When does order matter in probability

Question

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.

Answer 1

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.

Answer 2

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, $P(G|G)=0.5$.

Answer 3

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.

Answer 4

I know you didn't ask the solution to the problem but I am going to provide it just for the completness.

Let's understand this problem by keeping it simple. Because I have read many answers that delve too deeply into the problem statement, definitions of probability, structure of mathematics etc. making it very complicated to understand.

Before answering specific questions, let me first provide an explanation to create a structure in your mind to understand the problem. My approach will be more intuitive rather than relying heavily on very accurate mathematical language.

Now, it is easier for us to understand and solve a problem on probability if the sample space has uniform probability distribution, meaning all the outcomes in the sample space are equally likely.

Therefore to solve the problem, let us begin by creating a sample space. We will assume that there is an infinitely large number of couples, and we ask them to reproduce a child. After determining the gender of the child, the couples are divided into two groups which are "equal" in numbers based on the gender of their child. We then repeat the process and ask the couples to reproduce a second child, and divide each group again into two sub-groups which are "equal" in numbers based on the gender of the second child.

Universe of the given problem.

The scenario described in the above image is the universe for our problem. It is observed that there are more couples who have one boy and one girl child, compared to those who have two boys or two girls. To create a sample space with uniform probability distribution, it is necessary to take the order in which the children are born into consideration.

S = {BB, BG, GB, GG} each of this group has equal number of couples now.

As a problem poser, I have randomly selected a person from these groups, who turns out to be 'Alice', and revealed to you that one of her children is a girl. However, you do not know the order of the children or the gender of the other child. Therefore, based on the knowledge that at least one child is a girl, the Boy-Boy group can be eliminated, leaving a new sample space or sub-sample space with three groups: Boy-Girl, Girl-Boy and Girl-Girl, with equal number of couples.

Now the statement "Alice has 2 kids and one of them is a girl. What is the probability that the other child is also a girl?" is equivalent to asking "Alice has 2 kids and one of them is a girl. What is the probability that 'Alice' belongs to the Girl-Girl group?" which is clearly 1/3.

When calculating probability, the method of dividing the number of favorable outcomes by the total number of outcomes in the sample space assumes that all outcomes in the sample space have a uniform probability of occurrence. In this case, in order to use this method, it is necessary to create a sample space with a uniform probability distribution, taking the order of the events into consideration.

The fallacy is while making the sample space S = {BB, BG, GG}, in a way, you are assuming these 3 group to have equal number of couples, which is not true. Actually now, the BG group has couples from both the BG and GB group from our universe! (Please refer to the image for better understanding.). Therefore, the probability of Alice being from the new Boy-Girl (BG+GB) or Girl-Girl (GG) groups is not equal, or 50-50.