Sorry I haven't been in here in a while...
Is the answer something like this?
The expert had two normal sized cups, and the challenger had something smaller, maybe a shot glass of orange juice. He couldn't touch the experts glass at all until he set his down. So, the expert drank his first glass and put it over the challengers. Now, the challenger can't get to his glass because he would be touching the experts glass which is against the rules. So, the expert finished his second glass...
EDIT: I didn't read ronozzy_82's entire post because I got excited that I thought I had the right answer and just decided to post it; just letting you know that I didn't mean to make it appear that I cheated or anything.
Full Version: Riddles And Jokes
mm....I'd have to say a sieve. Is that right?
Eh -- litle bit confused, I can only agree and say seive or a strainer.
| QUOTE |
| What is round as a dishpan and no matter the size, all the water in the ocean can't fill it up? |
Its PIPS the answer is a sieve because a sieve is the size of a dishpan and because it has holes it will not allow water to stay in it for a long time
Yep, its a sieve, good job everyone! Daniel Potter got it first, so I guess you're up...?
XXXX
Anjali
(Sorry for the short post...)
XXXX
Anjali
(Sorry for the short post...)
Just as a general reminder to everyone who posts in this thread: one-liners are not allowed anywhere on the site--including here.
I understand that it's a little bit difficult to make longer posts at times, but if this thread turns into a string of one-liners broken up only occasionally by a longer, more thoughtful post, I'm afraid the mods will shut it down, so please let's protect the thread by making sure we're following all the forum rules.
Anjali, I think your short post was acceptable (I say that cautiously, though--when possible, confirming that a riddle has been correctly guessed should include a little more explanation about the riddle that may not have been explained very well by the guesser), but Daniel Potter and HP#1_wee_lil', both of you could have added a little explanation about why you thought sieve was the correct answer. As fun as it is to just guess answers, I think everyone benefits from seeing the thought process that leads to a correct answer--we all get better at getting riddles that way.
Anyways, Daniel Potter, you're up! I'm sure you'll give us a great riddle.
Carry on!
I understand that it's a little bit difficult to make longer posts at times, but if this thread turns into a string of one-liners broken up only occasionally by a longer, more thoughtful post, I'm afraid the mods will shut it down, so please let's protect the thread by making sure we're following all the forum rules.
Anjali, I think your short post was acceptable (I say that cautiously, though--when possible, confirming that a riddle has been correctly guessed should include a little more explanation about the riddle that may not have been explained very well by the guesser), but Daniel Potter and HP#1_wee_lil', both of you could have added a little explanation about why you thought sieve was the correct answer. As fun as it is to just guess answers, I think everyone benefits from seeing the thought process that leads to a correct answer--we all get better at getting riddles that way.
Anyways, Daniel Potter, you're up! I'm sure you'll give us a great riddle.
Carry on!
Does anyone think somebody should post a new riddle? Daniel Potter's last post was on the 10th at 8 P.M. So it has been almost 3 days. Just a thought.
Why don't you go ahead and post one, WhiteKnight. It has been a while and I don't think Daniel Potter's going to do it.
Actually, let's just open it up. The first person who has a riddle can be the next one to post. After that, it's back to normal rules.
Actually, let's just open it up. The first person who has a riddle can be the next one to post. After that, it's back to normal rules.
Oh sorry guys!! Im here...hmm...sorry..i didnt think I'd get it right..so since no one has posted I guess I'll go....
"The beginning of eternity
The end of time and space
The beginning of every end,
And the end of every place."
What am i?
"The beginning of eternity
The end of time and space
The beginning of every end,
And the end of every place."
What am i?
daniel potter is the answer the letter E.
beginning of eternety=e
end of space and time=e
beginning of every end=e
end of every place=e
beginning of eternety=e
end of space and time=e
beginning of every end=e
end of every place=e
Aha you got it!!! Auror4Life, you're up!!! Post a hard one!!!!!! 
cool this is my first one.
ok here it is
i have white coat with an inner layer of velvet and im gold at heart
what am i?
ok here it is
i have white coat with an inner layer of velvet and im gold at heart
what am i?
Hmm...I'm somewhat surprised no one has answered this one yet. Ah well, let's take a look.
White exterior, gold center, something soft in between. How about an egg?
