Is Probability Equally Distributed Among Unknown Options?

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Discussion Overview

The discussion revolves around the concept of probability distribution among unknown options, specifically in scenarios where choices are limited to a few alternatives. Participants explore whether it is valid to assume equal probability among options without additional information.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that if there are three options (A, B, C), one might assume a 33% probability for each, leading to a 66% probability for the combined outcomes of A and B resulting in Y.
  • Another participant counters that such an assumption of equal likelihood cannot be made without additional information confirming that each option is equally likely.
  • A participant provides a personal example from a cafeteria setting, indicating that their choice is heavily skewed towards one option, thus illustrating that probabilities can vary significantly based on context.
  • Further inquiry is made about the implications of lacking information to determine probabilities, questioning whether this makes the situation unsolvable.
  • It is noted that to arrive at a solution, one must have information about the likelihood of each alternative being chosen, or at least confirmation of equal likelihood.

Areas of Agreement / Disagreement

Participants generally disagree on the assumption of equal probability among options without additional context or information. Multiple competing views remain regarding how to approach the problem of unknown probabilities.

Contextual Notes

The discussion highlights the dependence on additional information to establish probabilities, as well as the potential for skewed distributions based on personal or contextual factors.

KarimSafieddine
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Hey, I am new to this forum and I wanted to ask on something important for my work concerned with sociology.

I believe I know this but I am just making sure.

Let's say you don't know what a person is going to do next, but you do know it's between 3 options: A, B and C.
Do we say, in this case, that there's a 33% for each to occur?
Moreover, A and B always lead to one result Y.
Hence, Y has a 66% to occur?

That's it! I know it's very simple but I just have to get this off my head, I am not really sure this is the right section to place it though.

Thanks.
 
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No, you cannot say this in general unless you have some additional information such as it being equally likely for the person to pick each of the alternatives. For example, in the cafeteria at my institute there are always three options for lunch, one meat dish, one fish dish, and one vegetarian. The probability that I pick the meat dish on any given day is of the order of 98%.
 
Orodruin said:
No, you cannot say this in general unless you have some additional information such as it being equally likely for the person to pick each of the alternatives. For example, in the cafeteria at my institute there are always three options for lunch, one meat dish, one fish dish, and one vegetarian. The probability that I pick the meat dish on any given day is of the order of 98%.

What if you don't have any information to prove that one is more likely than the other? Does that render this unsolvable?
 
KarimSafieddine said:
What if you don't have any information to prove that one is more likely than the other? Does that render this unsolvable?

Yes, in order to solve it you need to be supplied with the information that the alternatives are equally likely (or if not, what their probabilities are).
 
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