Bamboozling Birth, Part 2: Newsletter Challenge (November 2020)

50/50

I would say 1/3. Someone with 2 children can have BB, BG, GB, or GG. He must have BB, BG, or GB. 1/3 for 2 boys.

If you flip a coin 6 times in a row and it comes up heads every time, what are your odds your coin comes up tales on the next flip?

If you flip a coin 6 times in a row and it comes up heads every time, what are your odds your coin comes up tales on the next flip?

1/2, but that is a different situation.

I ask people at random if they have two children, and also if one is a boy born on a Tuesday. Finally, someone said yes. What is the probability that this person has two boys?

In the challenge question, you are given that the family chosen has 2 children, one of whom is a boy. A family with 2 children has an even distribution, 1/4 BB, 1/4 BG, 1/4 GB, and 1/4 GG. So the family selected has BB, BG, or GB. GG is eliminated. 1/3 of these are BB.

Of course, what day the boy was born on has no effect on the gender of his sibling.

I see your point, and it might play out statistically, but the odds are still 50/50 for all parties involved, if the couple has a boy the chances that the next child will be a boy is still 50/50....

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

I'm thinking the kicker is "...two children, and also if one is a boy...". This eliminates the "GG" families, and the "BB" families are 1/3 of the 3/4 that are left.

We seem to be comparing apples to oranges...before the event the odds are even, but compiling the results after the event, employs a different point of view....so the question is ambiguous ...

Your point is valid if the interview was conducted before any children were born, but that is not the case here.

So even though what you are trying to say is true, this particular situation does not take place "before the event," so it is also irrelevant. No apples. No oranges. Just one guy refusing to accept that he is wrong.

The 2 children have already been born. GG is not a possibility. 1/3 is absolutely the correct answer.

Not if you take it from the beginning...

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Information_about_the_child

Rixter,

SE is correct in his answer. You state the wrong list of possible answers to get the 1/3 answer. When you know one is a boy then you are only interested in the second. The GG (which has been eliminated) and the BG, BB, and GB choices are all predicated on looking at the probability of two events. However we are only looking at the probability of one event--since the other one has already been determined. Therefore your choices are -B, -G, B-, or G-. The answer is 2/4 which is the same as 1/2.

--JMM

Possible outcomes: Bb, Bg, Gg, Gb

Facts: There are 2 children. One is a boy. Therefore, Gg is eliminated.

The question is asking the probability that the sibling group is Bb. If it's not 1/3, it has 1/4, but I stand by my 1/3 choice.

In 2-children families, there are 4 equally likely possibilities: boy-boy, boy-girl, girl-boy, and girl-girl, assuming girls and boys are equally likely.

Notice that boys are twice as likely to have a sister than a brother, (2/3:1/3). Likewise, girls are twice as likely to have a brother than a sister, (2/3:1/3). That is because girl-boy and boy-girl account for 1/2 of the cases whereas boy-boy and girl-girl are only 1/4 each.

This is basically the root of this puzzle with all the extraneous red herrings removed.

If that were true than the 50/50 chance of a boy or girl would not hold true....possibilities ≠ probabilities....

..."1. Mrs Smith has two children. The eldest one is a boy. What’s the chance that both are boys.

2. Mrs Jones has two children. At least one is a boy. What’s the chance that both are boys?

3. Mrs Robinson has two children. At least one is a boy born on a Monday. What’s the chance that both are boys?

4. Mrs Taylor has two children. At least one is a boy called Oscar. What’s the chance that both are boys?

(Assume Mrs Smith, Jones, Robinson and Taylor are each chosen randomly from the population of families with exactly two children. The phrase ‘at least one is a boy’ is understood in the literal sense, i.e. in this case either one child is a boy, or both children are boys.)

The answers are:

1. 1/2 or 50 per cent,

2. 1/3 or 33.3 per cent,

3. 13/27 or about 48 per cent,

4. very very close to 1/2 or 50 per cent"...

https://www.theguardian.com/science/2019/nov/18/did-you-solve-it-the-two-child-problem

So are you sticking with 50/50? A fun exercise, but to be perfectly honest, I'm a bit less interested than I was yesterday, so I'm not even going to try to understand the math in #s 3 and 4, but #2 seems most relevant to the original question. I just don't see how day of birth or name could have any impact on gender, especially when the children are already born...

Frankly I don't get it either, but a clever mathematician can make anything look plausible...I'm sticking with 50/50...

Here are the results of the questionnaire they sent out....yes they actually checked this the old fashioned way...not a large enough sampling if you ask me....but still an impressive effort....