White exterior, gold center, something soft in between. How about an egg?
good job albus-wan. you are correct. wow that took 3 days to answer. it must have been a good riddle. looks like its your turn albus
Well done Albus-wan
I asked all of my friends and they couldn't guess it either.
Post a riddle soon
MJ
I asked all of my friends and they couldn't guess it either.
Post a riddle soon
MJ
I am fresh out of good riddles, so I'll post one that I may accept multiple answers for. Here goes:
What gets smaller as you put more and more into it?
I have a specific answer in mind, but I'll take anything that seems to work.
What gets smaller as you put more and more into it?
I have a specific answer in mind, but I'll take anything that seems to work.
Hmm...is it space? Not the moon and stars space, but space you have inside a box. If you put something into a box, part of that space will be taken up, and thus the space inside the box will get smaller and smaller if you continue putting things into it.
Your answer works, Hermione_Resilda--and is pretty close to the one I was thinking of--a hole or a pit. The more you put in a hole, the smaller it gets.
Okay, good work. Your turn, Hermione_Resilda.
Okay, good work. Your turn, Hermione_Resilda.
would you mind if I post a riddle?
What goes up but never goes down?
[Mod Edit] Hi, as Albus-Wan noted, I had to edit your signature to the first verse, if you have any questions please contact me.
What goes up but never goes down?
[Mod Edit] Hi, as Albus-Wan noted, I had to edit your signature to the first verse, if you have any questions please contact me.
Since it's Hermione_Resilda's turn to post and it hasn't been long enough since her guess was confirmed, it's her decision whether or not someone else can post a riddle.
Since I'm a **** for riddles I've thought of a few answers that work for your riddle. The first is entropy. A basic law is that entropy can never be decreased so it either stays the same or increases, and therefore goes up but never comes down.
A second answer I thought of his cumulative distribution functions (like the standard normal cumalative distribution function that everyone uses to state the probability that an observation is equal to or less than an observed deviation from a mean). CDFs are monotonic nondecreasing functions, therefore they go up but don't come down.
An answer that is probably closer to the answer you were looking for, though is age. Our age will never decrease no matter how much we want it to.
Despite the fact that I have attempted to answer your riddle, it really is still Hermione_Resilda's turn. If you would like to post a riddle in the future, try to correctly answer the riddle of the person who's turn it is to post.
Also, RemusLupin12, I'm afraid your signature is just too long. The rules that have been established for this site require that signatures contain no more than 5 lines of regular-sized text. Please edit your signature so that the mods don't descend upon you like locusts.
Since you're new here, I'd like to invite you to visit the Rules forum, and specifically, the Main Forums Rules, which have been established to help make Veritaserum one of the best Harry Potter fan sites on the web.
If you have any questions feel free to PM me or any of the moderators or prefects.
Since I'm a **** for riddles I've thought of a few answers that work for your riddle. The first is entropy. A basic law is that entropy can never be decreased so it either stays the same or increases, and therefore goes up but never comes down.
A second answer I thought of his cumulative distribution functions (like the standard normal cumalative distribution function that everyone uses to state the probability that an observation is equal to or less than an observed deviation from a mean). CDFs are monotonic nondecreasing functions, therefore they go up but don't come down.
An answer that is probably closer to the answer you were looking for, though is age. Our age will never decrease no matter how much we want it to.
Despite the fact that I have attempted to answer your riddle, it really is still Hermione_Resilda's turn. If you would like to post a riddle in the future, try to correctly answer the riddle of the person who's turn it is to post.
Also, RemusLupin12, I'm afraid your signature is just too long. The rules that have been established for this site require that signatures contain no more than 5 lines of regular-sized text. Please edit your signature so that the mods don't descend upon you like locusts.
Since you're new here, I'd like to invite you to visit the Rules forum, and specifically, the Main Forums Rules, which have been established to help make Veritaserum one of the best Harry Potter fan sites on the web.
If you have any questions feel free to PM me or any of the moderators or prefects.
the answer I was looking for was age...good job
Okay, cool, and sorry for the what-might-seem-late post. Well, this riddle might be easy, but it's the only one I can think of at the moment. It might look hard, but everyone goes 'oh, that's so simple!', once they figure out the answer.
How would you make the following equation true, by adding only one line to it?
