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Yes, but because it is random, the thief can not choose the 5 expensive ones. By dividing 5/94 we are assuming that at least one of them is the valuable one.

Yeah that is the odds of picking 1 valuable in 1 pick... The thief does have 5 picks, thereby much higher odds!

Another alternative would be 1 / (94-4). But I see it very idealistic. There is no randomness. Since I am ensuring that I choose only one face and leave the rest. But I can not know how to do it. And the rest is still part of the sample population