On Friday i presented everyone with this lovely puzzle…
Mrs Brown has two children. One of them is a boy called Albert. What is the probability of her other child also being a boy?
If you didn’t try to solve it, have a go now. For everyone else, the answer is after the break.
First of all, ignore the fact that Mrs Brown called her son Albert – that was just a red herring.
Second, many people think that the answer is 50%. This is wrong.
Let’s imagine that a woman has two children. Assuming that they were not born at the exact same time, there are four possible combinations….
First Second
Boy Boy
Boy Girl
Girl Boy
Girl Girl
Obviously, this will give rise to 50% boys and 50% girls. However, in the puzzle we know that one of Mrs Brown’s children (Albert) is a boy, therefore she cannot fall into the last category, thus leaving us with…..
Albert Boy
Albert Girl
Girl Albert
All of these options are equally likely, so the chances of her other child being a boy is one third.
It is a lovely puzzle because it shows how our intuitive sense of a situation is often at odds with the actual probabilities. This, in turn, explains the success of those really big casinos in Vegas.
Update: It seems that mentioning young Albert does change the puzzle. According to here, the various options are…
Boy Albert
Albert Boy
Girl Albert
Albert Girl
…making the probability a half, not a third. It seems wrong that simply naming the boy makes such a big difference. Any thoughts?
May 25, 2009 at 1:12 am |
Oooh there are gonna be some rows on here about that. It’s like the goats and doors thing. I’ve seen people get apoplectic over that one. No matter how many different ways you try to explain it…
May 25, 2009 at 1:12 am |
Aren’t “Boy Girl” and “Girl Boy” the same thing? I’m pretty sure order of the children doesn’t matter in this since nothing about age was announced.
May 25, 2009 at 11:08 pm
Probabilistically they are different, as you are considering 2 separate independent events (the gender of child A and the gender of child B).
Otherwise if you knew nothing about either child the probability of having 1 boy and 1 girl would be 1/3 rather than 1/2.
May 25, 2009 at 1:24 am |
If you say “I have 2 children,” the only possibilities you have are:
2 boys, 1 boy 1 girl, or 2 girls?
May 26, 2009 at 9:52 am
Yes, but you can have a boy and a girl in two different ways: First a boy and then a girl, or first a girl and then a boy. Having two boys or two girls can only happen in one way: First one and then another one. So the probability of having one of each is twice that of having two boys or two girls.
May 25, 2009 at 1:40 am |
I agree with samuel. Wouldn’t the possible combinations be 2 boys, 1 boy 1 girl (or 1 girl 1 boy), and 2 girls? Since Mrs Brown has one boy, its narrowed down to either 2 boys or 1 boy 1 girl. So if the logic is right, the chance of the other child being a boy would be 50%. Just my 2 cents.
May 25, 2009 at 1:45 am |
I think there is some false logic in the puzzle answer. The possible combinations does not affect the probability of an individual child being a boy or a girl. So the answer is 50%. Say you filpped a coin twice. Just becase the first coin was head does not make it more likely the second coin toss will be a tail. Its still 50/50
May 25, 2009 at 5:12 pm
Actually, is it not 51/49 of something like that? 50/50 apparently doesn’t exist :s
May 25, 2009 at 11:12 pm
You’ve actually given more information about the coin toss then we had for this puzzle (i.e. that it was the first coin which was a head).
If you specify that the eldest child is a boy then the only possible combinations are Boy Boy or Boy Girl (i.e. we elemenate both Girl Boy and Girl Girl), so the probability is 1/2.
Simiarly if you say you toss 2 coins an one of them is a head, the probability of the other being a head is 1/3 for the same reasons as above.
May 25, 2009 at 1:48 am |
I came up with the right answer, then thought about it some more and changed my mind. Given that there are 4 boys in the above scenario, couldn’t Albert be any of those 4, giving Albert boy; boy Albert; Albert girl; girl Albert – hence 50% chance. I expect the answer given in the post is correct, but I think I need some more explanation please.
May 25, 2009 at 11:18 pm
Being called Albert is a separate event to whether you are a boy or a girl.
Assuming there’s a 50/50 chance of any given child being a boy or girl, then the following 4 options are all equally likely
Boy Boy
Boy Girl
Girl Boy
Girl Girl
You could split Boy Boy into Albert Boy and Boy Albert, but these probabilities of these events would be half that of the others (assuming equally probabilities of Albert being either child).
May 26, 2009 at 9:54 am
The problem with this is that you’re assuming that the possibility of the sibling being a boy is equal to the possibility that it is a girl to begin with. If you set as a premise that the two possible outcomes are equally probable, that will be the conclusion too.
May 25, 2009 at 1:49 am |
I understand the answer (though tend to agree that boy then girl and girl then boy would generally be seen as the same) but I can’t go past the biological concept that no matter if she has 1 or 10 boys already, the chances or having a boy will always be 49-50% (and chances or having a girl will always be 50-51%). I guess I just don’t see the “trick” as clever enough to not want to insist on the boring scientific answer.
May 25, 2009 at 11:24 pm
Boy Girl and Girl Boy would be seen the same in many circumstances, but the point is that the 4 events (BB, BG, GB and GG) all have equal probability of occuring and we’re just ruling out the last one.
May 25, 2009 at 2:05 am |
I call shenanigans! Maybe I’m crazy, but I think there’s an experiment afoot. First, there was that set of optical illusions where there were first three, and then a fourth one popped up where Richard claimed that two shades were identical but in fact they were not. People called him out on it and it disappeared, and I don’t think there was an explanation. Now there’s this. It just doesn’t logically follow. I’m wondering if he’s testing how likely people are to trust information posted on a blog by a person with some authority, or something along those lines. It wouldn’t be all that different from the experiment in the mall where people were convinced that those objects were somehow enchanted, especially when dressed in lab coats. Again, I may be crazy, but its just a little theory of mine.
