Bugger! Well done KB, it's been about 25yrs since I did my statistics and I guess my rustiness shows in my inelegant solution.
Bugger! Well done KB, it's been about 25yrs since I did my statistics and I guess my rustiness shows in my inelegant solution.
OK geometry it is!Originally Posted by Kevin Bonham
Actually, I won't try composing any more problems but I do have a couple of favorites that like minded individuals might enjoy. The first one is a geometry one. It's cute and not very difficult.
I'll post it tonight. (Perhaps I should start a new thread).
So einfach wie möglich, aber nicht einfacher - Albert Einstein
Firstly let me congratulate all of you on excellent submissions. =D> =D> =D>
Before someone asks why I didin't bother entering there were two reasons.
Firstly I was/am busy doing ratings related stuff and secondly and more importantly I had seen this problem previously in a slightly modified form and therefore did not think it was fair to put in a response.
With regards to Matt's solution let me say I have seen this reasoning applied in similar cases, although to be honest I had forgotten about it until seeing Matt's response. #-o
One feature it has is its sheer simplicity. 8)
I commend him on it. Well done Matt. =D>
One interesting aspect is the following.
From the figures it can be seen that
((2^18 )-1)/((2^19)-1) = (2^-1) - (1/(2^20 - 2^1))
To the less mathematically inclined it is far from obvious that this is true. :-s :?
That's pretty much covered in my derivation of the elegant solution. It goesOriginally Posted by Bill Gletsos
P = (2^(n-2) - 1)/(2^(n-1) - 1)
multiply by 2/2
P = (2^(n-1) - 2)/(2^n - 2)
Separting the numerator into 2^(n-1) -1 and -1
P = (2^(n-1) - 1)/(2(2^(n-1) - 1)) - 1/(2^n - 2)
Simplifying the first term you get
P = 1/2 - 1/(2^n - 2)
QED
But you're right - it is not immediately apparent. And it is pleasing that two different lines of reasoning can reach mathematical equivalents which are not obvious.
Actually I prefer the first form for calculating the probabilities but I favoured the second form in my solution as it more clearly demonstrates the monotonical increasing and bounded at 0.5 behaviour of the solution for n marbles.
BTW as a further postscript had Dr Evil ensured the existence of marbles of both colours in the bag by a different method then different probabilities might result. EG say he looks after 19 marbles have been put in and if they are all the same colour then he adds a marble of the opposite colour, else he add one randomly. This is a similar problem and a similar answer (different colour is favoured) but with a slightly different probability.
Just goes to show, you can't assume nothing.
So einfach wie möglich, aber nicht einfacher - Albert Einstein
But you hadn't published your elegant solution.Originally Posted by Barry Cox
Then what's this?
Originally Posted by Barry Cox
So einfach wie möglich, aber nicht einfacher - Albert Einstein
Q4. The Three Circles Puzzle
Three conguent circles of radius r, colinear and centres are 2r apart (ie each are touching the adjacent circle at only one point).
The point A lies on the line passing through the centres of the circles and the first cirlce, but not the middle circle. The point D is tangential to the third circle. B and C are the points where the line AD passes through the middle circle.
Find the length of the chord BC in terms of r.
The following diagram should help clear up any confusion...
Send solution to me via PM before nexts Wednesday night, 21-Jan at 24:00 GMT+11.
So einfach wie möglich, aber nicht einfacher - Albert Einstein
Not even one solution received yet. You guys are allergic to geometry, aren't you?
Have a crack at it, you might be (pleasantly) surprised.
If anyone wants a hint, send me a PM.
So einfach wie möglich, aber nicht einfacher - Albert Einstein
Yes, I'm in IT, not a teacher.Originally Posted by arosar
Puzzles are a hobby though. Cryptic crosswords, maths puzzles, logicical puzzles, etc, doesn't matter much.
BTW I only got one response to the geometry puzzle. You know who you were. It wasn't right.
So here are some hints.
The answer is a rational coefficient of r.
It can be reasoned using Euclidean geometry and Pythagoras' theorem.
Key facts.
A line originating from a circle which is perpendicular to a chord of that circle, bisects the chord.
The same is true as the chord becomes diminishingly small (i.e. the perpendicular to a tangent passes through the centre).
So einfach wie möglich, aber nicht einfacher - Albert Einstein
When's the new deadline BJC?
When the Noachian deluge of entries abates.
The next puzzle will be pure number theory so the sooner I get the entries the better.
So einfach wie möglich, aber nicht einfacher - Albert Einstein
Is this over?Originally Posted by Barry Cox
No. Still no entries since the massive hints I dropped. Perhaps geometry is even less popular than I thought.Originally Posted by Matthew Sweeney
I'll post the solution when I find the text for the next puzzle which is much more difficult but related to number theory so might be more popular.
BTW Welcome back. You missed the start of the Club Rapid.
So einfach wie möglich, aber nicht einfacher - Albert Einstein
There are currently 1 users browsing this thread. (0 members and 1 guests)