What Are the Odds of an Event Occurring in Multiple Trials?

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Discussion Overview

The discussion revolves around calculating the probability of an event occurring in multiple trials, specifically focusing on the scenario of rolling a 100-sided die 100 times and determining the likelihood of rolling a specific number, such as a 9. The conversation includes both theoretical and mathematical reasoning related to probability distributions.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant poses a question about the probability of rolling a specific number (1/100) at least once in 100 iterations of rolling a 100-sided die.
  • Another participant suggests that the probability of rolling at least one nine can be calculated as 1 - 0.99^100.
  • A later reply confirms the focus on "at least one" and expresses interest in also knowing the probability of rolling "exactly" one nine.
  • One participant provides the formula for calculating the probability of rolling exactly one nine using combinations: C(100,1)(0.01^1)(0.99^99), and extends this to calculate for exactly two nines.
  • Another participant seeks a quick derivation for the probability of rolling at least one nine, indicating uncertainty about how to derive it from the combination formula.
  • One participant explains their reasoning by considering the complement of the event, stating that the probability of rolling at least one nine is 1 minus the probability of rolling no nines at all (0.99^100).
  • There is a suggestion that summing the series of combinations could yield the probability, but it is noted that this might be cumbersome.

Areas of Agreement / Disagreement

Participants generally agree on the approach to calculating the probability of rolling at least one specific number, but there is no consensus on the derivation methods or the exact formulas for different scenarios (e.g., exactly one versus at least one).

Contextual Notes

The discussion does not resolve the derivation methods for the probabilities, and assumptions about the independence of trials and the uniform distribution of outcomes are implicit but not explicitly stated.

Poop-Loops
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This has bugged me for quite some time now, but I've always shrugged it off. I've taken a course in statistics and have dealt with them in one way or another for a few years, so I KNOW I've seen the answer to this, but I can't remember it and don't know where to look for it.

Well, I could crack open my statistics book, but I figure this would go faster.

Anyway, my question is as follows: The probability of something happening is 1/100. You do 100 iterations. What is the chance of it happening?

So for example you have a 100 sided die, and you roll it 100 times. What's the chance of getting a 9? I know I can't be 1. But it has to get closer as you get to infinity, right? I just can't think off the top of my head how that formulation would go.
 
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What is the chance of getting exactly one nine or at least one nine?

I think at least one nine is 1-.99^100
 
That makes sense. And yeah, I meant "at least", but "exactly" would be a good one to know, too. Thanks.
 
I had to double check this one:

Exactly one nine is C(100,1)(.01^1)(.99^99)
then exactly two nines C(100,2)(.01^2)(.99^98)
and so on...
 
Oh, duh, it's a combination.

Okay, but do you have a quick derivation for getting at least 1 number? It's not obvious to me how you would get that from seeing the formula, but I suspect it has something to do with combinations anyway.
 
I didn't derive it. I just considered not no nines. So the probability of any result of 100 rolls (1) minus the probability of all the rolls being any number but nine (99^100). In this case nine, but it doesn't matter. The probability is the same for at least one of any single number 1-100.

I suppose you could sum the series of combinations. But that would be work I think.
 
Ahh I gotcha. Thanks. :)
 

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