May 25, 2009 at 11:21 pm
Sorry, in hindsight I realize that this was a bit out of line. It was semi-late at night when I posted this, and I probably shouldn’t have. I’m still not sure I buy the 1/3 approach, but my previous post makes me sound a bit like a conspiracy theorist. Sorry.
May 25, 2009 at 2:18 am |
I looked at it this way – the fact that Albert is a boy is also irrelevant. We are really only concerned with the gender of the other child and that would be determined by chromosomes.
The other child can be any of the following combos:
X from mom with X from dad
X from mom with Y from dad
therefore 50/50. How is the gender of the first child relevant when you are asking specifically about the gender of the 2nd child?
May 25, 2009 at 11:26 pm
The point is that we don’t know the order of the children, so we are not asking the gender of the second child.
If you specified which child Albert was then it would be 50/50.
May 25, 2009 at 2:29 am |
nice example when knowing less is better
May 25, 2009 at 2:30 am |
guys, if you don’t believe the result … read this http://en.wikipedia.org/wiki/Conditional_probability#Introduction
May 25, 2009 at 4:15 am |
“First of all, ignore the fact that Mrs Brown called her son Albert – that was just a red herring.”
Maybe it was intended to be, but it isn’t. If this was left out, your answer would be correct. As it stands, 50% is the right answer.
May 25, 2009 at 4:27 am |
Very well put Joanna. Straight to the point that I was kind of getting at in my earlier comment, but failed miserably at.
I can see that this debate is going to rage for a while. Should be fun.
May 25, 2009 at 6:11 am |
Yip, Richard missed the Boy Albert option. If the puzzle had read “Albert is the oldest boy”, then there would be a 33% chance of the other child also being a boy.
May 25, 2009 at 11:21 pm
Actually it’s the other way around.
As it’s stated, it’s 1/3
However, if you specify Albert as being the oldest then it’s 1/2.
May 25, 2009 at 6:29 am |
Here’s how I explained the problem to myself:
The question here is not “Mrs Brown is expecting her second child. What is the probability of it being a boy?”, where the answer would naturally be 50%. Instead, it’s rather like “What kind of two-child family does Mrs Brown have?”, or in mathematical terms, “Which pre-existing group does the Brown family belong?”. We are not anticipating a future event like a coin toss – the children are already born, and there is a certain number of certain kind of families already in existence.
When parents get their first child, the probability of it being a boy is 50% (for the sake of this problem, anyway). This divides the family into one of two more or less equal-sized groups – those having a boy, and those having a girl. When those families get their second child, they are further divided. We now have four groups, each with roughly the same amount of families – BB, GG, BG, GB.
Let’s take, say, a thousand two-child families selected at random. We now have (for the sake of simplicity) exactly 250 of each kind of them. One of the families is selected at random, and that happens to be Mrs Brown. One of the children is revealed to us at random, and that happens to be a boy (named Albert). The probability that this child is a boy is 50% – there are one thousand boys and one thousand girls in the group.
Now, which of these four groups does this family come from? It obviously can’t be the one where both children are girls. It _can_ be any of the other groups where the other child is a girl or a boy. However, there are 500 families left where the other child is a girl, and only 250 where it is a boy. Hence the probability 1/3.
The fact that the boy is or is not named Albert does not matter. If you think “Albert – boy”, you are thinking “Albert” as a third sex.
(Richard’s answer was a little misleading in this sense.) Note that here we do not know if Albert is the elder or younger sibling. If we _do_ know that, the problem is similar to coin toss, and the answer becomes 1/2.
Right or wrong, that’s my line of reasoning.
May 25, 2009 at 7:58 am |
yes, i got it right
May 25, 2009 at 8:13 am |
Nice clear explanation Captain, it all makes sense. Putting real numbers to the problem helps you to see how the probabilities work out.
And cheers Richard for bringing this puzzle to my consciousness for the first time. I knew the Monty Hall one, but this is a nice little variation of it.
May 25, 2009 at 8:15 am |
Well done Richard, you continued to fool many even after revealing the answer!
May 25, 2009 at 8:18 am |
Grr. I did not see this as a conditional probability problem at first and now it makes perfect sense. Never been good at probability, I have only recently been introduced to the Monty Hall problem.
Was confused in your answer when you said that the fact that the child’s name was Albert was a red herring. I guess maybe you mean that the Name was irrelevant but the sex wasn’t, I don’t know, bit odd.
I don’t think it matters if Albert was the eldest or youngest. The way I see it is, work out the probability for both scenarios and you get a 1/3 which surely equates to 2/6 -> 1/3. But like I said, I really know nothing about Probability. Need to keep my eye out for these probability problems
May 25, 2009 at 8:35 am |
Ah, I see now that I got it wrong in my earlier comment. Captain’s reasoning seems right. There is a 12.5% chance that Albert has a younger brother, a 12.5% chance that Albert has an older brother, a 25% chance that he has a younger sister, a 25% chance that he has an older sister and there is a 25% chance that there are two daughters.
Probabilities are definitely a weakness of mine.
May 25, 2009 at 8:49 am |
Michael: You’re correct that a second coin isn’t affected by the first result. However, if you flipped two coin there’d be a possibility of both being tails (the possibilities are HH, HT, TH, and TT with equal probability). Suppose I told you that there was at least one head in the results, that rules out TT. That obviously doesn’t affect the probability of the other 3 outcomes, so HH, HT, and TH are still equally. So there’s a 2/3 chance of a tail among there.
You can try this. Get two coins. Flip them both. If they are both tails then discard the results, since we know that didn’t happen. Otherwise label (one of) the heads as Albert and note whether t’other is heads or tails. Repeat several times. You’ll get twice as many tails as heads for the one that isn’t Albert.
