the singularity of being and nothingness

# Fun with Probabilities: Answers

So here’s our 2 probability puzzles:

**Puzzle #1:** I have two children. One of my children is a boy. What is the probability that I have two boys?

**Puzzle #2:** I have two children. One of my children is a boy who was born on Tuesday. What is the probability that I have two boys?

Let’s tackle the first one. By sheer intuition, the average person would probably answer that the probability of the second child being a boy is 50% (I know I did the first time). After all, the population is (seemingly) roughly divided equally between males and females, so it just feels right that the answer is somewhere around 50%.

However, there is a bit of a trick in the question. The question, after all, is not actually asking about the chances that any particular, individual child is male or female. Rather, it’s asking what the probability that both children are boys is, based on the knowledge that one is a boy.

So how do we work this out? First, we need to represent all the possible arrangements of children in this scenario. Since one is already known to be a boy, the possible arrangements are:

Boy – Boy (1/3)

Boy – Girl (1/3)

Girl – Boy (1/3)

As we can see, given the possible outcomes of children in this family, the probability that both children are boys is 1/3, or 33% (ish).

So after we plot this out, the answer seems a bit more reasonable, even though it might mess with our common sense a bit.

Now for the second:

“I have two children. One of my children is a boy who was born on Tuesday. What is the probability that I have two boys?”

Given our experience with the first question, it’s natural to feel like this question is merely a rehearsal of the last. After all, we know from past experience that the probability of having two boys in this scenario is 33% (ish)…and why should the day of the week matter?

Ah, but there’s another trick in the question. As with the first, this is not simply asking about the second child. Rather, it’s asking a question something like this:

What is the probability that I have two boys, one of which was born on Tuesday?

By the introduction of the additional variable of the birth-day-of-the-week, our approach to calculating the probability changes. Now, we must not only calculate the probability that the both children are boys, but must also take into account that one of them was born on a specific day of the week.

Therefore, we approach the solution in the same way as before: by breaking out our results into all possible scenarios. These look as follows:

Arrangement #1:

BTu – GM

BTu – GTu

BTu – GW

BTu – GTh

BTu – GF

BTu – GSa

BTu – GSu

————————–

Arrangement #2:

GM – BTu

GTu – BTu

GW – BTu

GTh – BTu

GF – BTu

GSa – BTu

GSu – BTu

————————–> 7 possibilities

Arrangement #3:

BTu – BM

BTu – BTu

BTu – BW

BTu – BTh

BTu – BF

BTu – BSa

BTu – BSu

————————–> 7 possibilities

Arrangement #4:

BM – BTu

BW – BTu

BTh – BTu

BF – BTu

BSa – BTu

BSu – BTu

————————–> 6 possibilities

In these scenarios, there are 7 possibilities: after all, if the boy we know about is born on Tuesday, then the other child can be born on any other day of the week…thus, 7 possibilities. The only divergence here is in arrangement #4. We leave out the overlap of both boys being born on Tuesday, for that has already been accounted for in arrangement #3.

Given these possibilities, there are a total of 27 (7+7+7+6) possible arrangements of children. Therefore, the probability that both children are boys is 13/27, or 48% (ish).

**Summary**

There are a couple of things I find fascinating about these exercises. First, in the original question, it’s quite counter-intuitive to our first guess that the answer is 33% (ish). While the probability methodology seems sound, the answer–although it’s logical–still doesn’t sit comfortably with our original assumptions.

But what’s even stranger is the working out of the second question. While 33% seems askew of common sense, the meaningfulness of “Tuesday” to probability is entirely out of bounds. Yet while the logic of the second answer seems the strangest of all, the actual answer brings us closer (48%) to our intuition than the more palatable logic of the first. And what if we introduced additional, rarer criteria (hair color, eye color, etc.)? Our probability would move even closer to 50%…wild.

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about 10 years ago

There is a better way to solve these puzzles. And

interestingly, it gets different answers. But they don’t have the

problem of making no sense whatsoever, by appearing to be different

for arbitrary reasons. And they match the correct solution method

for the Monty Hall Problem, as well (as opposed to your

explanations for how the correct answer happens). You have two

children. There are two facts of the various forms represented by

“one child is a who and ” that apply to them. These facts may be

the same. You want to tell me about one, but probably have to

choose one of two facts. What is the probability the fact you

didn’t choose (or the duplicate of what you did) includes “boy?”

The answer, regardless of what the additional facts are, is always

1/2. You can get this from your derivation of Question #2 by doing

it properly. Counting cases only works when each case has an equal

probability of presenting you with the information in the problem

statement. But if your arrangement is, for example, “GTh–BTu,”

there is only a 50% chance you will tell me about the boy. There is

also a 50% chance you will tell me about the girl. In fact, this

argument applies to all but one – “Btu-Btu” – of your

possibilities. The correct answer is not (7+6)/(7+7+7+6)=13/27, it

is (7/2+5/2+1/1)/(7/2+7/2+7/2+5/2+1/1)=(12/2+1)/(26/2+1)=7/14=1/2.

In Monty Hall, there is a similar 50% choice in all-but-one case.

It’s just hard to see this comparison because “all-but-one” means

exactly one. When your original door has shattered dreams, Monty

has no choice about what door to open. Just like a parent with two

children of the same fact has no choice. But when your door has the

prize, he has a 50% choice, just like most parents of two. So the

properly-solved probability that your door has the prize adds

probabilities rather than counting cases. That’s (1/2)/(1/2+1)=1/3,

not (1)/(1+1)=1/2. Most people get this wrong because they don’t

include the divisions by 2. And it belongs in all these

solutions.