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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?

The sample space consists of sequence where the last entry must be equal to six and no other entry can be six. Mathematically, we can describe the space as $$ S = \{(6), (x_1,6), (x_1,x_2,6), \dots, (x_1,x_2,\dots,x_i,6) | \text{ where } x_i=\{1,2,3,4,5\} \text{ and } i\geq 1\}$$

  • Let $E_n$ denote the event that $n$ rolls are necessary to complete the experiment. What points of the sample space are contained in $E_n$?

$E_n$ contains all sequences in the sample space of length $n$. Mathematically, we can $E_n$ as $$ E_n=\{(x_1,\dots,x_{n-1},6) | \text{ where } x_i=\{1,2,3,4,5\} \} $$

  • What is $\left(\bigcup\limits_{n=1}^\infty E_n\right)^c$?

By De Morgan's Law, $$ \left(\bigcup\limits_{n=1}^\infty E_n\right)^c = \bigcap\limits_{n=1}^\infty E_n^c $$

So here is my question: Is $E_n^c$ is the set of $n$ length sequences where the last entry is anything but 6?

I am having a hard time believing that because then those sequences would not belong in the sample space $S$?

Username Unknown
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  • Is it necessarily true that you always get a six? If not, what is your real sample space? – Thomas Andrews Aug 27 '22 at 19:57
  • I think the wording is a bit confusing, since the complement needs to be defined relative to something, but that something is never indicated, other than what seems to be defining the sample space to be such that there is no complement. To me, it would be natural to let the sample space consist of all possible outcomes of the experiment, including infinite sequences with no $6$s (which would then be the desired complement). The OP's proposal for $E_n^c$ is incorrect: $E_1^c$ contains $E_2$ and $E_3$ and so on, for example. – Greg Martin Aug 27 '22 at 20:06
  • @ThomasAndrews I thought so because then the experiment isn't complete. Or am I wrong on that? – Username Unknown Aug 27 '22 at 20:11

2 Answers2

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Your sample space isn't complete given how you've defined the process it describes. We need to allow for arbitrarily long sequences of "non-sixes".

We can express this as the following

$$S:= \left(\bigcup_{i=1}^{\infty}\{1,2,3,4,5\}^i \times \{6\}\right)\cup \{6\}$$

Given above:

$E_n:\{1,2,3,4,5\}^{n-1} \times \{6\}, n>1$ and $E_1 = \{6\}$

If we start with just $E_1$ we get:

$$E_1^c = S\setminus \{6\} = \left(\bigcup_{i=1}^{\infty}\{1,2,3,4,5\}^i \times \{6\}\right)$$

$$\left(E_1 \cup E_2\right)^c =S \setminus \{\{6\} \cup \{1,2,3,4,5\}\times \{6\}\} = \left(\bigcup_{i=2}^{\infty}\{1,2,3,4,5\}^i \times \{6\}\right)$$

In general,

$$S_n:=\left(\bigcup_{i=1}^n E_i\right)^c = \left(\bigcup_{i=n}^{\infty}\{1,2,3,4,5\}^i \times \{6\}\right),\;\;\forall n>0$$

We can see that the sequence of sets $S_i$ is strictly decreasing:

$$S_j \supset S_{j+1}\;\forall j>0$$

Therefore,

$$\lim_{n\to \infty} S_n = \{1,2,3,4,5\}^{\mathbb{N}} \times \{6\} = \{1,2,3,4,5\}^{\mathbb{N}}$$ up to isomorphism

If this seems weird I agree but you can totally do this — we just define $6$ to be at the special index $\infty$ of our sequence.

See here: https://math.stackexchange.com/a/979217/632875

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The $E_n$'s partition the sample space. For each $n\in\mathbb{N}$, $E_n$ is the set of $n$-tuples in $S$. Thus, $E_n^c$ is the set of $m$-tuples in $S$ where $m\ne n$. Also using this line of reasoning, $$\left(\bigcup_{n=1}^\infty E_n\right)^c=S^c=\emptyset$$

I am assuming that the sequences are finite and the experiment ends after finitely many steps.

David Dinh
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