I have a question that raises many debates around my coworkers which is the following:
You have to dices, one with 6 Sides (1,2,3,4,5,6) and a 4-sided one (1,2,3,4) You pick one randomly, and you throw it 2 times, the first throw is a 2. What is the expected value of the second one ?
The debates we had was about the dice picking, some thinks that you pick uniformly(p=1/2) some thinks that the 2 involve a different dice picking probability. Moreover this brainteaser is supposed to be done quickly so guessing the law is maybe too long, but i am curious to see what you guys think.
As always, thank you guys
For quick guess, I think the following logic is the simplest.
Expected value of d6 is $\frac{7}{2}$, and expected value of d4 is $\frac{5}{2}$, so prior expected value of random die of these two is $\frac{6}{2}$.
With probability $\frac{1}{6}$ ($\frac{1}{2}$ for having d6 and $\frac{1}{3}$ for rolling 5 or 6), we would get evidence that we have actually d6 and not d4. This would make our expected value $\frac{7}{2}$ - increase of $\frac{1}{2}$.
As expected change of our expectation is $0$ by law of total expectation, if one time in six we increase expectation by $\frac{1}{2}$, the rest five times we should decrease it by value $5$ times smaller - ie it will be $\frac{6}{2} - \frac{1}{2} \cdot \frac{1}{5} = \frac{29}{10}$.
To compare with Bayes formula, $P(\text{d4} | 2) = \frac{P(2 | \text{d4}) \cdot P(d4)}{P(2)} = \frac{\frac{1}{4} \cdot \frac{1}{2}}{\frac{1}{4} \cdot \frac{1}{2} + \frac{1}{6} \cdot \frac{1}{2}} = \frac{6}{10}$, so $P(\text{d6} | 2) = \frac{4}{10}$, and expected value is $\frac{6}{10} \cdot \frac{5}{2} + \frac{4}{10} \cdot \frac{7}{2} = \frac{29}{10}$.