"We’ll get to the workings later. First, the results of the questionnaire I asked parents of exactly two children to fill in.

1. Two child families where eldest child is a boy. Total submissions: 4949. Of these families, the percentage with two boys was 44.9 per cent.

2. Two child families where at least one child is a boy. Total submissions: 6060. Of these families, the percentage with two boys was 39.5 per cent.

3. Two child families where at least one child is a boy born on a Monday. Of these families, the percentage with two boys was 54.0 per cent.

4. Two child families where eldest child is a boy named Oscar. Total submissions: 1961. Of these families, the percentage with two boys was 55.6 per cent.

In an ideal world, in which two-parent families are selected genuinely randomly, in large enough numbers and respond truthfully, we’d expect the percentages of ‘families with two boys’ to approximate the answers to the questions."

Putting it in Matlab (MonteCarlo), I can confirm that it is option 3: 13/27, or about 48.15%.

Below is the Matlab Script, for fact-checkers and other hobbyists:

My interpretation of the question, in three steps: Create the relevant families (2 kids), select the relevant families (one boy born on a tuesday). Then calculate the chance for a brother.

% Create families with two kids:
nFams = 10000000;
nKidsPerFam = 2;
fams = double(rand(nFams,nKidsPerFam)>=0.5); % array of 0's and 1's

% check B/G ratio:

nGirls = length(find(fams==0));

nBoys = length(find(fams==1));
BGratio = nBoys/(nBoys+nGirls)

%Let the boys be born on day 1-7 of the week
fams(find(fams==1))=ceil(7*rand(nBoys,1));

for iD=1:7
nDay(iD)=length(find(fams(:)==iD));
end
% check distribution over days:
100*(nDay/nBoys-1/7)

% Find families with at least one boy which is born on a tuesday
[BtuesdayFams,~] = find(fams==2); % all families with
selectedFams = unique(BtuesdayFams); % remove doubles (fams with two boys born on tuesday)
brother = length(find((fams(selectedFams,1)>0) & (fams(selectedFams,2)>0)));

result = 100*brother/length(selectedFams)

Look at this from a "complete the logic" approach. Your result of 13/27 is for a boy born on a Tuesday and the second child being a boy. The remainder has to be a boy born on a Tuesday and the second child being a girl.

This must be the difference of 1-13/27 = 14/27.

So, according to your matlab simulation the probability of a girl vs. a boy being born is not 50%. Is it possible that in your simulation you accidentally introduced this unequal probability?

--JMM

Hi JMM,

I have verified that the B/G birth ratio in my simulation is 50/50 within 0.01%.

I think DMAZ gave a good explanation of the "option 3" in another post, resulting in the 13/27 ratio.

Greetings,

Karel

You are correct. Here is what is wrong with my reasoning in #20...

There are twice as many two-child families that have boy-girl than have boy-boy, but there are twice as many boys in the boy-boy families.

For example, if you have 2 boy-girl families and one boy-boy family, that's 4 boys, two with sisters and 2 with brothers.

So boys are just as likely to have a sister as a brother, but families with two children with at least one boy, are twice as likely to have a boy and a girl than 2 boys.

Rixter,

Your final sentence contains an impossibility. The first half of that sentence is correct but the second half (which contradicts the first half) is flat-out wrong. Once you have determined that one child is a boy then the second child has a 50% probability of being either a boy or a girl.

JMM

Friend,

This is not a version of the Monty Hall problem because the family does not have the ability to change their choice. They are stuck with the one child being a boy born on a specified day. I have given the correct answer in post #17. By eliminating the ability to change the choice we have limited the problem to the choice of only one child. This is now bB, bG, Bb, or Gb in which the "b" represents the child known to be a boy regardless of birth order. This is a 2/4 = 1/2 probability or 50%.

--JMM

Wrong again, now twice instead of just once.

Your mistake, along with all the others who got the Monty Hall problem wrong, is failing to account for multiple possibilities of one outcome.

In the case of two children, there are 4 equal possibilities: BB, BG, GB, GG. (Binary distribution equal to 1-2-1, the third row of Pascal's triangle.) Boy on Tuesday rules out GG, but the other 3 equally probable choices are still in play. Of these 3 remaining choices, only BB leads to another boy, but both BG and GB lead to another girl. I.e., 1 way for a boy vs. 2 ways for a girl.

Of course, there only two possible outcomes, but the chances are NOT equal. Those whose counting skills can't even distinguish between 1 and 2 will flunk this question.

