ponderousness

[Ben F-W]

In front of you are 100 bags, 99 of which contain 50 pieces of gold.
Gold. GOLD!

Ah, but there's a catch (of course). The 100th bag contains 50 pieces of
something that looks like gold, feels like gold, glisters like gold, but
is entirely worthless. The only difference between this stuff and real
gold is that this stuff is very slightly lighter. Not enough to be able
to tell by hand, but enough for, say, a machine to notice.

And, fortunately, just behind the bags is just such a weighing machine -
one of those digital push-down ones which will tell you the weight of
anything you put on it. Catch number two: it's coin-operated, and you
only have one coin. And no, the gold pieces don't fit in the coin slot.

Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Cheers,

Ben

P.S. if you've heard this before and think I've got an essential detail
wrong, think again. I've amended this slightly.

P.P.S. If you're going to give an answer, put a spoiler warning at the
top of your email.

[David Marshall]

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Yes. HTH.

Dave

[David Marshall]

Post by David Marshall

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Yes. HTH.

As you need a different number of coins from each bag, and there are more
bags than coins in each bag, why not just make some half-coins using a
Dremel?

Dave

[Ben F-W]

Post by David Marshall

Post by David Marshall

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Yes. HTH.

As you need a different number of coins from each bag, and there are more
bags than coins in each bag, why not just make some half-coins using a
Dremel?

You don't need to do that. No splitting of coins is necessary.

[David Marshall]

Post by Ben F-W

Post by David Marshall
As you need a different number of coins from each bag, and there are more
bags than coins in each bag, why not just make some half-coins using a
Dremel?

You don't need to do that. No splitting of coins is necessary.

Then you could rig up a plank with one end resting on the scale, the other
end resting on something fixed, and carefully place one coin from each bag
along the length...

Dave

[Ben F-W]

Post by David Marshall

Post by Ben F-W
No splitting of coins is necessary.

Then you could rig up a plank with one end resting on the scale, the other
end resting on something fixed, and carefully place one coin from each bag
along the length...

Clever! But nothing I haven't described is necessary.

[Matt Horsley]

Post by David Marshall

Post by David Marshall

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Yes. HTH.

As you need a different number of coins from each bag, and there are more
bags than coins in each bag, why not just make some half-coins using a
Dremel?
Dave

How does weighing a mixture of coins tell you which bag is fake?

All it'd tell you that one of the half coins is fake, and surely you
already know that much?

[David Marshall]

Post by Matt Horsley
How does weighing a mixture of coins tell you which bag is fake?
All it'd tell you that one of the half coins is fake, and surely you
already know that much?

With one coin from the first bag, two from the second, etc. the amount that
the mixture is underweight by will tell you which bag is fake. Ben's twist
is not enough coins in each bag.

Dave

[Timothy Packer]

Post by David Marshall
With one coin from the first bag, two from the second, etc. the amount that
the mixture is underweight by will tell you which bag is fake.

..if you knew how much the not-gold substance weighed (otherwise, as far as
I can see, you only know that the whole is 1 gram underweight, one coin is 1
gram under, or 2 coins are 1/2 a gram underweight, or 50 coins are 1/50th of
a gram underweight). I don't think we've been told we know that. Do we?

While you're at it, could you clarify the operation of the machine - does it
make exactly one measurement, or does it measure weight continuously until
there is no longer anything on it, or something else?

Tim
=-=

[Timothy Packer]

Post by Timothy Packer
..if you knew how much the not-gold substance weighed (otherwise, as far as
I can see, you only know that the whole is 1 gram underweight, one coin is 1
gram under, or 2 coins are 1/2 a gram underweight, or 50 coins are 1/50th of
a gram underweight). I don't think we've been told we know that. Do we?

Sorry, this was atrociously badly-phrased. The bit in brackets should be:

(otherwise, as far as I can see, you only know that the whole is, say, one
gram underweight; which could mean that one coin is one gram under, or two
coins are 1/2 a gram under, or 50 coins are 1/50th of a gram under)

Tim
=-=

[Ben F-W]

Post by Timothy Packer

Post by David Marshall
With one coin from the first bag, two from the second, etc. the amount that
the mixture is underweight by will tell you which bag is fake.

..if you knew how much the not-gold substance weighed (otherwise, as far as
I can see, you only know that the whole is 1 gram underweight, one coin is 1
gram under, or 2 coins are 1/2 a gram underweight, or 50 coins are 1/50th of
a gram underweight). I don't think we've been told we know that. Do we?

Yes, you're right. I should've said that you do know the weight of the
non-gold. Sorry everyone!

Post by Timothy Packer
While you're at it, could you clarify the operation of the machine - does it
make exactly one measurement, or does it measure weight continuously until
there is no longer anything on it, or something else?

It makes one discrete measurement.

Ben

[David Marshall]

Post by Ben F-W
Yes, you're right. I should've said that you do know the weight of the
non-gold. Sorry everyone!

You said on #durge, but not here, I think.

