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In most cases, I can intuitively understand when selection order is important in probabilistic inference. However, I've come across a few cases recently where I've come unstuck. Here's an example,

Alice has 2 kids and one of them is a girl. What is the probability that the other child is also a girl? You can assume that there is an equal number of males and females in the world.

The answer I am given is as follows,

The outcomes for two kids can be {BB, BG, GB, GG} Since it is mentioned that one of them is a girl, we can remove the BB option from the sample space. Therefore the sample space has 3 options while only one fits the second condition. Therefore the probability the second child will be a girl too is 1/3.

So if the answer is correct the order of selection matters, because both BG and GB appear as possible outcomes above. On the other hand, the way I tried to answer the question was to note that we are already given that Alice has 2 kids and one of them is a girl. My logic goes that we then have only two possibilities to choose from for the 2nd child G or B, which would lead me to answer 1/2.

If the question would have been, e.g. Alice wants to have two kids what is the probability that she will have a G and B. Then in my mind order would matter {BB, BG, GB, GG} but not in the above.

Can someone explain to me if/why I am wrong? and in any case, I'm interested to hear your general tips for establishing whether order matters in these types of questions.

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Merging BG and GB into a single case (i.e. ignoring order) doesn't change any probabilities. It's still twice as likely to get one boy and one girl compared to getting two girls.

So yes, you can think unordered, but then you lose the uniform probability distribution.

In fact, you often go the other way (from a context where order doesn't matter to pretending it does matter) exactly in order to get uniform probability. For instance, if you flip two identical coins, there is no "first" coin or "second" coin in any objective sense, but in order to make calculations we usually pretend there is a difference (one is blue and the other red, or one is thrown before the other, etc). That way it makes sense to talk about HT and TH, and all four outcomes are equally likely, even though there are really only three outcomes in the original setting.

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The order does not matter when the variables are independent random variables, i.e., when the probability of one variable does not depend on the outcome of the previous. The order when a previous even influences the current one.
In your example, Alice has two kids, one of which is a girl. The possible outcomes for two kids are {BB,BG,GB,GG}. Since one of Alice's children is a girl, we can cross out BB. If order does matter, i.e., one child was born before the other, then we can also cross out BG or GB. If the order does not matter, then there is not a difference between BG and GB. Therefore, the two options are (BG or GB) and GG. We know that $P(B)=P(G)=0.5$, therefore, $P(G|G)=0.5$.

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    $\begingroup$ "Since Alice's first child is a girl" This cannot be assumed. $\endgroup$ – eyeballfrog Nov 13 '18 at 21:53
  • $\begingroup$ @eyeballfrog I fixed my answer to comply with your premise. $\endgroup$ – HiMatt Nov 13 '18 at 22:08
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I don't really know if there is a legit definition out there for order in mathematical terms. I don't see how independence works here. However the problem at hand should define it good enough. The idea is that if you have some set ${A,B,C}$, then order is important when $\{A,B,C\} \neq \{B,C,A\}$ In Real Analysis, these are called tuples (kind of like a set but the way the elements are ordered is important so this makes it not a set). To make it more formal, a permutation counts the number of elements in the set of tuples which they themselves contain k elements. So a permutation counts, $\{(A,B,C),(A,C,B), (B,C,A), (B,A,C), (C,A,B), (C,B,A)\}$ which is 6. One way I can think of to test whether order is important is to take $(A,B,C) \cup (B,C,A)$ So say you are told you have 6 die and you are asked to roll 3 and then count the number of tails. So basically find out if $(C_1 = T, C_2 = T, C_3 = T) \cup (C_3 = T,C_1 = T,C_2 = T) = (C_2 = T,C_3 = T,C_1 = T)$, in this case it is. This is a Combination. Now for a permutation, order is important. So say you have to assign seats at random people will get the amount of money written on the seat. Assuming no two seats have the same amount of money written on them, then order is important here. Specifically, If you have 5 chairs, and 3 people, then there are 5 ways, to assign the first seat, 4 ways to assign the second. So the statement here is $(S_1, S_2, S_3) \cup (S_2, S_3, S_1)$. This should be understood as person 1 getting seat $S_1$ in the first case and in the second case, person 1 gets seat $S_2$ . So obviously when join these sets you don't get a set with 3 elements, now you get a set with 2 tuples and each has 3 elements such that $(S_1, S_2, S_3) \cup (S_2, S_3, S_1) = \{(S_1, S_2, S_3),(S_2, S_3, S_1)\}$. This is extending a set and I don't know of any ways to outright calculate this, but if you cannot join two different sets as the first method, then order is definitely important. So basically see if you can rearrange the elements of some event without having to count this new event. To me it this still seems without a proper definition. I'm guessing it just depends on the nature of the problem.

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  • $\begingroup$ While the word order has many distinct meanings in mathematics, in this context it refers to the sequence of events, and independence refers to the notion of (conditional) probability of events. Some reformatting of your Answer, even as minor as adding whitespace between paragraphs, would make it easier to grasp where you are going with the reasoning. $\endgroup$ – hardmath Jan 18 '19 at 18:01

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