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If I toss $2$ fair coins, the sample space is $\{TT, HH, TH, HT\}$.

But if a family has 3 children, the sample is $\{BBB, GGG, BBG, GGB\}$. Why don't we include cases like $\{GBB, BGB, GBG,\dots\}$? In the coins example, we included $\{TH, HT\}$ so why not here?

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  • $\begingroup$ It is your choice. In the $\{BBB, GGG, BBG, GGB\}$ example those events have different probabilities, possibly close to $\frac18,\frac18,\frac38,\frac38$ $\endgroup$
    – Henry
    Jul 8, 2020 at 8:13

2 Answers 2

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The short answer is: It is not the case that a given problem has one well-defined, fixed sample space that is associated with it. Often you actually choose the sample space in order to help think about the problem in the clearest way possible.

More generally though, beware, because some "elementary" probability puzzles (topics like coin tosses and sex of children are favourite themes!) are not clearly defined mathematically. I think it is really bad practice but probability is often - usually, even - taught in this confusing way. This leads to the silly puzzle (described here):

  • If a family has two children, at least one of which is a daughter, what is the probability that both of them are daughters?
  • If a family has two children, the elder of which is a daughter, what is the probability that both of them are the daughters?

The answers are sometimes reported as $1/3$ and $1/2$ because you might imagine a sample space $\{BB, BG, GB, GG\}$, where we write the eldest child first and each possibility occurs with equal probability. In the first case, conditioning on one of them being a girl reduces the sample space to $\{BG,GB,GG\}$. In the second case, conditioning on the eldest being a girl reduces the sample space to $\{GB,GG\}$.

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Let $n$ be the number of tosses/births, what is the number of cases in the sample space?

  • If the order of tosses/birth matters i.e. what came first- head/boy or tail/girl, then $2^n$ cases.
  • If it doesn't i.e. if one is interested in only the number of heads/boys, then $n+1$ cases.
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