Bright Sparks

[JohnW2]

When you look back at the monty hall it is very simple. Your first guess gives you a 1 in 3 chance of winning therefore a 2 in 3 chance of losing. Changing you mind after seeing the goat then gives the opposite of this therefore 2 in 3 chances of winning and 1 in 3 chances of losing. 1 in 3 becomes 2 in 3 and 2 in 3 becomes 1 in 3. (Just hope your first guess is wrong then change your mind) It is so simple when you kick yourself.

Let's leave Monty Hall on the sideline for at least a while, since it has the capacity to sidetrack us into dozens of pages of discussion! Suffice it to say that far more eminent statsiticians than either of us have, over the years, found it anything but 'simple'!

I thought the piles (here wrongly assuming again I am) had to stay in tens although it may be permitted to weigh more than one at any time ie 10,20,30, 40 or 50.
Hmm need a rethink now

As I said, you can have as many piles as you like, with as many coins in each pile as you want, provided only that there are at least as many coins in each pile as there are numbers of piles. You can weigh as few or as many coins as you want in the 'one weighing', and you don't have to keep them in their 'piles' for the weighing (i.e. if there are 10 in a pile, anything from zero to all 10 of them may be included in your one weighing).

Kind Regards, John.

 

[JohnW2]

Yes it is possible, and the answer confirms that for the method to work there must be more coins in a stack than there are stacks.

Not 'more than' but 'at least as many as'

Kind Regards, John.

 

[EFLImpudence]

Not 'more than' but 'at least as many as.

Surely there could be 'one fewer coins in each stack than the number of stacks'.

There could in my method. (I think)

 

[JohnW2]

Not 'more than' but 'at least as many as.

Surely there could be 'one fewer coins in each stack than the number of stacks'. There could in my method. (I think)

Yes, you're right - albeit with a variant of the answer that people usually give.

Kind Regards, John.

 

[bernardgreen]

Surely there could be 'one fewer coins in each stack than the number of stacks'.

There could in my method. (I think)

You are right, I was not considering zero as being a valid contribution to the result. If it isn't stacks 1 2 3 or 4 then it has to be stack 5

 

[JohnW2]

You are right, I was not considering zero as being a valid contribution to the result. If it isn't stacks 1 2 3 or 4 then it has to be stack 5

Indeed - and if that's not a good enough clue for those who have not 'got it' yet, I'm not sure what would be

Kind Regards, John.

 

[ebee]

Bingo got it - now I`ve been freed fro 10 coins remaining in each stack thereby each weighing being of multiples of ten coins.
Any number of coins you want on the pan makes it simple to do.

I have had my Archimedes Moment,


Yabadab a doo doo

 

[ebee]

It just goes to prove that to ASSUME makes an ASS of U and ME - Well it certainly made a ASS of me.
That will teach me not to make Assumptions in future folks

 

[ebee]

"all the coins in 4 of the stacks weigh exactly 10g, but all of those in the fifth stack weigh 11g."

Another assumption I made was the figures of 10g & 11g applied to each individual coin on its own and not the total number of coins in the piles (10) although this matters not providing you make the same assumption apply to all piles therefore all piles weigh either 10g or 11g per coin or per stack of 10 coins.

 

[ebee]

"ebee wrote:
The figure 0.4 ohms gives a length of 20.4918 metres for two conductors (R1 + R2) 12.1/1000 + 7.41/1000 = 19.52/1000 ohms per metre.Whoops 19.52 ? I shoulda said 19.51, silly me early in the morning . gives 20.5023.

Your arithmetic is correct, but I'm not sure what you think you're doing by adding together the resistivities of the two CSAs."


John,

I was adding two x 2.5mm (at 7.41 ) to get R1 + Rn and adding 12.41 to 7.41 to get R1 + R2 .

 

[ebee]

Dizz,
I am not sure why you seem so intent on using simultaneous equations.
Don`t get me wrong, they can be useful , the sheep question is a good example of finding the answer by using them although as a child I used trial & error to guess it.

 

[ebee]

Dizz,
anyway, what answers were you looking for in your opening question?
seems to me you were working on a negative length of cable for resistances to tally unless we got a typo creeping in.
BAS did ask how you got negative resistance

 

[bernardgreen]

I have had my Archimedes Moment,
Yabadab a doo doo

That reference to a bath bought a naughty idea into my mind. What would ( or could ) the plumbing section make of that question. ?

 

[JohnW2]

Bingo got it - now I`ve been freed fro 10 coins remaining in each stack thereby each weighing being of multiples of ten coins. Any number of coins you want on the pan makes it simple to do.
I have had my Archimedes Moment, Yabadab a doo doo

Good for you - I told you that the answer would become very obvious once you thought of it - as is usually the way with these things!

Kind Regards, John.

 

[JohnW2]

It just goes to prove that to ASSUME makes an ASS of U and ME - Well it certainly made a ASS of me. That will teach me not to make Assumptions in future folks

Indeed, but I don't think you shoud feel bad about it! Don't forget that many puzzles like this one rely on the fact that many/most people will probably make certain assumptions, so that lateral thinking is often needed, particularly with great care to not 'invent' conditions (e.g. that one has to weigh all 10 coins in a pile) that haven't actually been stated.

It's like this one, which probably does not work anything like as well today as it did a few decades ago when it first appeared. In very simplified (a lot of deliberately distracting irrelevant detail removed) form it is:

• A man and his son are travelling in a car which crashes. The man dies instantly and the son is seriously injured. When the son gets to hospital, the surgeon says "My god, that is my son". The question is ... 'what is the relationship between the surgeon and the patient?'

When this one first appeared, the great majority of people made a false assumption which caused them to struggle a lot before realising the obvious answer. There are very many puzzles based on such 'common assumptions'.

Kind Regards, John.