BB, GB, BG, and GG.
The only possible combinations that have a girl in it are GB, BG, and GG.
The only combination that has two girls in it is GG. Which is 1 combination out of the possible 3.
Now, why is this question misleading? Because the question never mentioned which child we're talking about. The only thing we know is that one of them is a girl. If the question was phrased, "If the first child was a girl, what is the probability that the second child is also a girl?" then the answer would indeed be 50%. But since you have to factor in the possibility the girl could be the first or the second child, it gets a bit complicated.
Another way to look at it is, "What is the probability that both children are girls given that there is zero possibility that both of them are boys?"
2. A and B are telling the truth. C, D and E are lying. The only tricky part here is picking up their 'and/or'-type of statements.
3. 50%. It doesn't matter how many other cards there are! The probability is the same whether or not there are 52, 100, or 1000 cards in the table, as long as you are only concerned with those two cards.
4. Stuffin'! Sorry if this was a sneaky question, I couldn't resist it ^^;
5. Her birthday lies on any of the dates between January 1 to January 5. The only possible birthday she can have is up to 5 days after New Year's Eve. I kinda messed this question up a bit so I'm allowing a range of answers for this one.
momayo
*sheep noises*
Thank you so much to everyone who gave an answer. I've always wanted to do a giveaway like this, so this was my great big chance.
Another interesting probability puzzle is the "Monty Hall" puzzle! http://en.wikipedia.org/wiki/Monty_Hall_problem
Another interesting probability puzzle is the "Monty Hall" puzzle! http://en.wikipedia.org/wiki/Monty_Hall_problem
