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Topic: A cube of friends (Read 736 times) |
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A cube of friends
« on: Feb 8th, 2006, 12:44am » |
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A person has N friends where N (0, positive infinity). THe probabilities are not evenly distributed. You can determine the probability of having N friends such that P(N) = 3^(-N).
A person can be said to have a square of friends if they have 9 friends whose Initials can be written in a 3x3 grid. They have a cube of friends if they have 27 friends such that they can write their initials in a 3x3x3.
Imagine initials to be distributed evenly -- from A to Z.
What is the probability of a person having a square of friends. What is the probability of having a cube of friends.
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Grimbal
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Re: A cube of friends
« Reply #1 on: Feb 8th, 2006, 12:55am » |
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Sum(3-N) is 1.5 If you want a probability, you should use P(N) = 2/3·3-N
And I don't understand what exactly is meant by the condition "initials can be written in a 3x3 grid". Do they have 2-letter initials? Should the initials match by row and by column? What about a cube of friends? Wouldn't it mean they have 3-letter initials?
I'd be happy you clarify.
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Icarus
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Re: A cube of friends
« Reply #2 on: Feb 8th, 2006, 6:56pm » |
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I agree. Since there is nothing mentioned that prevents an arbitrary group of 9 people from writing their initials in a grid, your answers would be P(9) = 2*3-10 and P(27) = 2*3-28, respectively. So as stated, the only tricky thing to it is recognizing that P(N)=3-N is not a valid probability.
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Eigenray
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Re: A cube of friends
« Reply #3 on: Feb 8th, 2006, 9:16pm » |
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I'm guessing the intention was:
While (rand(0..1) < 1/3): add a random (iid uniform) element of [A-Z]x[A-Z] to your (multi-)set S of friends.
What is the probability that S contains a subset of the form AxB, with |A|=|B|=3 (counting multiplicity, maybe)?
The probability of S, chosen in this way from a set of size N, failing to contain a given (nonmulti-)set of size n is
(1-p)[sum]r=1n (-1)r-1 (nCr)/[1 - p(1-r/N) ].
But one can't simply raise this to the (26C3)2 power, as the events of S containing two given squares are not independent. And if we have to factor in multiplicity . . . urgh.
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