Comment by skywhopper
Comment by skywhopper a day ago
50% chance of being right is equivalent to a coin-flip.
Comment by skywhopper a day ago
50% chance of being right is equivalent to a coin-flip.
No, if there are 100 order statuses there are 99 wrong choices and 1 right choice.
Additionally, the distribution of the choices is not guaranteed to be equal.
If you assume equal distribution, you have a 1% chance of being right and a 99% chance of being wrong.
There is sample space (choices) so for example 100 different status labels and event space (how the system grades your choice), so right and wrong.
My statement is true no matter how many choices are there, or how skewed the probabilities are. Your count of 99 incorrect labels is perfectly fine but it lives in sample space.
Arguing that there are 99 incorrect answers doesn't refute that evaluation is binary.
So counting 99 wrong labels tells us how many ways you can miss, but probability is assigned, not counted. Once a choice is made the system collapses everything to the two outcomes "correct" or "incorrect", and if the right label happens to have 50 % probability then the situation is mathematically identical to a coin flip, regardless of how many other labels sit on the die.
Example with a weighted die and 99 incorrect answers:
Die Faces: 100
Weights: Right status face = 0.50, the other 99 faces share the other 0.50
P(correct) = 0.50 -> exactly the coin-flip
The 1/N rule only applies when all faces are equally likely, once you introduce weights, the number of faces no longer tells you the probability.
You don't have a 50% chance of being right rolling an N-sided weighted die.