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Probability in Pi

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It is a well-known fact that the number pi is an irrational number. This means that its decimal representation is infinitely long.

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664...

What is the probability that pi's decimal representation contains a sequence of digits ...0123456789....?

What is the probability that it contains ...33333333333333333333...?

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Posted · Report post

digits appear randomly.
The first digit in a prescribed sequence can always be found.
The next digit ... Omg ... will also eventually occur.
And the third.

So the answer is NOT 1/10n-1 where n is the length of the sequence.

It's 1.

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Posted · Report post

Any other thoughts?

Would you agree with the following generalization?

Any event with a non-zero probability p>0 of occurring on a single trial (no matter how small) will occur with the probability of 1 over an infinite number of trials.

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Posted · Report post

The Ergodic Theorem?

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Posted · Report post

Maybe it is. I don't know as I've never heard of it before

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Posted · Report post

I don't know that it applies, although it sounds related. As I recall hearing it, it holds that a system will visit every available state given enough time. A weaker "quasi" ergodic theorem says that a system will visit a finite region around every available state (implying there is a distance function of some type) given enough time.

The resolution of your conjecture may not be covered by either of these.

The resolution may rest on certain conditions being satisfied. It's a zero-times-infinity puzzle that might require either a strong intuition or great mathematical skill to answer. I have some intuition, but it's often wrong. That fact keeps me coming back to this site. ^_^

So I wonder whether the decimal expansion for e can occur within the expansion for pi.

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Posted · Report post

I think this argument should hold water.

Consider an event with any given small, but finite and fixed, probability of occurring on any given trial. Call that probability p, and the number of trials that you run n.

No matter how many trials you run, you can never be 100% sure that the event will happen. But you can chose any probability g which is good enough, say 95% probability that the event will occur after n trials, and simply solve for the value of n that would be a sufficient number of trials to reach a probability g that the event occurs.

The probability that the event will NOT occur after n trials is (1-p)n, so you can solve:

(1-p)n < (1-g)

n > log(base 1-p) of (1-g)

For any value of g you want, like 99.9999999999%, you can solve for a value of n and say that the event will occur with probability g after n trials. And I think that's about as much of a guarantee as you can ask for.

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Posted · Report post

I think this argument should hold water.

Consider an event with any given small, but finite and fixed, probability of occurring on any given trial. Call that probability p, and the number of trials that you run n.

No matter how many trials you run, you can never be 100% sure that the event will happen. But you can chose any probability g which is good enough, say 95% probability that the event will occur after n trials, and simply solve for the value of n that would be a sufficient number of trials to reach a probability g that the event occurs.

The probability that the event will NOT occur after n trials is (1-p)n, so you can solve:

(1-p)n < (1-g)

n > log(base 1-p) of (1-g)

For any value of g you want, like 99.9999999999%, you can solve for a value of n and say that the event will occur with probability g after n trials. And I think that's about as much of a guarantee as you can ask for.

Agree.

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Posted · Report post

Nice explanation, plasmid. Marking as solved.

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