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#1 2006-01-19 13:07:20

asv
Novice

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A tricky problem to get you all thinking...

I was thinking, pi has been calculated to 1,241,100,000,000 decimal places. There are 6,670,903,752,021,072,936,960 possible valid sudoku grids. What is the probability that there are 81 consecutive digits that are a valid sudoku within the digits of pi we already know?

Hope that makes sense.

asv
(PS does anyone know a busier maths forum that might be better suited to problems like this?)

#2 2006-01-19 13:54:37

ryos
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Re: A tricky problem to get you all thinking...

That doesn't quite compute. Doesn't probability only apply to random situations? Pi isn't random, and neither, I presume, are sudoku grids. I mean, that's like saying, "What is the probability that 2 + 2 = 4?" It's 1 of course; we know it will always come out a certain way.

So, either there is a sudoku grid in the digits of pi we already know, or there isn't.

On the other hand, if pi really never ever repeats itself, then we can also be certain that a sudoku grid is bound to crop up eventually.

What's a sudoku?

El que pega primero pega dos veces.

#3 2006-01-19 14:38:05

Ricky
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Re: A tricky problem to get you all thinking...

"Pi isn't random, and neither, I presume, are sudoku grids."

It is random, but only once.

If you think of each digit of pi as a number, you cannot predict the next number by knowing the one you have.  That is part of the definition of random.

Of course, the second part of that is that it can't repeat when starting over, which pi doesn't.

Now it's more complex than this.  Many people think there are patterns in the digits of pi.  For example, a lot think that 9 will drop out at some point completely.  This is still under current study.

But let's just rephrase the question.  If there 1,241,100,000,000 numbers in a row, what's the chance of it being a sudoku grid, when layed out left to right, top to bottom?

This is fairly complex because when you look at the grid starting off at the first digit, it is mostly the same as the grid starting off at the second digit.  If the first digit contains an invalid number (like 99) at the very end, then so will the second, and the third, and the fourth, and so on.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."