Dave

[Tom York]

Post by Ben F-W
In front of you are 100 bags, 99 of which contain 50 pieces of gold.
Gold. GOLD!
Ah, but there's a catch (of course). The 100th bag contains 50 pieces of
something that looks like gold, feels like gold, glisters like gold, but
is entirely worthless. The only difference between this stuff and real
gold is that this stuff is very slightly lighter. Not enough to be able
to tell by hand, but enough for, say, a machine to notice.
And, fortunately, just behind the bags is just such a weighing machine -
one of those digital push-down ones which will tell you the weight of
anything you put on it. Catch number two: it's coin-operated, and you
only have one coin. And no, the gold pieces don't fit in the coin slot.
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Sell all 100 bags on ebay.

Tom

[n***@loowis.durge.org]

Post by Ben F-W
In front of you are 100 bags, 99 of which contain 50 pieces of gold.
Gold. GOLD!
Ah, but there's a catch (of course). The 100th bag contains 50 pieces of
something that looks like gold, feels like gold, glisters like gold, but
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

The 100th bag, obviously!
I *think* in the version of this I heard before, there were only 50
bags, not 100.

Andy Scheller

<can't be bothered with a sig>

[Chris Morris]

***@loowis.durge.org writes:
] Ben F-W wrote in alt.dur.general:
] > In front of you are 100 bags, 99 of which contain 50 pieces of gold.
] > Gold. GOLD!
]
] > Ah, but there's a catch (of course). The 100th bag contains 50 pieces of
] > something that looks like gold, feels like gold, glisters like gold, but
]
] > Given that you know how much gold weighs, can you work out which bag
] > contains the fake stuff?
]
] The 100th bag, obviously!
] I *think* in the version of this I heard before, there were only 50
] bags, not 100.

50 bags is much easier, of course. Presumably that's the amended detail?

[Ben F-W]

Post by Chris Morris
] > In front of you are 100 bags, 99 of which contain 50 pieces of gold.
] > Gold. GOLD!
]
] > Ah, but there's a catch (of course). The 100th bag contains 50 pieces of
] > something that looks like gold, feels like gold, glisters like gold, but
]
] > Given that you know how much gold weighs, can you work out which bag
] > contains the fake stuff?
]
] The 100th bag, obviously!
] I *think* in the version of this I heard before, there were only 50
] bags, not 100.
50 bags is much easier, of course. Presumably that's the amended detail?

Yes, that's the difference :)

[Nick Boalch]

On 2006-01-12, Ben F-W <***@private.invalid> wrote:
[...]

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Is the weighing machine tough enough to weigh 100 gold pieces at once? If
so, yes...

Cheers,

N.

[Ben F-W]

Post by Nick Boalch

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Is the weighing machine tough enough to weigh 100 gold pieces at once? If
so, yes...

It's a very tough weighing machine.

Anyone care to explain their method?

[Matt Horsley]

Post by Ben F-W

Post by Nick Boalch

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

Is the weighing machine tough enough to weigh 100 gold pieces at once?
If so, yes...

It's a very tough weighing machine.
Anyone care to explain their method?

Take all the bags and pay someone else to work it out.

[David Graham Knoll]

Post by Matt Horsley
Take all the bags and pay someone else to work it out.

And trust them not to nick any when you hand over 99 bags of real gold.

David

[n***@loowis.durge.org]

Post by Ben F-W
And, fortunately, just behind the bags is just such a weighing machine -
one of those digital push-down ones which will tell you the weight of
anything you put on it. Catch number two: it's coin-operated, and you
only have one coin. And no, the gold pieces don't fit in the coin slot.

Does the coin-operated weighing machine operate long enough to let me
make 99 measurements? ;)

Andy Scheller

<can't be bothered with a sig>

[Alasdair MacLeod]

Post by Ben F-W
In front of you are 100 bags, 99 of which contain 50 pieces of gold.
Gold. GOLD!
Ah, but there's a catch (of course). The 100th bag contains 50 pieces of
something that looks like gold, feels like gold, glisters like gold, but
is entirely worthless.

"The banker's offering you 12 pieces of gold, for that bag in front of
you. It might contain FIFTY pieces of gold, or it might contain NOTHING.

"So ... Deal or No Deal?"

Al .-.

[Ben F-W]

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

I assume no-one wants to tell me how to do it, then. Dave? Chris? Nick?
And no, Andy, it can only be used once :)

Ben

[Timothy Packer]

Post by Ben F-W
I assume no-one wants to tell me how to do it, then. Dave? Chris? Nick?
And no, Andy, it can only be used once :)

Ah, missed this.

[Ben F-W]

Post by Ben F-W

Post by Ben F-W
Given that you know how much gold weighs, can you work out which bag
contains the fake stuff?

I assume no-one wants to tell me how to do it, then. Dave? Chris? Nick?
And no, Andy, it can only be used once :)

Still no takers. Must admit, I'm slightly worried now that the solution
I made up is wrong - haven't actually tested it against anyone yet!

My answer: (scroll down)





















1) Take fifty of your hundred bags and put them into a different pile.
You now have fifty bags in in Pile One and fifty in Pile Two. Each bag
has 50 coins in it.

2) Swap one coin from Bag One Pile One with one coin from Bag One Pile Two.