The Monty Hall problem may not be directly isomorphic to this (different sample spaces), but it IS comparable. Dismissing it is your second mistake.

This is a version of the Monty Hall logic problem.

Zero. Coins never come up tales.

Both of your examples came up heads.

No, that's a head and a tale...

If you'd used this one:-

The connection would have been clearer.

Incidentally this one is on ebay in the condition seen for £1500 (about $2000).

Let n be the number of additional children after the first child, which is a boy. Would the probability of have one additional boy be (1/2)ⁿ / (1/2)^(n-1), which is 1/2?

No condition of birth order in the original statement...

"I would say 1/3. Someone with 2 children can have BB, BG, GB, or GG. He must have BB, BG, or GB. 1/3 for 2 boys."

This makes perfect sense until you consider the (often dismissed in these comments) last condition; Born on a Tuesday.

I agree that there will be an equal distribution of BB BG GB (1/3 each) in the viable population up to this point, (the GG group being eliminated) however you must meet the last condition. So there will be a 1 in 7 chance of any B member of both the BG and GB groups meeting this last condition, but there will be 2 in 7 chance of the BB group (because there are 2 candidates EITHER of which could be born on a Tuesday to qualify)

So the BB group will (at random) end up with double the representation in the final pool of potential candidates which has met all of the preconditions from which we can base our final probability. This leads to a 50/50 probability that the sibling will be a boy. Final group BB BB BG GB

It seems that all the conditions seem to cancel out leaving the final probability equal to the chance of B or G in the first place... ;-)

The problem does not say that "one is a boy" and "one was born on Tuesday." It says "one is a boy that was born on a Tuesday." These are known facts, not something we need to find a probability for.

They are known facts but they constrain the group of eligible families. This must be considered to determine any type of probability. To state a probability of the second child (other child, not birth order) being a boy or girl you must know the exact composition of the group from which you are making the selection.

That "potential" group is the set of all families who have 2 children where at least one is a boy who was also born on Tuesday. So from a set of large (or infinite) choices that conform to the statistical distribution of equal probability of having a boy or girl on any day of the week (as started in the initial problem) you must determine how many families conform to this requirement in the total set (in % proportion to each other as the actual total group size should make no difference). This constrained group is from where you must calculate the final probability.

That said, if you do not consider how the birth day changes the composition of the eligible group then you will have a whole bunch on unqualified participants being considered which will alter the calculated probability. This is exactly what stopping at the BB, BG, GB point does and why it understates the probability as being 1/3.

I confirmed option 3 with monte carlo above, but this is the best explanation :-)

Great answer

I agree with S.E. the answer must be BB or BG. Since the first was a B. then GG can't work nor GB. Because the first was a B. That's my 10c worth which is not worth the 10c.

No one forced the birth order into the conditions...

Birth order and Tuesday are irrelevant.

The chances it is a boy are approx. 105 to 100. Statistically, more boys are born than girls.

https://ourworldindata.org/gender-ratio

If you interpret this question as one and only one child is a boy born on Tuesday, then 42.86%. That is the 50% chance the second child is a boy minus the 1/14 chance the second child was also a boy born on a Tuesday.

We have the information that of two children, one is a boy born on Tuesday, the second child must be a boy or a girl...there are only these two choices, and the birth is a 50/50 chance of being a boy, and also a 50/50 chance of being a girl...anything other than tortured logic arrives at a 50/50 chance for the other child being a boy....I don't see why people think they can predict random occurrences...

1 in 4. 25 percent chance of two boys, 25 percent chance of two girls and 50 percent chance of one boy and one girl. (Of course this ignores the complicating effect of identical twins, triplets, etc. ).

I realise this is not as smart as matlab but it's much more accessible to most of us.

I did a spread sheet also with a sample of 1 million two child families.

Top of spread sheet

Spread sheet formulae

A3 =ROUND(RAND())

B3 =ROUND(RAND())

C3 =ROUND(RAND()*7-0.5)

D3 =ROUND(RAND()*7-0.5)

E3 =IF(((A3=0)AND(C3=2))OR((B3=0)AND(D3=2)))

F3 =IF((E3=1)AND(A3+B3=0))

E1000003=COUNTIF(E3:E1000002, 1)

F1000003==COUNTIF(F3:F1000002, 1)

I4 and I5 are just copies of the counts at the bottom of columns E and F

prob I7 =I5/I4

And the "several goes" columns are just Paste/Special/Number of prob I7 each time it re-calculates.

-----------------------------

Bottom of spread sheet

So it's clear that 13/27 is the answer.

Nice work!!!