Lux: You’re correct that if the mother already has one boy then the second one is equally likely (within 1%) to be either sex. However, the puzzle doesn’t say that the first one is a boy; the first one has an evens chance of being a girl (in which case the second one has to be a boy, since two girls has been ruled out).
The mother already having a boy is equivalent to doing the above coin-flipping but discarding the results if the first one comes out tails (before even flipping the second one). That will give 50-50 for the second one, demonstrating how this case is different from the question actually asked.
May 25, 2009 at 9:09 am |
Here is an equivalent problem, but redesigned in such a way as to encourage visual rather than algebraic thinking. How does this change influence people’s intuition about the solution?
There are four species of Alberts with equal populations but distinct colourings:
- all green
- all blue
- top half green, bottom half blue
- top half blue, bottom half green
I have a pet Albert, which is at least partly green. What’s the probability that my Albert is all green?
May 25, 2009 at 9:09 am |
Does it all come down to this? You have to understand that there is a difference between those two scenarios:
1. You flip a coin and it is tails. What are the chances of it being tails when you flip it a second time?
2. A coin has been flipped two times. One of those times was tails. What are the chances of the other flip also being tails?
I love the Monty Hall problem and I think I am very good at explaining it. But now I see again how our feeble mind just does not understand probabilities. Awesome.
May 25, 2009 at 9:29 am |
It boils down to how you interepret the question – are you looking at the probablility of an event giving rise to a particular group (boy girl, boy boy, girl boy, girl girl) or are you looking at the probability of an individual event (gender of a child). Depending which interpretation you use the answers are both correct.
May 25, 2009 at 9:39 am |
another way to come about it is:
P(any boys) = 3/4
P(two boys n any boys) = 1/4
n = intersection/AND
| = given
P(two boys | any boys) = P(two boys n all boys)/P(any boys)
= (1/4)/(3/4) = 1/3
This still took sleeping on it to get right, and I used to be good at probability.
sigh,
^_^
W R
May 25, 2009 at 9:57 am |
I managed to get it correct, and for the correct reasoning! Yay!
May 25, 2009 at 10:25 am |
If “Albert Girl” and “Girl Albert” count as 2 distinct possibilities, you must also count “Albert Boy” and “”Boy Albert” as 2 distinct possibilities, which puts the probability back at 50%. Either permutations matter or they don’t, but either way it’s 50%.
May 25, 2009 at 10:30 am |
Malte: Yes, that’s key.
Gordonrutter: For the question actually asked there’s only one correct answer.
Obviously if you misinterpret the question and answer a different one to that asked then you can get other answers, but that isn’t interesting — can get the answer “99·44%” like that (if I happen to misinterpret the question as being “What percentage of Ivory soap is pure?”)!
May 25, 2009 at 10:30 am |
That is not right maidden, because any couple has a 25% chance of having 2 boys and a 50% chance of having both a boy and a girl (BG & GB). Therefore, the chance that Albert has a younger brother is 12.5%, while the chance that Albert has a younger sister is 25% (if you include the chance that Albert does not exist due to there being 2 girls).
May 25, 2009 at 10:57 am |
You can’t include the chance that he doesn’t exist because we already know he exists. The point is that the puzzle has to be solved by either counting possible combinations (not caring about the order) or counting possible permutations (caring about the order).
If the puzzle had said “a woman has 2 children and one is a boy”, then yes, the answer would be 1/3 probability of another boy, because the boys would be completely indistinct from one another. But the puzzle named the boy Albert, so he is distinct from his possible brother. So, if you’re only considering combinations:
* Albert Boy (in whatever order) 50%
* Albert Girl (in whatever order) 50%
and if you’re considering permutations:
* Albert Boy (in this order) 25%
* Boy Albert (in this order) 25%
* Albert Girl (in this order) 25%
* Girl Albert (in this order) 25%
which gives 50% probability of another boy either way. Now, if the puzzle had just said “a boy” generically, you’d have these combinations:
* Boy Boy (in whatever order, but that doesn’t matter here) 50%
* Boy Girl (in whatever order) 50%
and these permutations:
* Boy Boy (change in order doesn’t matter) 1/3
* Boy Girl (in this order) 1/3
* Girl Boy (in this order) 1/3
which gives 2 different results depending on whether you’re counting permutations or combinations.
May 25, 2009 at 4:04 pm
Counting “Albert Boy” and “Boy Albert” seperately makes no difference – these probabilities must add up to 25% (or 33% when we rescale them after removing GG). Your probabilities are then:
Albert Boy – 1/6 ( 1/3 chance that it’s BB * 1/2 that Albert is the younger)
Boy Albert – 1/6 ( 1/3 chance that it’s BB * 1/2 that Albert is the older)
Albert Girl – 1/3 ( 1/3 chance that it’s BG. Albert must be younger)
Girl Albert – 1/3 ( 1/3 chance that it’s GB. Albert must be older)
The various combinations you give aren’t equal – you must also take into account the probability of each particular possibility occurring – that is, the sum of the probabilities of each combination can’t exceed the slice of probability that you took it from.
May 25, 2009 at 11:13 am |
Be careful of missing the bigger picture. A one child family has a chance at the following profiles
B – 50%
G – 50%
The second child has a 50% chance of being either a boy or girl, so this yields the following probabilities:
BB – 25%
BG – 25%
GB – 25%
GG – 25%
Thus, if there is a boy named Albert, he is either part of a BB family (1/3 chance), part of the BG family (1/3) or the GB family (1/3) if you disregard the GG family. There is a 1/3 chance he has a older sister, a 1/3 chance he has a younger sister (total of 2/3 chance he has a sister), a 1/6 chance of having a younger brother and a 1/6 chance of having an older brother.
May 25, 2009 at 12:21 pm |
Only this puzzle doesn’t have a “BB” family, it has a “BA” or an “AB”, since the boys are distinguishable (one of them is named Albert). The reasoning you’re using is perfectly correct for the case of permutations with indistinguishable objects, but this isn’t the case of this puzzle.
