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Is there a standard way to view this? The problem is,

In an experiment, die is rolled continually until a 6 appears, at which point the experiment stops. What is the sample space of this experiment?

My first instinct was to say that it was the set of all finite sequences with exactly one 6, which is in the final position. But this neglects the (zero-probability) event that a 6 never comes up. Should this be in the sample space? Does it matter?

crf
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  • I generally think of it as ${1,2,3,4,5,6}^{\Bbb N}$; the event of getting a $6$ on the first roll, for instance, can then be identified with the set of sequences whose first term is $6$. – Brian M. Scott Jul 04 '13 at 04:44
  • There is a lot of flexibility, Your suggestion works fine, also same thing with strings of type $N^k6$, where $N$ stands for non-six. There are also infinite sequence choices. What is not clear is whether (if this is a homework problem) the grader has similar flexibility. – André Nicolas Jul 04 '13 at 04:53
  • @AndréNicolas Not a homework problem, just self study with a textbook that doesn't have solutions. I was just curious as to whether there is a "right" way to do this. It seems like the answer is no. – crf Jul 04 '13 at 04:56
  • There are often practical considerations. Some sample spaces make probabilities easier to compute. – André Nicolas Jul 04 '13 at 05:00

2 Answers2

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Since you (might) need to account for the hypothetical event that a 6 is never rolled, I would go ahead and say that Brian M. Scott is more or less correct - in plain terms, the sample space is the set of all finite sequences terminating in a 6 and all infinite sequences containing no 6. If the zero-probability event that a 6 is never rolled is irrelevant, then it doesn't matter too much, as Ross Millikan points out in another answer.

Nick
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For most questions, it won't matter. The event of a $6$ never coming up has probability zero. Unless you multiply by a function that goes up to infinity too fast it won't matter. But there is the Saint Petersburg paradox that shows that if the payoff grows too fast, the argument fails.

Ross Millikan
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