5 + 5 + 5 = 550
Oh, and RemusLupin12, please elaborate more on your future posts!
How would you make the following equation true, by adding only one line to it?
5 + 5 + 5 = 550
Oh, and RemusLupin12, please elaborate more on your future posts!
That's easy:
5 + 5 + 5 ≠ 50
Just a line, between equal...
Ever~
5 + 5 + 5 ≠ 50
Just a line, between equal...
Ever~
Nope, sorry..I'm looking for a different answer. I guess I should've mentioned that the line can't cross out anything, it can be as long as you want it to be, and it can't have any curves.
Ok, I have it now!!
I hope it is...
If you add to the symbol '+' a ray, you get a '4'...
So, you'll get : 545 + 5 = 550...
Is that it?...
Greetings,
Ever~
I hope it is...
If you add to the symbol '+' a ray, you get a '4'...
So, you'll get : 545 + 5 = 550...
Is that it?...
Greetings,
Ever~
I hope Hermione_Resilda doesn't mind too much if I step in for her. I held off on answering this because I had seen it before.
That said, Everseer, your most recent answer is the answer I had seen before, therefore you are correct and can post the next riddle.
If for any reason Hermione_Resilda says that Everseer's post is incorrect, then we'll resume guessing on her riddle. Otherwise Everseer's up! Let's see what you've got.
That said, Everseer, your most recent answer is the answer I had seen before, therefore you are correct and can post the next riddle.
If for any reason Hermione_Resilda says that Everseer's post is incorrect, then we'll resume guessing on her riddle. Otherwise Everseer's up! Let's see what you've got.
OK i really hop ethis is right but anyway this is my answer:
you draw a diagonal line across the last + sign making it look like 5+545=550
you draw a diagonal line across the last + sign making it look like 5+545=550
Ok, I'll see if you understand this one...
Is easy if you understand the procedure...
You have a number of 10 digits...
__ __ __ __ __ __ __ __ __ __ <---- This is the 10-digit number...
The first digit of the number is the quantity of '0' the number has...
The second digit of the number teel the quantity of '1' the number has...
The third digit of the number marks the quantity of '2' the number has...
It will be like this:
Here (in the lines) you put the quantity of the numbers showed below...
__ __ __ __ __ __ __ __ __ __
.0..1..2..3...4..5..6..7...8...9...
You must say how many of these numbers above are in total in the number and write it in the corresponden (sp?) line...
Note: the dots I added were only to separate the numbers and put them in their corresponding line, because the spaces doesn't work...
Now, you must make a 10-digit number that matchs with the characteristics you write...
For example, if you write '4' as the first digit, there must be four numbers zero'0'...
If you write '7' as the fourth digit, there must be seven numbers three '3'...
I hope I've explained well, if I didn't, feel free to ask your question either here or via PM...
Good Luck,
Ever~
Is easy if you understand the procedure...
You have a number of 10 digits...
__ __ __ __ __ __ __ __ __ __ <---- This is the 10-digit number...
The first digit of the number is the quantity of '0' the number has...
The second digit of the number teel the quantity of '1' the number has...
The third digit of the number marks the quantity of '2' the number has...
It will be like this:
Here (in the lines) you put the quantity of the numbers showed below...
__ __ __ __ __ __ __ __ __ __
.0..1..2..3...4..5..6..7...8...9...
You must say how many of these numbers above are in total in the number and write it in the corresponden (sp?) line...
Note: the dots I added were only to separate the numbers and put them in their corresponding line, because the spaces doesn't work...
Now, you must make a 10-digit number that matchs with the characteristics you write...
For example, if you write '4' as the first digit, there must be four numbers zero'0'...
If you write '7' as the fourth digit, there must be seven numbers three '3'...
I hope I've explained well, if I didn't, feel free to ask your question either here or via PM...
Good Luck,
Ever~
At first I thought this was going to be one of those problems that I was going to have to write code to solve, but I think I got it without any computer assistence.
Here's the process I went through:
First, there can't be zero zeros, so I thought that maybe there would only be one zero:
1_ _ _ _ _ _ _ _ _
Which would mean there is a 1:
11 _ _ _ _ _ _ _ _
But that means there are two ones which means there is one two:
121_ _ _ _ _ _ _
If all the rest could be zeroes we'd be good, except that then we'd have to change the first number to 7:
7210000000
This is wrong because no we only have one 1, but that's good because we would have to put a 1 at the 7 location to allow there to be 7 zeros.