May 25, 2009 at 12:35 pm |
I didn’t get it. Am hitting myself!
Great puzzle, thanks!
May 25, 2009 at 1:56 pm |
The fact that the boy is called Albert does make a difference, it is not a red herring. It changes the probability from 1/3 to approx 1/2. For the full explanation read the Boy / Girl paradox article on wikipedia, in particular the answer to the Third Question…
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
May 25, 2009 at 2:24 pm |
I agree with those who say the answer is 50%, although without most of the math. The answer is very simple:
it doesn’t matter in which order the children are born. She has two children, and the sex of one of them is given. This child is now out of the equation. Any other children she may have in the future, children she has had in the past and are no longer with us, or pets are also equally irrelevant. The question boils down to: “what is the sex of Mrs. Brown’s other child,” and that is roughly 50/50.
May 25, 2009 at 2:47 pm
Graey: Nope, the one whose sex is given isn’t out of the equation, because he wasn’t chosen randomly.
Suppose somebody putting objects into a bag is equally likely to pick a sprout or a bar of chocolate. We get him to put two items into a bag, to share. I get to look in the bag first and take out a bar of chocolate. You get the other item. Is that fair?
By your logic the item I remove from the bag is irrelevant, so you’re equally like to get chocolate or a sprout. In which case picking first wouldn’t be an advantage.
May 25, 2009 at 2:46 pm |
If we’d been given us the name of the other child then we’d have solved it a lot quicker!
May 25, 2009 at 2:49 pm |
Ignore error in last post
May 25, 2009 at 2:54 pm |
I’ve managed to confuse myself again, using a related problem in which numbers replace names. I hope the following survives formatting in a sufficiently readable form.
Suppose two people have independently selected a random number between 1 and 8.
Possibilities:
1&1 1&2 1&3 1&4 1&5 1&6 1&7 1&8
2&1 2&2 2&3 2&4 2&5 2&6 2&7 2&8
3&1 3&2 3&3 3&4 3&5 3&6 3&7 3&8
4&1 4&2 4&3 4&4 4&5 4&6 4&7 4&8
5&1 5&2 5&3 5&4 5&5 5&6 5&7 5&8
6&1 6&2 6&3 6&4 6&5 6&6 6&7 6&8
7&1 7&2 7&3 7&4 7&5 7&6 7&7 7&8
8&1 8&2 8&3 8&4 8&5 8&6 8&7 8&8
Using ‘e’ for even and ‘o’ for odd:
oo oe oo oe oo oe oo oe
eo ee eo ee eo ee eo ee
oo oe oo oe oo oe oo oe
eo ee eo ee eo ee eo ee
oo oe oo oe oo oe oo oe
eo ee eo ee eo ee eo ee
oo oe oo oe oo oe oo oe
eo ee eo ee eo ee eo ee
Summary:
16 occurences of oo
16 occurences of ee
16 occurences of eo
16 occurences of oe
Now suppose it’s revealed that one person selected the number 2 (but not who). This leaves:
— 1&2 — — — — — —
2&1 2&2 2&3 2&4 2&5 2&6 2&7 2&8
— 3&2 — — — — — —
— 4&2 — — — — — —
— 5&2 — — — — — —
— 6&2 — — — — — —
— 7&2 — — — — — —
— 8&2 — — — — — —
Summary:
7 occurences of ee (namely 2&2 2&4 2&6 2&8 4&2 6&2 8&2)
4 occurences of eo (namely 2&1 2&3 2&5 2&7)
4 occurences of oe (namely 1&2 3&2 5&2 7&2)
So the probability that both numbers are even is 7/15
If the numbers were between 1 and 4, the answer would be 3/7
If the numbers were between 1 and 6, the answer would be 5/11
If the numbers were between 1 and 100, the answer would be 99/199
The probability approaches 1/2 as the range of numbers increases.
So its 1/2, not 1/3. Um…
June 1, 2009 at 3:11 pm
Fortunately or unfortunately, parents do not first name their children, and then expect the name to determine the gender of the child!
May 25, 2009 at 3:13 pm |
What melts my head on this is that if Albert is the older of the two, then the chance is 1/2 he has a brother, and if Albert is the younger of the two, then, again, the chance of him having a brother is 1/2. As either scenario *must* be true, one would expect the odds of Albert having a brother would be 1/2. But it’s 1/3.
May 25, 2009 at 3:39 pm |
If I had been paying attention and seen the word “probability” for what it was the first time I read the problem, I would have been a lot better off. My mind parsed it into “chance”, which led me down the wrong path.
May 25, 2009 at 3:44 pm |
Many people here suggest the “boy boy” scenario needs to be counted twice, but then neglect that in that case the “girl girl” scenario does too. That brings us to 2/6 = 1/3.
May 25, 2009 at 4:01 pm |
Can someone post something from a site that is not wikipedia for their examples?
May 25, 2009 at 4:04 pm |
Individual chance of child being a boy is 1/2. Combined probability of second child being a boy when first one is boy is 1/3. Richard’s decision tree from the original answer is correct.
May 25, 2009 at 4:15 pm
Ah, now I get the issue with the named boy. It changes the decision tree…
If looked at by possible birth order, we get…
Albert – Girl
Albert – Boy
Boy – Albert
Girl – Albert
Which gives 50:50 chance for the sibling being a boy or girl.
May 25, 2009 at 4:18 pm |
I don’t know if this helps but I created a google docs spreadsheet that hopefully models the problem.
http://spreadsheets.google.com/ccc?key=rielzrngrhtVWzWV2OQixFA
Column A: Child 1 has a 50/50 chance of being Boy/Girl
Column B: Child 2 has a 50/50 chance of being Boy/Girl
Column C (Is there at least one boy?, i.e. could this be Mrs Brown’s family?): If both Child 1 and Child 2 are Girls then No, otherwise Yes.
