I watched this Youtube clip, TED-Ed’s Frog Riddle Is Wrong, from Presh Talwalkar’s Youtube channel, MindYourDecisions. I enjoy and appreciate the posts on the channel, but in this case I do not completely agree with setup presented, and so I decided to write my assumptions in an even more formal way.

### Introducing the paradox

The paradox is a set of two (possibly different) probability problems. The solution to problem two is compared to the solution to problem one to see how similar the solutions seem.

There is a part of the problems that is common to both problems. In both problems we imagine that __“Mr. Jones has two children”__.

In problem one, (A), we are asked, what is the probability that Mr. Jones has a girl and a boy, __given that we know that at least one of the two children is a boy__?

In problem two, (B), we are asked, what is the probability that Mr. Jones has a girl and a boy, __given that we are told that one of the children is a boy born on a Tuesday__?

You may have noticed that I have __not__ copied the premise text exactly as stated in the clip’s riddle. This is a conscious decision on my part. I feel that the fact that the two problems are formulated directly beneath each other might convey interpretational information about problem two, (B), that might be of relevance to the solution.

### Solution to problem one (A)

Most people agree on how to solve problem one.

If all we knew was that Mr. Jones has two children, we would say that the four possibly gender combination possibilities, “GG”, “GB”, “BG” and “BB”, was equally probable. In two of the four cases Mr. Jones would have a girl and a boy, so our answer would have been 50 % (= 2/4).

The notation here being that “GG” means both children are girls, “GB” means the elder is a girl and the younger a boy, “BG” means the elder is a boy and the younger a girl and “BB” means both children are boys.

If we already know that at least one of the children is a boy most of us agree to use the (conditional) probability that out of the three possible, and equally probable, outcomes, “GB”, “BG” and “BB”, two of them mean that Mr. Jones has a girl and a boy. So we answer 67 % (= 2/3).

### Elaborate solution to problem one (A)

We expect trouble ahead, in problem two, so why not elaborate some more on the example that we can agree on.

Let us denote the gender and birthday of a child with:

- Gt – for a girl born on a Tuesday.
- Go – for a girl born on a day other than a Tuesday.
- Bt – for a boy born on a Tuesday.
- Bo – for a boy born on a day other than a Tuesday.

Any pair of siblings can be described by one of the following sixteen combinations:

[Gt:Gt] 1/196 |
[Gt:Go] 6/196 |
[Gt:Bt] 1/196 |
[Gt:Bo] 6/196 |

[Go:Gt] 6/196 |
[Go:Go] 36/196 |
[Go:Bt] 6/196 |
[Go:Bo] 36/196 |

[Bt:Gt] 1/196 |
[Bt:Go] 6/196 |
[Bt:Bt] 1/196 |
[Bt:Bo] 6/196 |

[Bo:Gt] 6/196 |
[Bo:Go] 36/196 |
[Bo:Bt] 6/196 |
[Bo:Bo] 36/196 |

I bet you already guessed what my notation is supposed to mean. For instance “[Bt:Go]” means that the first-born child is a boy born on a Tuesday and the younger child is a girl born on a day other than a Tuesday. The (unconditional) probability of “[Bt:Go]” is 6/196 (1/2 * 1/7 * 1/2 * 6/7 = 3 %) as noted in the table.

If we wanted to, we could tally up the number of cases (out of the original 196) with at least one boy (147 cases) and the number of cases with a girl and a boy (98 cases). The answer derived in this way, 67 % (= 98/147), match the answer we already know.

No dark magic so far, so let us see if we can’t stir up a bit of trouble …

### Solution to problem two (B) – classical

One way to solve problem two is to look at the above table of the sixteen possible combinations for a pair of siblings, remove the combinations that we know cannot apply to problem two, (B), and check how often you will encounter a boy and girl sibling pair among the combinations that are left.

That solution, presented in the Youtube clip, will have us look at a table not unlike the following, where only combinations satisfying “at least one of the children is a boy that is born on a Tuesday” remain:

[Gt:Gt] – |
[Gt:Go] – |
[Gt:Bt] 1/27 |
[Gt:Bo] – |

[Go:Gt] – |
[Go:Go] – |
[Go:Bt] 6/27 |
[Go:Bo] – |

[Bt:Gt] 1/27 |
[Bt:Go] 6/27 |
[Bt:Bt] 1/27 |
[Bt:Bo] 6/27 |

[Bo:Gt] – |
[Bo:Go] – |
[Bo:Bt] 6/27 |
[Bo:Bo] – |

The number 27 is just the sum of the number of cases left in the table (i.e. 1 + 6 + 1 + 6 + 1 + 6 + 6).