3) Swap two coins from Bag Two Pile One with two coins from Bag Two Pile
Two. Repeat this fifty times: for the fiftieth bag, you are simply
swapping both bags across.

4) Weigh one of your piles.

Now in the 100/100 solution, you could tell which bag was the culprit
because it was lighter. Here, you're going two ways around the average -
heavier and lighter. You take the total weight of the 100 bags (taking
into account that one is lighter), and divide it by two. If your scale
reading is less than that weight, the bag is in the pile you weighed. If
it's heavier, you should look in the other pile. The number of points
away you are from that weight is the number of the guilty bag.

Let me know if I'm wrong!

Cheers,

Ben

[David Graham Knoll]

Post by Ben F-W
Let me know if I'm wrong!
Cheers,
Ben

Urgh, maths. Just been to a formal. Don't really care right now.

David

[Trevor Roberts]

Post by Ben F-W
1) Take fifty of your hundred bags and put them into a different pile.
You now have fifty bags in in Pile One and fifty in Pile Two. Each bag
has 50 coins in it.
2) Swap one coin from Bag One Pile One with one coin from Bag One Pile Two.
3) Swap two coins from Bag Two Pile One with two coins from Bag Two Pile
Two. Repeat this fifty times: for the fiftieth bag, you are simply
swapping both bags across.
4) Weigh one of your piles.
Now in the 100/100 solution, you could tell which bag was the culprit
because it was lighter. Here, you're going two ways around the average -
heavier and lighter. You take the total weight of the 100 bags (taking
into account that one is lighter), and divide it by two. If your scale
reading is less than that weight, the bag is in the pile you weighed. If
it's heavier, you should look in the other pile. The number of points
away you are from that weight is the number of the guilty bag.
Let me know if I'm wrong!

You're wrong. HTH, HAND.

*ahem*

OK, to explain...

First of all, averaging isn't important here. I don't know why you've
done that at all, so I'll just work with the total weight, and how far below
the 'all gold' weight it would be.

Consider the situation where you have 4 bags with 2 coins in, A, B, C, D.
For the sake of argument, bag A is the fake. You follow the same regime,
and have bags A and B in one pile, and bags C and D in another.

Swap coins: (letter signifies a coin from that bag)

AC CA
DD BB

You now weigh the piles. Aha! You are under by the equivalent of one
fake coin. Therefore the fake bag must be bag A. Except, of course, you
can't say that. You only know one coin is fake - which means it could just
as easily be the coin from bag C. You still do not know which bag has the
fake coins.

Extend this out to six bags:

AAD DDA
BEE EBB
FFF CCC

Pick a different bag, this time bag B, as the fake. You weigh pile one,
and are under the amount expected by the equivalent of one coin. Aha! It
must be bag B! Except, no, hang on, it could also be bag D, that only
contributed one coin to that pile. Even supposing you weighed pile two, you
would be under by the equivalent of two fake coins - which could have come
from bag B or D.

You can keep extending this out, and come up against similar problems -
there are almost always two possible fake bags. You have narrowed it down
to two possibilities from 100, so you have 98 bags of gold, but not all 99.

The special cases are when the bag of fake gold is bag 50 or bag 100. If
the fake gold is in bag 50, you're in luck - you will be bang on the right
weight, know there are no fake coins, and the only bag which didn't
contribute is bag 50. If the fake is bag 100, you'll be under by the
equivalent of 50 fake coins, and again know exactly which one it is.

Unfortunately, this method cannot guarantee you the answer. You are
guaranteed to knowing for sure 98 bags of gold, so you're ahead in that
sense. However, the odds of finding the fake gold are 50-1 - you've halved
this from 100-1 from just picking a bag at random and weighing, but it
doesn't seem like good odds to me.

When we talked about this on IRC the other night, this was the solution I
came up with, and then refused to tell you once I realised it was wrong.
So, yes, I did get your answer, but then I noticed it was wrong... ;)

If anyone spots somewhere I've gone wrong, please let me know so I can
kick myself.

[Ben F-W]

Post by Trevor Roberts

Post by Ben F-W
Let me know if I'm wrong!

You're wrong. HTH, HAND.

Hum. Was scared of that.

Post by Trevor Roberts
When we talked about this on IRC the other night, this was the solution I
came up with, and then refused to tell you once I realised it was wrong.
So, yes, I did get your answer, but then I noticed it was wrong... ;)

Yes, and that's why I wanted to know - just in case it didn't work for
some reason I hadn't worked out! That's the trouble with not discussing
a solution with people - I don't get to think it through properly.
Sorry, everyone!

Ben

[Ben F-W]

Post by Trevor Roberts

Post by Ben F-W
Let me know if I'm wrong!

You're wrong. HTH, HAND.

Hum. Was scared of that.

Post by Trevor Roberts
When we talked about this on IRC the other night, this was the solution I
came up with, and then refused to tell you once I realised it was wrong.
So, yes, I did get your answer, but then I noticed it was wrong... ;)

Yes, and that's why I wanted to know - just in case it didn't work for
some reason I hadn't worked out! That's the trouble with not discussing
a solution with people - I don't get to think it through properly.
Sorry, everyone!

Ben