Column D: (Are there two boys?): If both Child 1 and Child 2 are boys, then yes.
The “Yes” count in Column D is about a third of that in Column C, so for all the families that have at least one boy, about a third of them have two boys. Mrs Brown’s family has one boy, so there’s a roughly 1/3 chance she has two boys.
(To simulate a different set of 1000 families, highlight columns A and B and press CTRL+R to regenerate the random numbers.)
Hope this helps!
May 25, 2009 at 4:37 pm |
@Smylers
I’m looking at it this way: The guy puts either a chocolate bar or a spout into a bag, which Mrs Brown takes. It happens to be a sprout which she names Albert. The guy puts another item into the bag. Mrs. Brown then takes the bag, with the hidden item in it, and starts the puzzle. (She could have done it in reverse too.) When she opens the bag, there is an even chance it will be a sprout or a chocolate bar.
I understand the probabilities if you start from the beginning with no children, but the way I see the puzzle is different. She has one known child named Albert. She has another child. What are the odds that it’s a boy?
May 25, 2009 at 5:56 pm
Gracie: Yes, if Mrs Brown first had Albert and then had a second child then what you says is true, and the chance of the second child being a boy is 1/2. But the puzzle doesn’t say that.
It merely says that _one_ of the children is Albert, without stating which. So there’s a chance that the first child is female (or that the first item put in the bag is chocolate); that’s what changes things.
If you still don’t believe me, simply try it! Get some sprouts and chocolate, or flip some coins like I outlined above.
May 25, 2009 at 4:44 pm |
For me, this makes it really clear:
The event that the first child in a family is a boy and the event that the second child in a family is a boy are INDEPENDENT. Thus, assuming that P(boy)=P(girl)=0.5 the probabilities are:
P(Boy,Boy) = 0.25
P(Boy,Girl) = 0.25
P(Girl,Boy) = 0.25
P(Girl,Girl) = 0.25
We have to count P(Boy,Girl) and P(Girl,Boy) separately and if you really look at demographics, families with one boy and one girl actually do account for about 50% of families with 2 children!
Given that Albert is a boy, clearly out of (Boy, Boy), (Boy, Girl) and (Girl Boy) possibilities that probability that his sibling is a boy is 1/3
Now, if we knew that Albert was either the younger or the older, the probability would rise to 1/2 because we would eliminate one of the (Boy, Girl) options (depending on whether Albert was younger or older).
As I said, sometimes knowing less is better!
May 25, 2009 at 5:21 pm |
Okay, I heard, and may well be completely wrong, that it is genetically more likely to conceive a male on your first pregnancy than a female*. So, knowing if Albert is younger or older would affect the puzzle if we were being anal enough to include this statement.
*I’m not sure were I heard this, I may have just made it up
May 25, 2009 at 5:26 pm |
Alex, these types of issues are covered in the Wikipedia article that Christian linked to.
“These assumptions have been shown empirically to be false. It is worth noting that these conditions form an incomplete model. By following these rules, we ignore the possibilities that a child is intersex, the ratio of boys to girls is not exactly 50:50, and (amongst other factors) the possibility of identical twins means that sex determination is not entirely independent. However, one can see intuitively that the occurrence of each of these exceptions is sufficiently rare to have little effect on our simple analysis of the general population.”
May 25, 2009 at 5:27 pm |
It makes a difference whether the boy is named.
If he’s not named, the question is simply “In a family with two children, if one is a boy, what’s the chance the other is a boy” (ie it’s a question of conditional probability). The possibilities are therefore Boy Boy, Boy Girl and Girl Boy, so the chance of 2 boys is 1/3 (counterintuitively).
If he’s named, the odds are changed. We can call the different outcomes Girl, Boy-A (ie boy called Albert) and Boy-NA (boy not called Albert). Given we know we have at least one boy (who’s called Albert) and we can reasonably ignore the chance of a family in which two sons are called Albert, the possibilities are:
Girl Boy-A
Boy-A Girl
Boy-NA Boy-A
Boy-A Boy NA
Bringing us back to 1/2. Which is again counterintuitive if you’ve just got your head round the first question.
There’s a very clear (and funny) explanation of all this in Leonard Mlodinow’s book “The Drunkard’s Walk”
HTH.
May 25, 2009 at 6:09 pm
Lizzie, you can’t tabulate it like that: being called Albert or not isn’t something which has a 50:50 chance like being a boy is.
Telling you that one of Mrs Brown’s children is called Albert is mathematically equivalent to telling you that at least one of Mrs Brown’s children is male. Again, try the coin-flipping thing: if it comes out HH you get to pick which one is Albert, but it doesn’t make HH any more likely.
May 25, 2009 at 7:04 pm
But the probability of Boy-A Boy-NA is not the same as Boy-A Girl.
Boy-A Boy NA can occur iff there are 2 boys AND the younger is called Alfred. Similarly Boy-NA Boy-A occurs iff there are 2 boys AND the older is Alfred. This means that each of these probabilities must be divided in half, then added to get the probability that the other is a boy – they have only half the weight as Alfred-Girl or Girl-Alfred.
May 25, 2009 at 5:35 pm |
@Lizzie, I think I disagree on that with you … the probability is the same, the two groups of families, one with (Boy-NA Boy-A) and the other with (Boy-A Boy-NA) are the same for our purposes. This is not the same situation as (Boy, Girl) and (Girl Boy) which needs to be counted twice.
May 25, 2009 at 5:39 pm |
Scibuff, I agree it’s counterintuitive. The explanation in the book I mentioned above is excellent- if I get a chance later I’ll summarise it. It helps in understanding it just to write out all the combinations (inc those which aren’t correct (eg GG) then cross out those which don’t apply. A lot of typing though
May 25, 2009 at 5:54 pm |
Took me a while to think about the “named / not named Albert” -issue and how it affects the outcome, but I think I finally got it. Probably.
