SOLUTION: In a class on probability, a statistics professor rolls two balanced dice. Both fall to the floor and roll under his desk. A student in the first row informs the profes

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Question 1071373: In a class on probability, a statistics professor rolls two balanced dice. Both fall to the floor and roll under his desk. A student in the first row informs the professor that he can see both dice. He reports that at least one of them shows a “five”. What is the probability that the other die is also “five”? (Beware the obvious.)
Found 2 solutions by KMST, ikleyn:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
One obvious issue is that the “five” in the question does not spell "five",
but that is unicode characters, not math.

There are 36 ordered pairs of numbers that could represent the outcome.
That would be easy to visualize if you use my pair of dice (one red, and one green).
There are 11 of those ordered pairs where the top face of one or both die shows a 5,
and only 1 of those ordered pairs have both top faces showing a 5.
That is 1%2F11 .
So, if the top face of one of the dice shows a “five”,
the probability that the other die is also “five” is 1%2F11 .
If the professor had used my set of dice, and the green die was a "five",
then the probability that the red die was also a "five" would be 1%2F6 .

Of course, the “five” the student reports could be
a lie,
a 5 of spades playing card,
a misread number of dots on the top face of one of the dice,
the number of dots in a visible face of one of the die that is not the top face, etc.

Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.
This problem (or, better to say, the KMST's solution) is about the accurate definition of the space of events (outcomes). 

1.  If the experiment (and the question) is about the probability of having "5" on the second dice at multiple repetition 
    of rollings the two dices, then the answer is 1%2F6, since the outcomes on different dices are independent events.


2.  If the experiment (and the question) is about the probability of having "5" on the second dice 
    UNDER THE CONDITION THAT THE FIRST DICE SHOWS "5", then the answer is 1%2F11, as KMST calculated; 
    but this time we have ANOTHER space of events/outcomes.


3.  Asking about the probability regarding a unique rolling (or a unique rolling of two dices) is nonsense.


So, the answer depends on accurate formulation the problem.

Different formulations can lead to different answers, but it is not amazing, because the experiment's conditions are different in this case.