A wine taster claims she can distinguish four vintages of a particular Cabernet. What is prob this can be done by guessing? She is confronted with four unlabeled glasses.
I have a solution to this question, but I also saw a solution elsewhere and I am confident that my solution is actually the correct one.
My Solution
Since there are four unlabeled glasses, the taster can either guess correctly or guess wrong for each glass. As such there are $2^{4}$ possible sets of four guesses. There is also only one way of getting all four glasses correct (call this event $A$). As such the probability will be:
$$P(A) = \frac{1}{2^{4}} = \frac{1}{16}$$.
Other Solution
Now thinking about it while writing this post, I think in my solution I considered the glasses as "ordered" whereas in the other solution they treated them as unordered hence difference in sample space size. Could somebody confirm that is actually the case?
While your approach could work in considering each glass to be the right wine or the wrong wine, the probability for each is not $1/2$. Instead, the probability is $1/4$ for the first glass, $1/3$ for the second glass (as, $1/2$ for the third glass, and $1/1$ for the last glass. The probability for each is chosen by assuming that all previous glasses were correctly assigned.
It is easier to look at the first glass as having $4$ possible states: Wine $A$, $B$, $C$, or $D$. Naively, this would make it seem like there is $4^4 = 256$ possible states, but this reduces to $4! = 24$ possible states (with only one right one), since each glass must map one-to-one to the guess of a wine.