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Postby Beardy » Mar 8th, '09, 05:48



hmmmmmm........

touch two adjacent lids and close them
*if not completed, spins*
touch two diagonally opposite lids and close them.
*if not completed, spins*
There is only one open lid left. Grab two lids. If one is open, shut it - you win!
If, however, both are shut, you only open one of them. You now know that there are two open and two closed lids
*spins*
You grab two lids that are opposite each other diagonally. If they are both the same (open or closed) you win by changing them to what the other two are (either open or closed). If they are different, you leave them be, knowing that two matching ones are next to eachother

(diagram at this point)

x o

x o

*spins*

You grab two lids. Now, one of two things will happen. You grab a matching pair, and flip them thus matching the other two, allowing an alarm to sound.

*or*

You grab two that are next to eachother but not matching. If this occurs, you flip them both. This means that the resulting table will look like this:

x o

o x

meaning that a matching pair is opposite eachother. So next time the table is spinned, just grab diagonal opposite, knowing it will match - and you win!

Thus, if my reasoning is correct, and if my explanation isn't too b*ll*cks, at worse case scenario it would take, I think, 6 moves to complete? Or at best one move, or even none, depending on the randomisation of the table

Love

Chris
xxx

"An amazing mind manipulator" - Uri Geller
"I hope to shake your hand before I die" - Derren Brown
"That was mightily impressive - I have absolutely no clue how you did that" - Tim Minchin
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Postby moonbeam » Mar 8th, '09, 11:58

Oooooh so close Beardy, but this is where you're going wrong:

Blapsing_Beard wrote:There is only one open lid left. Grab two lids. If one is open, shut it - you win!
If, however, both are shut, you only open one of them. You now know that there are two open and two closed lids



You are correct with the first 2 moves that enables you to reach the point where you know that you can only have one open lid left. On your next move you say, "grab 2 lids". You need to be more specific as to which lids (in relation to each other) you're gonna grab, ''cos it is possible from here on in, to keep track of where the open/closed lids are in relation to each other.

The minimum number of turns is five by the way :wink: .

QUESTION:
If we can sue McDonalds for making us fat and cigarette companies for giving us cancer; why can't we sue Smirnoff for all the ugly gits we've sh*gged ??
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Postby Beardy » Mar 9th, '09, 02:04

surely it makes no difference?

If I have the layout at this point, with 3 open and one closed:

o o

x o

Then whether you grab diagonal or next to eachother, and get two open ones, you still have no idea as to where the closed one is out of the remaining two?

Love

Chris
xxx

"An amazing mind manipulator" - Uri Geller
"I hope to shake your hand before I die" - Derren Brown
"That was mightily impressive - I have absolutely no clue how you did that" - Tim Minchin
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Postby moonbeam » Mar 9th, '09, 09:57

Blapsing_Beard wrote:surely it makes no difference?

If I have the layout at this point, with 3 open and one closed:

o o

x o

Then whether you grab diagonal or next to eachother, and get two open ones, you still have no idea as to where the closed one is out of the remaining two?


Not entirely correct.

If you grab an adjacent pair and they are both open, you know that one of these must be opposite the closed hole and the other must be adjacent to it - not very helpful.

However, if you grab a diagonal pair and they are both open - these must both be adjacent to the closed hole. Close one of these holes and we now know that we have 2 adjacent closed holes and 2 adjacent open holes.

Next move ??

QUESTION:
If we can sue McDonalds for making us fat and cigarette companies for giving us cancer; why can't we sue Smirnoff for all the ugly gits we've sh*gged ??
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Postby Beardy » Mar 9th, '09, 11:20

my mistake! So basically, from there on in you just follow my original instructions from the diagram above!

Jeez that was a brain teaser!

Love

Chris
xxx

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"I hope to shake your hand before I die" - Derren Brown
"That was mightily impressive - I have absolutely no clue how you did that" - Tim Minchin
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Postby Part-Timer » Mar 9th, '09, 13:43

Blapsing_Beard wrote:my mistake! So basically, from there on in you just follow my original instructions from the diagram above!


Yes (although you must go for an adjacent pair on the next go).

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Postby Beardy » Mar 9th, '09, 16:58

Two blokes are talking on the street and John says to Dave, "Guess how old my three daughters are." Dave, being a typical bloke, is desperate for attention, so he agrees.
John says, "Their ages add to the address of that house," he points at a house across the street, "and their ages multiply to 36."

Dave works on the riddle for a while but eventually he tells John that he can't solve the riddle. John tells Dave that his oldest daughter has red hair. Dave thinks about this, and then he tells John the correct ages of his three daughters.



The riddle is: What are the ages of the three daughters?

That's all the information you need to solve it. Assume that the ages are integers (i.e. no 23-and-a-half stuff). There is only one correct answer

Love

Chris
xxx

"An amazing mind manipulator" - Uri Geller
"I hope to shake your hand before I die" - Derren Brown
"That was mightily impressive - I have absolutely no clue how you did that" - Tim Minchin
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Postby Nightfall » Mar 9th, '09, 17:23

Blapsing_Beard wrote:Two blokes are talking on the street and John says to Dave, "Guess how old my three daughters are." Dave, being a typical bloke, is desperate for attention, so he agrees.
John says, "Their ages add to the address of that house," he points at a house across the street, "and their ages multiply to 36."

