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Event $A$ represents my belief that a three-sided dice is fair, i.e: PMF of $P(A)$ is given by:

$f_A(w_k) = \frac{1}{3}$

where $w$ represents a single distinguishable outcome and $k$ represents one of the face of the die.

Some new evidence comes along (represented by event $B$) that suggests that my dice is not fair.

Given that the distirbution of P(B) is:

$w_1=\frac{1}{3}$, $w_2=\frac{1}{6}$, $w_3=\frac{1}{2}$

Calculate the actual probabilties of my dice.

Using Bayes:

$P(A | B) = \frac{P(A, B)}{P(B)}$

Step 1: Calculate likelihood:

$P(A,B) = P(A \cap B)$

Assuming A and B are indepdent then:

$P(A=w_1 \cap B=w_k) = \frac{1}{3}*\frac{1}{3}+\frac{1}{3}*\frac{1}{6} +\frac{1}{3}*\frac{1}{2} = \frac{1}{3}$

$P(A=w_2 \cap B=w_k) = \frac{1}{3}*\frac{1}{3}+\frac{1}{3}*\frac{1}{6} +\frac{1}{3}*\frac{1}{2} = \frac{1}{3}$

$P(A=w_3 \cap B=w_k) = \frac{1}{3}*\frac{1}{3}+\frac{1}{3}*\frac{1}{6} +\frac{1}{3}*\frac{1}{2} = \frac{1}{3}$

This implies that $P(A \cap B)= [w_1= \frac{1}{3},w_2= \frac{1}{3},3=w_3= \frac{1}{3}]$

Step 2: Calculate Marginal likelihood:

$P(B) = [w_1=\frac{1}{3}, w_2=\frac{1}{3}, w_3=\frac{1}{3}]$

This implies that $P(A | B) = [1,1,1]$!!!!


What am I doing wrong?

Thanks!

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  • $\begingroup$ Maybe I'm not getting this, but shouldn't $A \cap B = \emptyset$? Think of $P$ as being defined on $\mathbb{R}^3$, where $(x_1, x_2, x_3)$ represents bias of dice. $\endgroup$ – grndl Jan 18 '18 at 23:38
  • $\begingroup$ What is event $B$? What is the new evidence that comes along? $\endgroup$ – BallBoy Jan 19 '18 at 1:42
  • $\begingroup$ New roles of the dice, with different probabilties $\endgroup$ – SFD Jan 19 '18 at 10:35
  • $\begingroup$ @SFD I'm sorry, I still don't think you've defined your event $B$ precisely. Were there particular rolls with particular outcomes? How did we (the dice-rollers) come to believe the new probabilities? $\endgroup$ – BallBoy Jan 19 '18 at 15:26
  • $\begingroup$ @SFD Also, just a tip -- if you're responding to someone's comment, use the "@" symbol followed by their name so they see it. $\endgroup$ – BallBoy Jan 19 '18 at 15:27
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To make sense of the OP's question, let us agree to strike out the 2nd part of his question that begins with "Using Bayes:". Also, we feel it is necessary to frame the question in a way that makes more sense.

We are interested in obtaining statistical estimates for a discrete random variable $X$ with outcomes in $\{1,2,3\}$.

Statistician $A$ comes up with

$\tag A (\frac{1}{3},\frac{1}{3},\frac{1}{3})$

and statistician $B$ comes up with

$\tag B (\frac{2}{6},\frac{1}{6},\frac{3}{6})$.

We trust each of the statisticians the same amount, so we simply take the average, $C = \frac{A+B}{2}$, to estimate $X$:

$\tag C (\frac{4}{12},\frac{3}{12},\frac{5}{12})$

I am wondering where the OP found this question. But no matter, it is now necessary for us to commission another statistician, $C$, to put this matter to rest (or for the OP to work on clarifying the question).

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