Why Bayes' Theorem gives me a logically wrong answer

Question

I will first talk about the example at hand, and then provide why I feel the answer is logically wrong.

Say, there are 4 dices in a bag. One dice with 2 sides. Two dice with 4 sides. One dice with 6 sides. Given that, what is the probability of the first number rolled being 2?

The answer would be simple Since all 4 dices can have 2, and the total number of possibilities is 16. The Probability of rolling a 2 would then be $\frac{4}{16} = \frac{1}{4}$

So a follow-up question to that is, Given that the rolled number is 2, what is the posterior probability of the dice being 6 sided? For which I approached with Bayes' Theorem

D = the Side, N = the number.

$P(D=6|N=2) = \frac{P(N=2|D=6)*P(D=6)}{P(N=2)}$

$P(D=6|N=2) = \frac{\frac{1}{6}*\frac{1}{4}}{\frac{1}{4}} = \frac{1}{6}$

Hence, from Bayes' Theorem there is a $\frac{1}{6}$ probability of it being a 6 sided dice. However, realistically/logically speaking. Shouldn't the probability of it being a 6 sided dice, still be $\frac{1}{4}$. As the number 2 exists in all 4 of the dices.

So I was wondering if I missed a concept from Bayes' theorem? Or is my math/formula wrong? Or is it that Bayes' theorem's answer is right and my logic is wrong?

Answer 1

You are asking two different questions.

For the first part, it is incorrect to assume that each of the $16$ die faces are equally likely to be observed. This is because the observed value is the result of a two-step random process:

1. You pick one of the four dice out of the bag at random, each choice being equally likely.
2. You roll the chosen die and observe its value.

So, because the number of sides on the dice are different, once you pick a die out of the bag, the set of outcomes you can observe depend on which die was chosen. The probability of observing a $2$ is not just $4$ faces out of $16$, because not all of the faces are equally likely to occur. The face with $2$ on the six-sided die will only have a probability of $(1/4)(1/6) = 1/24$ of being observed because you have to pick the six-sided die first, then roll a $2$ with it; whereas the face with $2$ on the two-sided die has a probability of $(1/4)(1/2) = 1/8$ of being observed, since you pick that die with probability $1/4$ and then you have a $1/2$ probability of getting the $2$.

For the second part of your question, allow me to frame the question this way:

Suppose instead you have just two dice, one that gives a number in $\{1, 2\}$ with equal probability, and another one that gives a number in $\{1, 2, \ldots, 1000\}$ with equal probability. You pick a die at random, roll it, and the outcome is $2$. What is the probability that the die that was picked was the thousand-sided die?

Clearly, the fact that you obtained a $2$ is highly informative of the posterior probability that the die that was picked was the two-sided die. Just because both dice have $2$ does not mean that it is equally likely that either die could have been picked; it is much, much more unlikely that you picked the thousand-sided die and rolled a $2$, than having picked the two-sided die and rolled a $2$.

Answer 2

Your calculation of $P(N=2)$ is wrong. There are 3 possibilities, either you get $D=6, D=4, D=2$ with probabilities $\frac{1}{4}, \frac{1}{2}, \frac{1}{4}$ respectively. Now $P(N=2|D=6) = \frac{1}{6}$, $P(N=2|D=4) = \frac{1}{4}$, $P(N=2|D=2) = \frac{1}{2}$

Finally we get $P(N=2) = P(N=2|D=6)P(D=6) + P(N=2|D=4)P(D=4) + P(N=2|D=2)P(D=2) = \frac{1}{6}\cdot \frac{1}{4} + \frac{1}{4}\cdot \frac{1}{2} + \frac{1}{2}\cdot \frac{1}{4} = \frac{1}{24} + \frac{1}{8} + \frac{1}{8} = \frac{7}{24} > \frac{1}{4}$

This makes sense logically because for example the $P(N=5) = P(N=5|D=6)P(D=6) = \frac{1}{24}$ as we can only roll a $5$ if we first pick $D = 6$