7210000100
Which can't be, since now there are only 6 zeros, so if we change the first number to 6 and move the 1 to the 6 location we get:
6210001000
Which is my final answer.
Here's the process I went through:
First, there can't be zero zeros, so I thought that maybe there would only be one zero:
1_ _ _ _ _ _ _ _ _
Which would mean there is a 1:
11 _ _ _ _ _ _ _ _
But that means there are two ones which means there is one two:
121_ _ _ _ _ _ _
If all the rest could be zeroes we'd be good, except that then we'd have to change the first number to 7:
7210000000
This is wrong because no we only have one 1, but that's good because we would have to put a 1 at the 7 location to allow there to be 7 zeros.
7210000100
Which can't be, since now there are only 6 zeros, so if we change the first number to 6 and move the 1 to the 6 location we get:
6210001000
Which is my final answer.
Albus-Wan has solved the riddle!...
Very well, 6210001000 is the only possible answer to this problem, and you've done well in the process you made...
Great job!...
Now is your turn to add a riddle; I'll be waiting because I have many good riddles of math tricks ^^ ...
Was it a good riddle?...
Ever~
Very well, 6210001000 is the only possible answer to this problem, and you've done well in the process you made...
Great job!...
Now is your turn to add a riddle; I'll be waiting because I have many good riddles of math tricks ^^ ...
Was it a good riddle?...
Ever~
All right, let's try this one:
What's eleven thousand eleven hundred and eleven divided by three
I know it's mostly easy, but I'm fresh out.
What's eleven thousand eleven hundred and eleven divided by three
I know it's mostly easy, but I'm fresh out.
I hate that sort of question Albus Wan! 
Anywho, my calculator says: Nothing, i typed and it went.... 4 thousand and something.loads of sixes then a seven!
Anywho, my calculator says: Nothing, i typed and it went.... 4 thousand and something.loads of sixes then a seven!
My first guess is...
What's eleven thousand eleven hundred and eleven divided by three...?
..........<------ 11011 ----------><-------111---------><------ /3 ----->
So, 11011 times 111 is 1222221...
Then, 1222221 divided by 3 is 407407...
Also, there's another option...
What's eleven thousand eleven hundred and eleven divided by three...?
We separate them as 11000 (eleven thousand), 1100 (eleven hundred) and 11 (eleven)...
We make the operation and it is...
11000
+ 1100
. . . 11
12111
Next, 12111 divided by 3 is 4037...
Is any of them correct?...
Ever~
What's eleven thousand eleven hundred and eleven divided by three...?
..........<------ 11011 ----------><-------111---------><------ /3 ----->
So, 11011 times 111 is 1222221...
Then, 1222221 divided by 3 is 407407...
Also, there's another option...
What's eleven thousand eleven hundred and eleven divided by three...?
We separate them as 11000 (eleven thousand), 1100 (eleven hundred) and 11 (eleven)...
We make the operation and it is...
11000
+ 1100
. . . 11
12111
Next, 12111 divided by 3 is 4037...
Is any of them correct?...
Ever~
Is it 4037?
I interpreted it as 11 thousand, 11 hundred and 11, which is 12111, as 11 hundred is actually 1 thousand 1hundred.?
I interpreted it as 11 thousand, 11 hundred and 11, which is 12111, as 11 hundred is actually 1 thousand 1hundred.?
Looks like Everseer slipped in before you, Pix. The correct answer is 4037.
The trick was just to recognize that eleven thousand and eleven hundered overlap.
Nick, I'm guessing you typed 11111 into the calculator and divided by 3.
It looks like Everseer's up again. She says she has math tricks, so everyone go run and get your calculators.
The trick was just to recognize that eleven thousand and eleven hundered overlap.
Nick, I'm guessing you typed 11111 into the calculator and divided by 3.
It looks like Everseer's up again. She says she has math tricks, so everyone go run and get your calculators.
Ok, now a hard one...
It's not that hard if you think slowly and with very careful detail...
You have the following numbers:...
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
Using ANY mathematic symbol, sign or feature you must make a TRUE EQUATION... Use anything you know to reach that answer (6)...