Again, the real point here is not if the boy has a name or not. It’s rather either:
“We consider the Brown family, who have a boy named Albert, who live at Math Street 15, and who are unique in the whole universe.”
or
“We consider _any_ family that has a boy named Albert.”
If you look at Wikipedia’s “Third question”, it does say “A (random family) has…”, instead of giving a name to the family, like the other questions do. Richard’s original question talks about Mrs Brown, which in my mind implies we talk about a real, single, chosen family, not just any family that fits the parameters. In this case, giving more information about the family (other than the gender of the other child) does not change anything. In the other case (any family that has a boy named Albert), everything becomes more complex, like Wikpedia’s example shows..
May 25, 2009 at 7:24 pm
In other words – is the information that the family has a son named Albert something that:
- is given to us after a single family has been selected
or
- is used as a search filter when selecting a family
May 25, 2009 at 7:20 pm |
@Captain – I think you’re right. It depends on how the family is arrived at. In the Wikipedia question, it’s making the assumption that we’ve started with the condition “Family with a son named Albert”, found a matching family and used it.
With this question, that isn’t the case – we’ve started from an existing family, and there’s no reason to assume that we’ve searched for an Albert. ie. the information “Has a child named Albert” is not independent of the selected family, but is chosen after the family was selected. If Mrs Brown’s son had been called Charlie the question would have read “One of them is a boy called Charlie”, thus it doesn’t give any more information than “Has a son with some name”
May 25, 2009 at 7:27 pm
Exactly, you beat my second explanation with a couple of minutes
May 25, 2009 at 10:47 pm |
Okay, after reading all of these comments my brain is pretty much broken. I still think it’s 50/50.
May 25, 2009 at 11:11 pm |
Simple! The odds that the other child is a boy are 0%. We know Albert is a boy, and we know “Mrs. Brown, you have a lovely daughter”.
May 25, 2009 at 11:20 pm |
From wikipedia about coin example:
(A) If I throw 2 coins and let you see one, have I given you any information about the 2nd (hidden) coin? – Obviously not, its probability of heads or tails remains .5/.5
May 26, 2009 at 12:18 am
You’ve not told us whether the hidden coin is a head or tail.
But you have not only told us one coin is a head, but crucially you have told us which coin is a head.
We are not told here which child is Albert, just that one of them is.
May 26, 2009 at 12:18 am |
My biggest problems was that I was thinking about a parent with a single child and then her giving birth to a second. Hence it would be 50/50.
I see how not specifying which of the two children are boys makes a difference.
As to Sam’s coin tossing analogy… being shown one of the coins is not the same as being told that one of the coins is a head. In the first case you know *which* of the coins is a head and in the second you do not. The two situations are different.
May 26, 2009 at 10:05 am |
Here’s a bit of code that should hopefully throw some light on the naming / not naming question.
It performs the following process:
1. Randomly generate a 2-child family.
2. Any children that are boys are randomly assigned a name, with equal probability from 50 possible names (ie Albert = 2% chance for boy)
3. The name of a RANDOMLY CHOSEN BOY is revealed (assuming there is a boy)
4. If there are no boys, or the revealed name is not Albert, ignore this selection
5. Otherwise, count whether the other child is a Boy or Girl.
The way this differs from the situations Lizzie and Christian have pointed out is in whether the question is independant of the name.
If we changed point 3 to “If the name of EITHER BOY is Albert, use that name”, discarding other results at that point, we would reach the approx 50% chance
(actually more than 50%, but approaching 50% as the probability of the name approaches 0)
However, this is different from the question as stated – we’ve no reason for assuming Mrs Brown has decided to preferentially reveal sons called Albert – it’s more sensible to assume she has revealed a name at random. If the child she had picked had been called Bob or Charlie, she would have used that name in the question, rather than picking a potential Albert named child over them.
ie.
– Pick a family where a RANDOMLY CHOSEN child is named Albert : 66%
– Pick a family with ANY child named Albert : ~ 50%
The mindbending aspect of these problems is how such seemingly simple changes in picking a sample alter the results.
Here’s the code (in python):
import random, itertools
boy_names = ['Albert'] + ['name%s' % i for i in range(49)] # Albert + 49 other names - assume equal probability
def get_child():
sex = random.choice(['boy','girl'])
if sex != 'girl':
return random.choice(boy_names)
return sex
def families():
while 1:
c1, c2 =get_child(), get_child()
if c1 == 'girl' and c2 == 'girl': boy_name = None # No boys
elif c1 == 'girl': boy_name = c2
elif c2 == 'girl' : boy_name = c1
else: boy_name = random.choice([c1,c2]) # Choose name from either boy randomly
yield c1,c2, boy_name
num_BB = 0
num_BG = 0
count =0
SAMPLE_SIZE = 5000
for c1,c2, name in families():
if name =='Albert': # At least one boy, called Albert
if 'girl' in [c1,c2]: num_BG += 1
else: num_BB += 1
count +=1
if count == SAMPLE_SIZE: break
print "%d boy/boy, %d boy/girl = %d%% other child is a girl" % (num_BB, num_BG, 100*num_BG / float(num_BB + num_BG))
May 26, 2009 at 10:07 am
Oops – that lost all the formatting – here’s the code again in a form that should hopefully work:
import random, itertools boy_names = ['Albert'] + ['name%s' % i for i in range(49)] # Albert + 49 other names - assume equal probability def get_child(): sex = random.choice(['boy','girl']) if sex != 'girl': return random.choice(boy_names) return sex def families(): while 1: c1, c2 =get_child(), get_child() if c1 == 'girl' and c2 == 'girl': boy_name = None # No boys elif c1 == 'girl': boy_name = c2 elif c2 == 'girl' : boy_name = c1 else: boy_name = random.choice([c1,c2]) # Choose name from either boy randomly yield c1,c2, boy_name num_BB = 0 num_BG = 0 count =0 SAMPLE_SIZE = 5000 for c1,c2, name in families(): if name =='Albert': # At least one boy, called Albert if 'girl' in [c1,c2]: num_BG += 1 else: num_BB += 1 count +=1 if count == SAMPLE_SIZE: break print "%d boy/boy, %d boy/girl = %d%% other child is a girl" % (num_BB, num_BG, 100*num_BG / float(num_BB + num_BG))May 26, 2009 at 12:15 pm
Apologies for following up myself again, but I’ve thought of a better way of illustrating the difference in questions. Consider these two questions:
1. You meet Mrs Brown. She points to someone beside her, and says “This is my son, Albert. My other child isn’t here today.”
