Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.
Binomial proof...hmm, I guess zero is divisible by p.
Actually, that expression comes up during the proof for Fermat's Little Theorem, spoiler here:
Note x|0 (x divides 0) is true for all non-zero x.
EDIT: For questions such as yours, the answer is usually something to do with Fermat's Little Theorem or its corollary, Wilson's Theorem.
The problem with your method is that you essentially make a yearly average of the first quarter, a yearly average of the three last quarters, and then simply average the two. This is wrong, because that makes the average of the first quarter become just as important as the average of the three last quarters when calculating the overall average.
Back home in Sci/Tech!
I suppose the move to OT was good for thread publicity.
On topic: you shouldn't really use "average" as a term in the maths thread - the wikipedia page lists 14 different types of averages
originally it is someone else's question
but it is per year.
Seriously, forget about the quarters, they don't matter. Think of it as two unspecified timeperiods which together is one full year.
So what? When you add (average expenses per employee for jan-mar) + (average expenses per employee for apr-dec), you will automatically get (average expenses per employee for jan-dec) which is the number you're looking for. There's no need to "scale" anything.
When we apply Fermat's Little theorem we get (1 + n)^p - n^p - 1 == 1+n -n -1 ==0 mod p
where == denotes congruence modulo p.
Just from some scribblings on a piece of paper, doesn't this work for p is any natural number (ie not just primes..)
Oh and I didn't use fermat's little theorem, just expanded binomially, then simplified.
EDIT: actually scratch that, I made a bad assumption
(1+1)^4 - 1^4 - 1 = 16 - 1 - 1 = 14 == 2 (mod 4)
Dutchfire is visiting ParadigmShifter in Liverpool to play Scrabble with him. They're going to determine the starter of the first game by tossing a coin, and just as soon as Dutchfire has reached a one euro coin from his pocket, ParadigmShifter stops him and says: "No, we can't use that coin, we're in the UK, we must use pound sterling", and takes a coin out of his pocket. He calls tails and is about to toss the coin when Dutchfire grabs his hand. "I saw what you did there!", he ejaculates. "The coin isn't fair. Due to your queens fat face on the other side, it's bound to land tails more often". ParadigmShifter is infuriated by the insult, but swallows it because of the embarrassment of getting caught: the coin indeed lands more often tails.
Can you help them to use PS's coin to determine the starter of the game in a fair way?
Do you know how many times it falls on tails vs heads? If so then you can just toss it 10 times and if tails comes up more than the expected number times then tails wins.
What comes to mind is to spin the coin instead of tossing it. Then hit the coin from the side to knock it over. This way the side it falls on should depend on the silhouette of the coin, instead of the weight, and silhouette is the same from both sides. A problem with this approach is that it would not work if you hit the coin from the top, as in that case the heavier side is likely to be face up.
They have no knowledge of the probability of heads vs. tails.
They also refuse to use any methods that aren't reducible to coin toss since they want a method that can be generalized to other mediums too (die for example).
You're on the right track that they are going to toss the coin multiple times.
Well you could "calibrate" it by flipping 100 times or something, to figure our roughly the probabilities, then flip it 10 times and see which beats the calibrated probabilities.
That won't do either, they want an exact method. There is at least one (if we admit that there in general is a fixed probability for the coin (and that's the reason I find this thing fascinating)).
Can you flip it 100 times, and then each player guesses whether after 100 more tosses, the number of tails comes out higher or lower than the first 100?
What if they both want to guess the same?
How do they determine which one gets to guess first?
Hint: It isn't predetermined how many times the coin is tossed. Except that the number is even.
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