Well, I must say the only symbol you can't use is ' ≠ ', because the sense would be taken off...
Show me your knowledge in math by using all the possible signs of operations in this riddle...
For example, if I have ' 10 10 10 = 10 ', I must make the following....
10 + 10 - 10 = 10, and it makes the equation true...
By the way, Albus-wan, I'm a guy...
However, it's not the first time I'm confused with a girl; in another forum I was asked if I was a girl because of my way of writing...
Anyway, why did you think I was a girl?... It's really funny to know...
Ever~
It's not that hard if you think slowly and with very careful detail...
You have the following numbers:...
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
Using ANY mathematic symbol, sign or feature you must make a TRUE EQUATION... Use anything you know to reach that answer (6)...
Well, I must say the only symbol you can't use is ' ≠ ', because the sense would be taken off...
Show me your knowledge in math by using all the possible signs of operations in this riddle...
For example, if I have ' 10 10 10 = 10 ', I must make the following....
10 + 10 - 10 = 10, and it makes the equation true...
By the way, Albus-wan, I'm a guy...
However, it's not the first time I'm confused with a girl; in another forum I was asked if I was a girl because of my way of writing... Anyway, why did you think I was a girl?...
It's really funny to know...Ever~
| QUOTE (Everseer) |
| Anyway, why did you think I was a girl? |
Mostly the female eye, and the fact that girls outnumber guys here.
I'm sitting here waiting for a simulation to finish, so I guess I have too much time to think about these things.
I guess I'll do them in the order I solved them:
2+2+2=6
3*3-3=6
6+6-6=6
7-7/7=6
5+5/5=6
4+4/sqrt(4)=6
9-9/sqrt(9)=6
8-sqrt(sqrt(8+8))=6
(1+1+1)! = 6
The last one was really stumping me until I looked up a list of mathematical operators and remembered the factorial.
So, let me know if I missed any. If I didn't, why don't you go ahead and post another one since I've run out of riddles.
You're pretty good...
I also stucked on the factorial one ^^... Snd you did well with the square roots, but you can also use the cube (third) roots (sp?) (³√)...
For example, ³√8 + ³√8 + ³√8 = 6, because ³√ = 2...
But it's still correct...
As Albus-wan's petition, this is the next riddle...
Bird Dad asked to Bird Mom: "Well, how many worms do I have to bring for our Bird Children?"
Bird Mom answered: " Wait, let me think; the smallest of our Bird Children eats more than 100 and less than 1000 worms; and the biggest too, but he eats a bit more than the smallest Bird. I don't remember exactly how many worms does each one eat, but yesterday I multiplied (sp?) the worms that each one eats, and the result was 555 555"...
Quickly, Bird Dad made his mental calculations and after a few seconds he went flying to the deepest part of the forest, after he said to Bird Mom: "Oh, now I know how many I must bring".
How many worms does Bird Dad had to bring for each one of his Bird Children?...
Note: They are only 2 Bird Children...
Ever~
<offtopic> Well, about the eye. As I've exaplined on the Thread where we explain the subject of our names, I told something about it. I'm 'Everseer, the Foreseer', a prophet and clarivident, that's the why of the eye... I have a huge collection...
Also, I loved Luna Lovegood instead of Harry or Cedric </offtopic>
I also stucked on the factorial one ^^... Snd you did well with the square roots, but you can also use the cube (third) roots (sp?) (³√)...
For example, ³√8 + ³√8 + ³√8 = 6, because ³√ = 2...
But it's still correct...
As Albus-wan's petition, this is the next riddle...
Bird Dad asked to Bird Mom: "Well, how many worms do I have to bring for our Bird Children?"
Bird Mom answered: " Wait, let me think; the smallest of our Bird Children eats more than 100 and less than 1000 worms; and the biggest too, but he eats a bit more than the smallest Bird. I don't remember exactly how many worms does each one eat, but yesterday I multiplied (sp?) the worms that each one eats, and the result was 555 555"...
Quickly, Bird Dad made his mental calculations and after a few seconds he went flying to the deepest part of the forest, after he said to Bird Mom: "Oh, now I know how many I must bring".
How many worms does Bird Dad had to bring for each one of his Bird Children?...
Note: They are only 2 Bird Children...
