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

1. Mar 11, 2016

nolxiii

Like exponentially less likely each plank unit of time, but never reaches 0 probability. Will it ever happen?

2. Mar 11, 2016

micromass

Staff Emeritus
Best you can do is find the probability of it happening. For that, we'll need to know the specific distribution.

3. Mar 11, 2016

nolxiii

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.

4. Mar 11, 2016

micromass

Staff Emeritus
Depends on the specific numbers. It might, it might not.

5. Mar 11, 2016

nolxiii

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)

Last edited: Mar 11, 2016
6. Mar 12, 2016

Stephen Tashi

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.

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 ?

7. Mar 12, 2016

nolxiii

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?

Last edited: Mar 12, 2016
8. Mar 13, 2016

Stephen Tashi

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/

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.

9. Mar 13, 2016

nolxiii

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.)

Last edited: Mar 13, 2016
10. Mar 14, 2016

Stephen Tashi

Ok, that's understandable.

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

$\lim_{n \rightarrow \infty} { \sum_{i=1}^n \big( \frac{1}{2} \big) ^{xy^{i-1}} }$
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. Mar 16, 2016

nolxiii

doesn't sound like anyone over there knows either. will i get one of those fields prize things if i solve this?

12. Mar 23, 2016

Stephen Tashi

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".