I will roll a dice, and ask you, what is your probability of guessing the roll at random. The answer is 1/6, obviously. Then I will look at the result, and tell you, if the answer is 5 or not. What is the probability that you will guess the number right afterwards?

In Bayesian, you can encode the extra info into the prior, saying that the prior p = 1/5 if result is not 5, and 0 otherwise. Then the posterior being the normalized product between normal dice roll and the prior will be the same as the prior, which makes sense, you should have 1/5 chance to guess the dice correctly, if, for example, you know it is not 5.

Are there other ways of solving this problem. In particular, I am looking for a formal frequentist solution.


The main idea here is that we had $6$ possibilities all with equal weight, but one was eliminated, so now we simply have $5$ possibilities all with equal weight.

Generalizing this idea, let there be a set of $n$ probabilities $P_i$ with $$\sum_{i=1}^n P_i=1$$ Now let's subtract the final probability $P_n$, thus eliminating it from the list, we get $$\sum_{i=0}^{n-1}P_i=1-P_n$$ In order to get the new probabilities $Q_i$, the new probilities must still sum to one. We must therefore divide each side of the equation by $1-P_n$ as follows. $$\sum_{i=0}^{n-1}\frac{P_i}{1-P_n}=1$$ Thus each new probability after eliminating $P_n$ is $$Q_i=\frac{P_i}{1-P_n}$$

  • $\begingroup$ Hey Isaac. Your derivation makes sense, but my question is a bit deeper than that. In particular, what is the formal law that you apply when you re-scale the probabilities by their sum, after one has been eliminated? Intuitively it makes sense, but I want a little more theoretical explanation $\endgroup$ – Aleksejs Fomins Apr 10 '17 at 14:01
  • $\begingroup$ I think it's just the rule that probabilities must add up to one. Bayesians and frequentists would both agree on this matter $\endgroup$ – Isaac Browne Apr 10 '17 at 14:25
  • $\begingroup$ I don't disagree with that. I disagree that it is obvious that in order to force the probabilities to sum up to 1 it is correct to normalize them. $\endgroup$ – Aleksejs Fomins Apr 10 '17 at 15:28
  • $\begingroup$ Probabilities are just weights that something will happen. Thus, the only way to change the numbers while keeping the same weights is to multiply each by the same value. $\endgroup$ – Isaac Browne Apr 10 '17 at 17:50

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