Ever~
<offtopic> Well, about the eye. As I've exaplined on the Thread where we explain the subject of our names, I told something about it. I'm 'Everseer, the Foreseer', a prophet and clarivident, that's the why of the eye... I have a huge collection...
Also, I loved Luna Lovegood instead of Harry or Cedric
</offtopic>
I thought I'd give others a chance to answer, but now it's been long enough I think.
My answer is 1492, which is 715 for the smaller bird and 777 for the larger bird. These are the only two combinations of the factors of 555,555 (which are 3, 5, 7, 37, 143) that both fall between 100 and 1000.
If this is right and you have more, Everseer, go ahead and post your next one.
My answer is 1492, which is 715 for the smaller bird and 777 for the larger bird. These are the only two combinations of the factors of 555,555 (which are 3, 5, 7, 37, 143) that both fall between 100 and 1000.
If this is right and you have more, Everseer, go ahead and post your next one.
Once again, Albus-wan, you're correct...
If we descompose the factors of 555 555, they are 3, 5, 7, 11, 13, 37...
Now, if in both cases the numbers are of 3-digits, we must only make mixes and combinations of these factors until we get the answer...
The only combinations as possible and correct are, as Albus said, 715 and 777, making a total of 1492...
The next riddle...
After many weeks of deep meditations, Ruy Pérez Tamayo - the famous doctor and investigator from México - finnaly got an answer that he would ask as an admission exam to Julieta Fierro - famous mexican astronomer - to the Mexican Language Academy...
A bit nervious and without have slept in 4 days of so much study, Julieta went to the place where she'd take her exam. But she felt so relaxed when the exam was only one answer to the following question:...
"In the following series with pairs of letters, ¿which letter is missing on the last pair?: M-Y, V-S, E-H, M-S, J-R, S-N, U-S, N-E, P-__." ...
Julieta Fierro is now a member of the Mexican Language Academy, ¿what was the letter she gave as answer and why did she selected that letter?...
I see you like my riddles, or is just my imagination?...
Ever~
If we descompose the factors of 555 555, they are 3, 5, 7, 11, 13, 37...
Now, if in both cases the numbers are of 3-digits, we must only make mixes and combinations of these factors until we get the answer...
The only combinations as possible and correct are, as Albus said, 715 and 777, making a total of 1492...
The next riddle...
After many weeks of deep meditations, Ruy Pérez Tamayo - the famous doctor and investigator from México - finnaly got an answer that he would ask as an admission exam to Julieta Fierro - famous mexican astronomer - to the Mexican Language Academy...
A bit nervious and without have slept in 4 days of so much study, Julieta went to the place where she'd take her exam. But she felt so relaxed when the exam was only one answer to the following question:...
"In the following series with pairs of letters, ¿which letter is missing on the last pair?: M-Y, V-S, E-H, M-S, J-R, S-N, U-S, N-E, P-__." ...
Julieta Fierro is now a member of the Mexican Language Academy, ¿what was the letter she gave as answer and why did she selected that letter?...
I see you like my riddles, or is just my imagination?...
Ever~
I've been the only one guessing recently, so I was going to let someone else get this one, but no one else has jumped in, so I'll guess again.
The letter is "O". The pattern is the first and last letter of the planets ordered by their average distances from the sun. P and O are the first and last letter of the planet Pluto.
The letter is "O". The pattern is the first and last letter of the planets ordered by their average distances from the sun. P and O are the first and last letter of the planet Pluto.
Once again, my great Albus-wan, you're correct...
Just as Albus wisely said, they're the planets ordered in their distances from the sun from the closest to the farthest (sp?)...
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
And it certainly makes the following series as...
'M-Y, V-S, E-H, M-S, J-R, S-N, U-S, N-E, P-O'...
So, Albus-wan is the right one to post the next riddle. Unless he decides to give me the opportunity to place another riddle...
He didn't tell me this time to place it, so I may presume he's got a good one to share with us...
Ever~
Just as Albus wisely said, they're the planets ordered in their distances from the sun from the closest to the farthest (sp?)...
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
And it certainly makes the following series as...
'M-Y, V-S, E-H, M-S, J-R, S-N, U-S, N-E, P-O'...
So, Albus-wan is the right one to post the next riddle. Unless he decides to give me the opportunity to place another riddle...
