I As time goes on, a given event becomes less and less likely

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The discussion centers on the probability of an event occurring over time, emphasizing that while the likelihood decreases exponentially, it never reaches zero. Participants explore whether the probability of an event approaches 1 as time approaches infinity, depending on specific distributions and initial conditions. A coin-tossing analogy illustrates the complexity of defining the event and calculating probabilities, particularly when outcomes lead to further iterations. The conversation also touches on mathematical series and convergence, questioning how to accurately model these probabilities. Ultimately, the challenge lies in precisely defining the event and understanding the underlying mathematical principles.
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Like exponentially less likely each plank unit of time, but never reaches 0 probability. Will it ever happen?
 
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Best you can do is find the probability of it happening. For that, we'll need to know the specific distribution.
 
micromass said:
Best you can do is find the probability of it happening. For that, we'll need to know the specific distribution.

So I guess to be more specific, will the probability of it ever happening approach 1 as time goes to infinity? The probability being so unlikely that you can't comprehend it and becoming even more unlikely as time moves forward, the specifics of exacly how improbable it is (i think) being irrelevant as long as its greater than 0.
 
nolxiii said:
So I guess to be more specific, will the probability of it ever happening approach 1 as time goes to infinity?

Depends on the specific numbers. It might, it might not.
 
Ok, probability at t = 0 is 1/(10^^^^^^^^^^^^^123)^1, at t = 1 it is 1/(10^^^^^^^^^^^^^123)^2, at t = 2 is 1/(10^^^^^^^^^^^^^123)^3, and so on.

Is there a way to find the threshold where it no longer approaches 1?

Edit: Is this equivalent to asking if the series converges using a root test? (http://www.mathwords.com/r/root_test.htm)
 
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nolxiii said:
As time goes on, a given event becomes less and less likely

Will it ever happen?
You need to use more precise language in order to arrive at a specific mathematical question.

Let's consider a physical model. Suppose we have a coin and we execute a sequence of tosses of the the coin. For some reason, the probability of heads p_n on the n -th toss changes each time we toss the coin - perhaps after each toss we hit the coin with a hammer to bend it.

What "event" are you asking about ?

For example, are you asking about the probability that a head will occur on some toss ? (That probability won't be zero if the probability of a head on the first toss is not zero)

Are you asking about the probability that "eventually" no heads will occur ? To define what that means with mathematical precision still needs some work. Can you define it as a limit ?
 
Ok, gotcha I think. Here goes..

Say I have 10 coins and I flip all of them. If any of them land on tails each coin splits into 10 new coins, and then I flip all of them again, and continue on doing this until a round where every single coin lands on heads. What is the probability that I will ever stop flipping coins as the number of flips goes to infinity?

Or to generalize, I have X coins and if any are tails they split into Y new coins. (or even instead of coins I have dice with Z number of sides and I roll until each die lands on 1)

Edit: Been a while since I've taken math or stats but will try to write this out, probably incorrectly. (And not sure how to put the ∞ above the Σ)

Σi=1 i =(1/z)xyn-1

^ is that the right thing to solve /in any way intelligible?
 
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nolxiii said:
Edit: Been a while since I've taken math or stats but will try to write this out, probably incorrectly. (And not sure how to put the ∞ above the Σ)

Using the editor's features, I'm not sure how to do that either. It's worth learning about LaTex. Many math oriented forums on the internet use it. https://www.physicsforums.com/help/latexhelp/

Σi=1 i =(1/z)xyn-1
is that the right thing to solve /in any way intelligible?

I don't think the idea behind that is correct. For example, suppose you start with 5 coins and split each coin in two each time it lands tails. What's the probability that you throw all heads on the the second attempt ? On the first toss you might have tossed 5 tails and after splitting, you toss 10 coins on the second toss. Or perhaps on the first toss you tossed 3 heads and 2 tails and thus have 3 + (2)(2) = 7 coins to throw on the second toss. So you can't assume that on the i-th toss that there is a single known number of coins to be thrown.
 
Sorry, to be clear and to keep it simple, each coin splits no matter what it lands on, but I keep going until all are heads.

Would I expect the example with 10 coins splitting into 10x more each round to ever conclude?

(And I do appreciate you taking the time to even partially entertain my random curiosities.)
 
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  • #10
nolxiii said:
Sorry, to be clear and to keep it simple, each coin splits no matter what it lands on, but I keep going until all are heads.

Ok, that's understandable.
What is the probability that I will ever stop flipping coins as the number of flips goes to infinity?

The probability that you stop flipping after N or fewer tosses
= \sum _{i=1}^n pr( \ stop\ after\ exactly\ i\ tosses)
= \sum_{i=1}^n pr( \ toss\ heads\ with\ each\ of\ xy^{i-1}\ coins
= \sum_{i=1}^n \big( \frac{1}{2} \big) ^{xy^{i-1}}

Your ask the value of
\lim_{n \rightarrow \infty}<br /> { \sum_{i=1}^n \big( \frac{1}{2} \big) ^{xy^{i-1}} }
which is also denoted by \sum_{i=1}^\infty \big( \frac{1}{2} \big) ^{xy^{i-1}}

The general problem is how to deal with series like \sum_{i=0}^\infty C^{D^{\ i}}

If that question were asked in the Calculus section, someone would probably have some ideas!
 
  • #11
doesn't sound like anyone over there knows either. will i get one of those fields prize things if i solve this?
 
  • #12
nolxiii said:
doesn't sound like anyone over there knows either. will i get one of those fields prize things if i solve this?

We might get a W. C. Fields prize if we don't start out correctly.

I didn't compute the probability you asked about. The probability of stopping exactly on the N th step is the probability of: Not stopping on the previous N-1 steps and also tossing all heads on the N th step. I didn't account for the condition "not stopping on the previous N-1 steps".
 
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