What is the probability of her other child being a girl?
Answer: 66%
2. You meet Mrs Brown one day, and get into conversation. You happen to mention your name is Albert. “That’s a coincidence.” She says, “I have a son called Albert”. What is the probability of her other child being a girl?
Answer: A little over 50%, depending on the popularity of the name Albert among children of that generation.
The difference in (2) is that we’ve chosen the name before revealing the information. Mrs Brown is now giving us information about a child chosen from “sons of Mrs Brown called Albert”, rather than “sons of Mrs Brown”
Without any further information about why Mrs Brown reveals the child, I think the most reasonable assumption is that it is random and independent (ie. she picks a child at random and tells you its name and gender)
May 26, 2009 at 1:09 pm |
You’ve really put your finger on the issue with this comment. Even leaving aside the boy’s name and just taking ‘mrs brow has 2 kids, one is a boy’, the answer can be 1/3 or 1/2 depending on if she was selected from a population of 2 kid families or a population of 2 kid families with one boy.
As I see it there are (at least) 3 possible equally valid interpretations of the genuinely ambiguous problem.
1) Out of a population of 2 kid families, a single family is chosen. A kid is randomly selected, it happens to be a boy. His name happens to be Albert. Probability of other kid being boy: 1/2. This is often considered to be a wrong answer, and many people pick 1/2 as the right answer for the wrong reasons. But it is valid answer, even if the boys name is not specified in the question.
2) Out of a population of 2 kid families, a sub-population of families where at least one kid is a boy is selected. A single family is chosen from this sub-population. A random boy (there is at least one) is chosen from this family, his name happens to be Albert. Probability of other kid being boy: 1/3. This is how the problem was originally interpreted by Richard, and is often considered the only correct answer.
3) Out of a population of 2 kid families, a sub-population of families where at least one kid is a boy called Albert is selected. A single family is chosen from this sub-population. Probability of other kid being boy: 1/2.
In other words: context is vital!
May 26, 2009 at 1:16 pm |
Brian, the answer to your #1 is 50%. Not because of the name but because you are identifying the particular child who’s gender you are trying to estimate.
Or to look at it another way, you are selecting the family from the population of all two kid families, not just families with 2 kids including at least one boy. In other words, Mrs Brown might have had a girl with her.
May 27, 2009 at 1:33 pm
Oops, you’re right – I should have left that question as worded in the article. Thinking further on it, its actually still pretty ambiguous, and all depends on exactly how we learned Albert’s name. A sequence of events that reaches your situation (2) is actually pretty unlikely – we need to learn she has at least one boy, and also determine that a randomly selected boy is named Albert, wherease the most likely scenario for learning both pieces of information would be revelation of a son called Albert, so 1/2 may actually be a better answer to the question as posed (if not as intended).
May 26, 2009 at 2:04 pm |
Am I going mad here or is the probability 50%?
This is not the same as the Monty Hall or the Bertrands Box problems.
The problem with thinking the probability is 33% is that you are repeating one of the options (Boy/Girl or Girl/Boy) by somehow thinking that the order they are born in is part of the equation.
In a 2 child family where one is a boy, there are 2 options: (a) both boys; or (b) boy and girl.
If you say there are 3 options [(a) both boys; (b) boy and girl but girl born first; (c) boy and girl but boy born first], you may as well bring other factors into the equation:
a) both boys; (b) boy and girl but girl born first; (c) boy and girl but boy born first ; (d) boy and girl but girl born first but boy was first to take up smoking; (e) boy and girl but girl born first and also first to take up smoking…etc.
To further illustrate my point lets go into the naming thing:
Mrs Brown has two children. One of them is a boy called Albert. If her other child is a girl lets call her Lucy. If her other child is a boy lets call him Jack. What is the probability of her other child also being a boy?
The Options:
Albert & Jack Brown
Albert & Lucy Brown
The proposed options if you think probability is 33%:
Albert & Jack Brown
Albert & Lucy Brown
Lucy & Albert Brown
… hold on why is Lucy getting preference?
If it turns out I’m a bit slow and incorrect please explain it to me gently, but until then I agree with Samuel, Riff and Sage at the top of the page.
Rant over.
May 26, 2009 at 2:20 pm |
dangle, put the name aside coz that confuses things slightly, do you get why it would be 33% if the name isn’t specified?
May 26, 2009 at 2:36 pm
No, coz I don’t see why the order the children are born makes any difference.. and if it does why not take it into account when both boys?
May 26, 2009 at 2:43 pm |
It has be to 50%. If you have 1 child there is a 50% chance its a boy, 50% chance its a girl. This never changes regardless of what comes before or after. If they had 8 boys in a row the probabily that the 9th kid is a boy is 50%. Past performance has no bearing on current probablity…
May 26, 2009 at 3:09 pm
That applies to all unborn children (predicting the outcome of a future event), but that is not what this problem is about – please read my first comment (for example).
May 26, 2009 at 2:59 pm |
4 equally probable cases of families with 2 kids are:
G G
G B
B G
B B
We know it’s not case 1 so that leaves the 3 equally probably cases
G B
B G
B B
Only one of these has 2 boys. So answer = 1/3. This is basically repeating Richard’s explanation.