Dave works on the riddle for a while but eventually he tells John that he can't solve the riddle. John tells Dave that his oldest daughter has red hair. Dave thinks about this, and then he tells John the correct ages of his three daughters.



The riddle is: What are the ages of the three daughters?

That's all the information you need to solve it. Assume that the ages are integers (i.e. no 23-and-a-half stuff). There is only one correct answer


I had solved it before let's see if I remember how to solve it.
All the age combinations are :

1 1 36 (which adds to 38 )
1 2 18 (21)
1 3 12 (16)
1 4 9 (14)
1 6 6 (13)
2 2 9 (13)
2 3 6 (11)
3 3 4 (10)

When he sees the House number we would have been able to say that he knows the solution only if there where no duplicate additions (if all the additions where unique he could choose one of the answers according to the house number). Therefore the solution is one of the duplicates, so we narrow it to :

1 6 6 or
2 2 9

Saying that the oldest daughter has red hair, he implies that there is ONE older child (the daughter) s owe can rule out 1 6 6 and the solution is :

2 2 9

PS:In case I'm correct can somebody grab the riddle since I'm about to leave work.

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Postby moonbeam » Mar 9th, '09, 18:49

Blapsing_Beard wrote:my mistake! So basically, from there on in you just follow my original instructions from the diagram above!

Jeez that was a brain teaser!


Full answer:

1. Pick a diagonal pair and open them.
2. Pick an adjacent pair and open them. (If you're not done, there's just one closed hole).
3. Pick a diagonal pair. If one of them is closed, open it and you're done. If both of them are open, close one and the two closed holes are now adjacent.
4. Pick an adjacent pair and flip them. If they were the same you're done; if not, then the closed holes are now diagonally across from each other.
5. Pick a diagonal pair and flip them - and now you're done :wink: .

QUESTION:
If we can sue McDonalds for making us fat and cigarette companies for giving us cancer; why can't we sue Smirnoff for all the ugly gits we've sh*gged ??
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Postby moonbeam » Mar 17th, '09, 23:12

Okay it's been over a week now - time to ask a riddle that usually stirs up quite an argument, as some people cannot grasp/understand the answer.
Even when it's fully explained, some people refuse to believe the answer .........

Riddle me this:

A couple have two children (and they are a normal couple with a 50:50 chance of having a girl or a boy).
You are introduced to one of their children who is a girl. What is the probability that the other child is also a girl?

QUESTION:
If we can sue McDonalds for making us fat and cigarette companies for giving us cancer; why can't we sue Smirnoff for all the ugly gits we've sh*gged ??
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Postby Ian The Magic-Ian » Mar 17th, '09, 23:34

The probability that the other child is a girl is 1/4? Because if it was 50:50 chance of it being a boy or a girl that would mean that if the first one was a girl than that's already 50:50 or 2/4. If it's another girl it's a 1/4 chance. There's a 3/4 chance however it's a boy so...it's a boy most likely?

Right or...way off? :P Or is it that chance has no memory? Hence it can't be foretold?

Barton: Have you read the Bible, Pete?
Pete: Holy Bible?
Barton: Yeah.
Pete: Yeah, I think so. Anyway, I've heard about it.
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Postby moonbeam » Mar 18th, '09, 00:07

Ian the Mental-Ian wrote:The probability that the other child is a girl is 1/4?


........... wrong.

This does seem like a logical answer, but it is incorrect ......... :?

Answer on a postard please ............. :shock:

Last edited by moonbeam on Mar 18th, '09, 00:30, edited 1 time in total.
QUESTION:
If we can sue McDonalds for making us fat and cigarette companies for giving us cancer; why can't we sue Smirnoff for all the ugly gits we've sh*gged ??
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Postby IAIN » Mar 18th, '09, 00:16

50%?

IAIN
 

Postby EckoZero » Mar 18th, '09, 00:19

I'm with Iain on this one.

The chance of their first child being a girl is 50%.

The chance of their second child being a girl has no relation to the chance of their first child being a girl.
I mean, let's say that the chance of having a human child is 100% with 50% of that being a boy and 50% being a girl.
When they have a girl they got the 50% that it would be a girl but still from the 100% chance that it would be human.

The second child has a new 100% chance of being human so the chance of it being a girl is STILL 50% - am I right?

You wont find much better anywhere and it's nothing - a rigmarole with a few bits of paper and lots of spiel. That is Mentalism

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Postby moonbeam » Mar 18th, '09, 00:27

IAIN wrote: 50%?
.......... and ....... you're still wrong ..........

(This is what most people say the answer is by the way).

Note the wording of the puzzle.

You only know that one of the children is a girl (you don't know which one - i.e. the youngest or the eldest).

If I change the original wording
You are introduced to one of their children who is a girl. What is the probability that the other child is also a girl?


.....to:

You are introduced to the eldest child, who is a girl. What is the probability that the other child is also a girl?


....... then the answer is 50%.
But that is not what I'm asking .............

Try again ???

QUESTION:
If we can sue McDonalds for making us fat and cigarette companies for giving us cancer; why can't we sue Smirnoff for all the ugly gits we've sh*gged ??
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