He didn't tell me this time to place it, so I may presume he's got a good one to share with us...
Ever~
Good riddle. Very clever The order of planets and the first and last letter were good challenge. I shall try to answere the next one because I have a good riddle I want to share. Oh well I will have to guess the answer to get a turn
MJ
MJ
It's getting harder and harder for me to find new riddles these days, so maybe I'll refrain from answering any more until I have one of my own to post (I made that rule for myself a while ago, but I keep breaking it).
EDIT: Since I have a nearly infinite store of probability problems, I hope everyone can forgive me if I put another probability problem on here. If no one gets it within the week, then I'll post another one or let someone else go.
A casino has developed a simple new game based on flipping a coin where the probability that the coin lands heads is 0.5 and the probability that the coin lands tails is also 0.5. The gambler will flip the coin until it lands on heads and his prize will double for each time it lands on tails before it lands on heads, starting at 1 coin. So if he got tails then heads he would get 1 coin, and if he got tails, tails, tails, heads, he would get 4 coins, and so on. What is the minimum number of coins that the casino should charge people to play in order to guarantee that they will make a profit on the game?
Good luck!
EDIT: Since I have a nearly infinite store of probability problems, I hope everyone can forgive me if I put another probability problem on here. If no one gets it within the week, then I'll post another one or let someone else go.
A casino has developed a simple new game based on flipping a coin where the probability that the coin lands heads is 0.5 and the probability that the coin lands tails is also 0.5. The gambler will flip the coin until it lands on heads and his prize will double for each time it lands on tails before it lands on heads, starting at 1 coin. So if he got tails then heads he would get 1 coin, and if he got tails, tails, tails, heads, he would get 4 coins, and so on. What is the minimum number of coins that the casino should charge people to play in order to guarantee that they will make a profit on the game?
Good luck!
Umm let's see 1 coin tails before heads is one coin to the dealer/flipper. I think 100 for some reason. I know it is wrong answer but oh well I gave it a shot
Mj
P.s: Are you posting the answer on 8 may
Mj
P.s: Are you posting the answer on 8 may
Okay, Albus-wan, probability is one of my mathematical weak spots. I suppose that I have to give it a try, though. I know that I will never get it right.
I would say that the casino should charge 2 coins per "coin flip." I am not entirely sure how to work the 50/50 probability into the answer, but given that ratio, it is assumed that one would naturally get an equal amount of 'heads' as 'tails', even if it does not always turn out that way.
amt. tails= 1; cost= 2 coins; payment= -1; casino profit =1
That, due to the 50/50 chance, should be how most of the games would end, resulting in the casino making 1 coin for every turn somebody takes. Of course, if the gambler gets more than one tails in a row, the casino would still be making a profit until the gambler reaches 4 coin flips landing on tails in a row. That, if my math is correct, is a 1/16 chance. Even still, the casino would be even: no loss in profit, nor gain. Once the person reaches 5 or more in a row, then the casino would lose a profit. This is a 1/32+ chance of happening, though, which is pretty slim in my opinion.
Well, I apologize if nothing that I just said made sense. I warned you in the beginning that I am not very good at probability problems
...
_daviD
I would say that the casino should charge 2 coins per "coin flip." I am not entirely sure how to work the 50/50 probability into the answer, but given that ratio, it is assumed that one would naturally get an equal amount of 'heads' as 'tails', even if it does not always turn out that way.
amt. tails= 1; cost= 2 coins; payment= -1; casino profit =1
That, due to the 50/50 chance, should be how most of the games would end, resulting in the casino making 1 coin for every turn somebody takes. Of course, if the gambler gets more than one tails in a row, the casino would still be making a profit until the gambler reaches 4 coin flips landing on tails in a row. That, if my math is correct, is a 1/16 chance. Even still, the casino would be even: no loss in profit, nor gain. Once the person reaches 5 or more in a row, then the casino would lose a profit. This is a 1/32+ chance of happening, though, which is pretty slim in my opinion.
Well, I apologize if nothing that I just said made sense. I warned you in the beginning that I am not very good at probability problems
..._daviD
good thinking David I don't think I am write.
Did you see my post. I am so dumb
Did you see my post. I am so dumb
No correct answers so far.
