Why Do We Use the Ratio of Favorable to Possible Outcomes in Probability?

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Discussion Overview

The discussion revolves around the rationale behind the definition of probability as the ratio of favorable outcomes to total possible outcomes. Participants explore whether this is merely a definitional choice or if it holds deeper significance, particularly in the context of various examples such as coin tosses and dice rolls.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether the ratio of favorable outcomes to possible outcomes is simply a definition or if it has special significance.
  • Others suggest that "favorable" outcomes are context-dependent and that probability relates designated outcomes to all possible outcomes.
  • A participant provides examples of coin tosses and dice rolls to illustrate their confusion about why favorable outcomes are in the numerator and total outcomes in the denominator.
  • Some argue that defining probability as favorable outcomes over possible outcomes allows for meaningful calculations, such as summing probabilities to equal one.
  • There is mention of an alternative concept of "odds," which could be defined differently but serves a different purpose than probability.
  • A historical reference is made to Laplace, who defined probability in this manner, indicating it is a foundational definition.
  • Participants express that if the definition were reversed, it would lead to nonsensical interpretations of probability.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the definition of probability as favorable outcomes over possible outcomes is merely a definition or if it has deeper significance. Multiple viewpoints remain regarding the implications of this definition.

Contextual Notes

Some participants highlight that the definition of probability relies on the context of favorable versus unfavorable outcomes, which may not be universally applicable. There are also unresolved questions about the implications of defining probability differently.

Juwane
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Why do we take the "probability of an event" ratio as favorable / possible outcome

If A is an event, then probability that A will occur is given by

P(A)= no. of favorable outcomes / total no. of possible outcomes

Is this just a definition, or is there some special significance in taking the number of favorable outcomes as the numerator and total no. of possible outcomes as the denominator? Is it because since the numerator will always be less than the denominator, it will be easier to work with the ratio?
 
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Definitions.

Given a sample space, the probability of the entire sample space is 1, the probability of any element in the sample space is [0,1].
 


farful said:
Given a sample space, the probability of the entire sample space is 1, the probability of any element in the sample space is [0,1].

And how does this relate to the numerator being the number of favorable outcomes and the denominator being the total number of possible outcomes?
 


Juwane said:
And how does this relate to the numerator being the number of favorable outcomes and the denominator being the total number of possible outcomes?

"Favorable" is a judgment within some context. A probability simply relates the number of designated outcomes to all possible outcomes. The designated outcomes could be favorable, unfavorable or neutral depending on your point of view.
 


My question is that why the favorable outcomes is in the numerator and total possible outcomes is in the denominator? Why not vice versa? Is it just a definition or does it have a significance?
 
Hi Juwane! :smile:

Because for two independent events, you want both P(A + B) = P(A) + P(B), and P(AB) = P(A)P(B), and favourable/possible is the only way of doing that. :wink:
 


I still don't understand. I like to understand through an example:

A coin is tossed. The probability that we will get a head is 1 over 2.

Another example:

A die is cast. The probability that we will get a 3 is 1 over 6.

Now my question is why, in the first example, it is 1 over 2 and not 2 over 1; and why in the second example it is 1 over 6 and not 6 over 1?
 
Juwane said:
… why in the second example it is 1 over 6 and not 6 over 1?

You could define it as 6 over 1, but then the probability of a 4 would be 6 over 1 also. and the probability of a 4 or a 3 would be 6 over 2, = 3 over 1.

And the probability of throwing the dice twice and getting a 4 first time and a 3 second time would be 36 over 1.
 


tiny-tim said:
You could define it as 6 over 1, but then the probability of a 4 would be 6 over 1 also. and the probability of a 4 or a 3 would be 6 over 2, = 3 over 1.

And the probability of throwing the dice twice and getting a 4 first time and a 3 second time would be 36 over 1.

So it's just the definition, right? Because it makes sense to say that the probability of getting a head in a coin throw is half, so it's 1/2, right?
 
  • #10
Juwane said:
So it's just the definition, right? Because it makes sense to say that the probability of getting a head in a coin throw is half, so it's 1/2, right?

Right. :smile:

Mathematicians can define anything they like.

But a definition, even if perfectly valid, is only useful, if it successfully models the real world. :wink:
 
  • #11


Well, you could take instead favorable outcomes / unfavorable outcomes and call it "odds".

So odds of 10 : 1 mean two probabilities of 10/11 and 1/11.
 
  • #12


By Gee!, by Gosh! It was Laplace who DEFINED the probability of an event as the favorible outcomes divided by the total outcomes! It is the basic definition.
 
  • #13


Example: What is the probability that heads will fall on one throw of a coin? From theoretical probability, P(heads)=1/2. This means that 1/2 or 50% of your tosses of a coin should land on heads, theoretically. The numerator must be the favorable outcomes and the denominator must be the total possible outcomes. This ratio is 1 to 2 or 1:2, meaning, 1 out of every 2 tosses of the coin should result in heads. If you had possible outcomes in the numerator the ratio would be 2 : 1 , which would mean that you could toss 2 heads in 1 throw with one die. Impossible.
 
  • #14


bcramton said:
Example: What is the probability that heads will fall on one throw of a coin? From theoretical probability, P(heads)=1/2. This means that 1/2 or 50% of your tosses of a coin should land on heads, theoretically. The numerator must be the favorable outcomes and the denominator must be the total possible outcomes. This ratio is 1 to 2 or 1:2, meaning, 1 out of every 2 tosses of the coin should result in heads. If you had possible outcomes in the numerator the ratio would be 2 : 1 , which would mean that you could toss 2 heads in 1 throw with one die. Impossible.

2 possible outcomes doesn't mean they are both heads.
 
  • #15


Juwane said:
A die is cast. The probability that we will get a 3 is 1 over 6.

Now my question is why... is it 1 over 6 and not 6 over 1?

1] What would it accomplish to have it defined as 6/1? Are there any ways this would be better?

2] There are a number of very useful reasons for defining it as 1/6. For example, you can add up all the possbilities and arrive at 1 again: 1/6+1/6+1/6+1/6+1/6+1/6 = 1. i.e. all individual possibilities add up to the total possible outcomes.

How would you do this if it were represented as 6/1?
 

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