The solution, when the problem is interpreted this way, becomes that in 14 cases out of 27 cases Mr. Jones has a girl and a boy. The 14 cases are the combinations [Gt:Bt], [Go:Bt], [Bt:Gt] and [Bt:Go], and the rest of the 27 cases are the cases [Bt:Bt], [Bt:Bo] and [Bo:Bt].

### Interpretation of problem two (B)

The ‘paradox’ is that when we compare the solution to problem one (A), 67 % (= 2/3 = 98/147), with the (naive or classical) solution to problem two (B), 52 % (= 14/27), we see that the results differ.

There is a quite serious caveat to the interpretation though!

In the way the ‘paradox’ is presented to us it seems to me that the argument goes something like this: “You are in an almost similar situation to problem one, (A); only this time you are told that the boy (in question) was born on a Tuesday”. I imagine that this information is told to me by someone who knows Mr. Jones well. This interpretation leads me to a different calculation and another answer to problem two, (B).

### Solution to problem two (B) – interpreted

So someone that knows Mr. Jones well is giving me the information that at least one of the two children is a boy and that the boy is born on a Tuesday. I quite naturally wonder what other possible pieces of information this person might have given me in case the situation had been another. If for instance the boy was born on a Wednesday, what had I been told then?

We can of course imagine all kinds of behaviors with regard to what we will be told. Why not just consider one (more or less arbitrary) strategy that I find reasonable and see what kind of answer we come up with then.

I imagine that the person giving me information about the children does it like this:

- If both children are girls I will be told that “Mr. Jones does not have any boys”.
- If the first-born is a boy I will be told that “Mr. Jones has at least one boy” and I will be told the birthday of the first-born, i.e. Tuesday, Wednesday or what ever day he is born. I will not be told that the boy in question is the first-born.
- If the first-born is a girl and the younger child is a boy I will be told that “Mr. Jones has at least one boy” and I will be told the birthday of the younger child, i.e. Tuesday, Wednesday or what ever day he is born. I will not be told that the boy in question is the younger child.

Let me identify the possible, mutually exclusive, pieces of information with roman numerals:

- (I) No boys.
- (II) At least one boy and the day mentioned
__is a Tuesday__. - (III) At least one boy but the day mentioned is
__not__a Tuesday.

With the above algorithm in mind we can repeat the above table showing the 16 possible combinations and assign roman numerals indicating the resulting information we would be told in each case:

[Gt:Gt] (I) 1/196 |
[Gt:Go] (I) 6/196 |
[Gt:Bt] (II) 1/196 |
[Gt:Bo] (III) 6/196 |

[Go:Gt] (I) 6/196 |
[Go:Go] (I) 36/196 |
[Go:Bt] (II) 6/196 |
[Go:Bo] (III) 36/196 |

[Bt:Gt] (II) 1/196 |
[Bt:Go] (II) 6/196 |
[Bt:Bt] (II) 1/196 |
[Bt:Bo] (II) 6/196 |

[Bo:Gt] (III) 6/196 |
[Bo:Go] (III) 36/196 |
[Bo:Bt] (III) 6/196 |
[Bo:Bo] (III) 36/196 |

This time when we do our removal trick we get the table below, where only information option (II) cases are allowed to remain:

[Gt:Gt] (I) – |
[Gt:Go] (I) – |
[Gt:Bt] (II) 1/21 |
[Gt:Bo] (III) – |

[Go:Gt] (I) – |
[Go:Go] (I) – |
[Go:Bt] (II) 6/21 |
[Go:Bo] (III) – |

[Bt:Gt] (II) 1/21 |
[Bt:Go] (II) 6/21 |
[Bt:Bt] (II) 1/21 |
[Bt:Bo] (II) 6/21 |

[Bo:Gt] (III) – |
[Bo:Go] (III) – |
[Bo:Bt] (III) – |
[Bo:Bo] (III) – |

As last time around, the number 21, is just the sum of the remaining number of cases.

The solution to problem two, when interpreted this way, then becomes that in 14 cases out of 21 cases Mr. Jones has a girl and a boy. This is indeed the well-known answer, 67 % (= 2/3 = 14/21), that we recognize from problem one, (A).

### The morale of the story

The difference between the two candidate answers stems from the “Bo;Bt” cases. We might phrase the observation as a question (or two):

“What would we be told if Mr. Jones had two boys and they were born on different days of the week? Which boy or which day of the week would be assigned more importance?”.

Sure, if, for instance, you are willing to accept that Tuesdays somehow, in the context of the paradox, are particular important days, then yes possibly the answer might be 52 % (= 14/27). On the other hand, the answer might as well be 67 % (= 14/21).

To me the interesting statistical tidbit to take away from this paradox is that, if you regard the information about the day of the week as pointless then that information indeed turns out to be insignificant, whereas if you believe you are given important information about the birthday then that bit of information is quite significant. The good news is that you will always be right!

Robert Nielsen, 2016-jun-03.