Can I Predict the Correct Dice Using Probability Distributions?

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Discussion Overview

The discussion revolves around predicting which of two dice was rolled based on their differing probability distributions after observing the outcome of a single roll. Participants explore the application of probability theory, particularly Bayesian inference, to determine the likelihood of having rolled a specific die given the result.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant describes a scenario involving two dice with unknown probability distributions and seeks to determine which die was rolled based on the outcome.
  • Another participant suggests that Bayesian inference is a suitable approach for this problem.
  • The original poster expresses familiarity with Bayes' theorem and attempts to apply it to their scenario, defining the relevant probabilities.
  • A later reply confirms the correctness of the original poster's application of Bayes' theorem in this context.

Areas of Agreement / Disagreement

Participants generally agree on the applicability of Bayes' theorem to the problem, but the discussion does not resolve the broader question of how to express the probability of being correct for arbitrary probabilities.

Contextual Notes

The discussion does not fully address the complexities involved in calculating the probability of being correct for arbitrary probabilities, leaving some assumptions and mathematical steps unresolved.

aaaa202
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Say I have two dice which have different probability distributions but I don't know which probability distribution belongs to what dice.
I now throw one of them exactly one time and the result is x. I know that die A has probability pA(x) for rolling x and die B has probability pB(x).
I want to find out if the die I rolled was die A or B. Obviously my best strategy is to guess the die for which the probability for rolling x is the bigger. But what is my probability for being correct? I have been trying to figure out this problem for days. Obviously if pB(x)=0 while pA(x)>0 I have 100% chance of being correct. Moreover if pB(x)=pA(x) then I have 50% chance of being correct. But is there an expression for arbitrary probabilities and how do I find it?
 
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Nice question. This would be a perfect application for Bayesian inference.

Are you familiar with Bayes theorem?
 
Yes P(XlY) = P(YlX) * P(Y)/P(X)

So in my case P(Alx) is the probability that the die is A given that it shows x, while P(xlA) is the probability that the die shows x given that it is die A and P(A) is the probability that it is die A given no prior information, i.e. P(A)=0.5 and P(x) is then the probability for tossing an x given no prior information, which would be 0.5*(pA(x)+pB(x)).. hmm.. is that correct?
 
Yes. I think that is correct.
 

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