David, keep in mind that the person pays once and flips the coin until he or she gets heads.
Let me make a slight modification so that it may be easier to understand. The player will receive the 2^x coins, where x is the number of tails they flip before getting heads. This means that if they get heads on the first flip they get 1 coin, tails then heads they get 2 coins, tails, tails, then heads they get 4 coins, tails, tails, tails, then heads they get 8 coins, and so on.
So that's it. Good luck!
David, keep in mind that the person pays once and flips the coin until he or she gets heads.
Let me make a slight modification so that it may be easier to understand. The player will receive the 2^x coins, where x is the number of tails they flip before getting heads. This means that if they get heads on the first flip they get 1 coin, tails then heads they get 2 coins, tails, tails, then heads they get 4 coins, tails, tails, tails, then heads they get 8 coins, and so on.
So that's it. Good luck!
So tails before heads is 2 coins? So if they pay one they could win say 10.
Hmmm. I think they need at least 5 beacuse The gambler will flip the coin until it lands on heads and his prize will double for each time it lands on tails before it lands on heads, starting at 1 coin. So if he got tails then heads he would get 1 coin, and if he got tails, tails, tails, heads, he would get 4 coins/amt. tails= 1; cost= 2 coins; payment= -1; casino profit =1.
I know it is wrong but I had a go.
MJ
Hmmm. I think they need at least 5 beacuse The gambler will flip the coin until it lands on heads and his prize will double for each time it lands on tails before it lands on heads, starting at 1 coin. So if he got tails then heads he would get 1 coin, and if he got tails, tails, tails, heads, he would get 4 coins/amt. tails= 1; cost= 2 coins; payment= -1; casino profit =1.
I know it is wrong but I had a go.
MJ
So it looks like no one is going to get this. I'll explain the solution, then mjane95 can go.
If we say that the probability of getting heads is p and the probability of getting tails is q = (1-p) and x is the number of tails before getting heads, then the probability of getting exactly x is
p*q^x
for x = 0, 1, 2, ...
If we sum over all the possible values of x we can show that the probability is 1 (Note: each time sum is written from here on out, it's referring to the sum over all x = 0, 1, 2, ..., infinity):
sum(p*q^x) = psum(q^x) = p(1 - q^(inifinity+1))/(1 - q) = p/(1 - q) = p/p = 1
That last part wasn't necessary to get the answer--I was just showing that my function was correct.
Now we can substitute the values for p and q, which are each 1/2, to get Probability of x = (1/2)^(x + 1) for x = 0, 1, 2, ...
The payout for each x is 2^x, so to get the expected payout for this game we need to multiply each payout by the probability of having to pay that payout and sum over all possible values of x:
sum(2^x*(1/2)^(x+1)) = sum((1^x)/2) = sum(1/2) = infinity
This means that the expected payout is infinite, so the casino would have to charge an infinite fee in order to guarantee that they would at least break even, so the game should probably be scrapped since no one can afford to play.
All right, I hope I haven't caused too many headaches. Like I said, I know mjane95 has a riddle, so go ahead.
If we say that the probability of getting heads is p and the probability of getting tails is q = (1-p) and x is the number of tails before getting heads, then the probability of getting exactly x is
p*q^x
for x = 0, 1, 2, ...
If we sum over all the possible values of x we can show that the probability is 1 (Note: each time sum is written from here on out, it's referring to the sum over all x = 0, 1, 2, ..., infinity):
sum(p*q^x) = psum(q^x) = p(1 - q^(inifinity+1))/(1 - q) = p/(1 - q) = p/p = 1
That last part wasn't necessary to get the answer--I was just showing that my function was correct.
Now we can substitute the values for p and q, which are each 1/2, to get Probability of x = (1/2)^(x + 1) for x = 0, 1, 2, ...
The payout for each x is 2^x, so to get the expected payout for this game we need to multiply each payout by the probability of having to pay that payout and sum over all possible values of x:
sum(2^x*(1/2)^(x+1)) = sum((1^x)/2) = sum(1/2) = infinity
This means that the expected payout is infinite, so the casino would have to charge an infinite fee in order to guarantee that they would at least break even, so the game should probably be scrapped since no one can afford to play.
All right, I hope I haven't caused too many headaches. Like I said, I know mjane95 has a riddle, so go ahead.
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