Actually Flesh-eating Dragon’s analogy above is really clever:
http://richardwiseman.wordpress.com/2009/05/24/answer-to-the-friday-puzzle-2/#comment-4587
May 26, 2009 at 3:29 pm |
Sorry, I still don’t get it. You are again saying that the order of birth is a factor..
Here’s my version:
3 equally probable cases of families with 2 kids are:
*One of each sex
*Both Boys
*Both Girls
Ruling out the last one makes it 50/50.
Flesh-eating Dragon’s analogy makes sense because he is specifying coloured lizards where the order of body parts is a differentiating factor (i.e. a lizard with green top and blue bottom is not the same as vice versa). In this case, a family with a boy and a girl is just the same as a family with a girl and a boy.
And to go back to coin flips which someone earlier mentioned:
Mrs. Brown flipped a pound coin using the thumb of her right hand. The next day, she flipped the same pound coin using the thumb of her left hand. One of the coin flips resulted in ‘Heads’. What is the probability the the other coin flip also resulted in ‘Heads’?
You could equally say:
HH
HT
TH
TT
but why does the order of the head or tails matter (ie why include both HT and TH in the options)?
May 26, 2009 at 3:49 pm
The mixed-gender group is two times the size of the single-gender groups. So, if you want to classify a pool of 1000 randomly selected two-child families like this:
*One of each sex
*Both Boys
*Both Girls
…then the first group has 500 families, and the others 250 each. That’s why you can’t group them like that and treat them like they have equal probabilities.
May 26, 2009 at 3:45 pm |
It doesn’t matter that we know Albert’s name. Most kids have names, it doesn’t change the probabilities whether you know them or not. I’m somebody out there could design a computer simulation as an experiment.
May 27, 2009 at 1:54 pm
I’ve written one up above, though I’ve now changed my mind about the conclusion.
The simulation samples random families with at least one son, ignoring all except those where a randomly selected son is named Albert. This will give a 66% probability.
However, that “randomly selected son” part is a bit unnatural. To get there, we’d need to do the equivalent of asking Mrs Brown:
Q. Do you have at least one Son?
A. Yes
Q. Pick one of your sons randomly. What is his name?
A. Albert
However, change the code to be “Are ANY of your sons named Albert?”, by changing the line:
to
and you will get a probability close to 50%. This is what you’d get if you randomly sampled 2 child families until one answered “Do you have a son named Albert?” with “Yes”
So it all depends on how exactly we learned this information, and the name can indeed change the result – Liam has a good description of the distinctions a bit further upthread.
May 27, 2009 at 2:11 pm
Brian – How does it affect your simulation if you ask, “Are any of your children boys with a name that is different from all of your other children?” Certainly the fact that his name is Albert (as opposed to Stuart or Michael, say) is irrelevant. But I’d say that knowing his name is no different from knowing that his name is unique among his siblings (which is a reasonable assumption for most, though not all, families).
May 27, 2009 at 3:49 pm
@jenniferm:
That’s going to be roughly equivalent to “are any of your children boys”, assuming the liklihood of duplicated names is negligable, and gives 66% by my simulation. If its possible that children get the same name (and boys and girl names are disjoint), then the probability will actually increase *above* 66%, as families which don’t contain duplicate names are more likely to have a girl. Eg. if 50% of boys were called Albert, regardless of siblings, the chance that a family containing a boy and no duplicate names has a girl rises to ~80%.
(If you want to try this, the check I used was replacing the “if name==’Albert’:” check with “if c1 != c2:” which only looks at non-identically names children (and happens to also ignore girl/girl since all girls are currently named the same))
The actual name (Albert vs Michael etc) does matter, but only to a very small extent, since most names have fairly low probabilities. I’ve been simulating assuming a name occurring in 2% of the population, which is pretty much unnoticable. Here’s some stats to illustrate:
Family contains >=1 boy, A randomly picked boy’s name is called Albert:
– probability of a girl is always 66%
Family contains >=1 boy, Family contains a boy called Albert:
– 2% of boys are called Albert : ~50%
– 10% of boys are called Albert : ~52%
– 50% of boys are called Albert : ~57%
– 90% of boys are called Albert : ~64%
– All boys are called Albert : 66%
May 26, 2009 at 4:13 pm |
dangle:
Those cases are not equiprobable. You are twice as likely to have one of each sex than two boys. Consider the ages of the kids to be the differetiating factor. You do not know if the one you know about is the older or younger of the two. So the older is to the top bart of the lizard as the younger is to the bottom part.
Treats:
That’s a really good idea, you could design a computer simulation to do the experiment but this about how it would work:
1. create a random population of families with 2 randomly named and gendered kids.
2. randomly select a family from it
3. check if this family is suitable for the problem, if not reject and start again.
4. if it is, check if there’s 2 boys, and then repeat back to step 1
5. after repeating lots of times, count the fraction of times there was 2 boys.
The issue is step 3. What rile do you use to reject families? Do you reject a family chosen at this point if there are no boys? My guess is you say yes. Do you reject a family if there is no boy named Albert? My guess is you think no. But why not? The problem states it must be an Albert, right?
If you do reject all families that don’t have an Albert the (slightly surprising) result will be a little less than a probability of 1/2 that the other kid is a boy, depending how common the name Albert is.
Let me know if you follow my argument here!
May 26, 2009 at 4:16 pm |
Sorry Treats, typos, I should read what I type!. I meant ‘but think about how it would work’ and ‘What rule would you use’.
May 26, 2009 at 4:21 pm |
OK now I think I got the ultimate simple explanation:
There are more two-child families where a boy has a sister than those where a boy has a brother.
This explains the asymmetry which I guess many people are looking for – why the answer (in some interpretations of the problem) is not the beautiful, simple 1/2, but 1/3.
May 26, 2009 at 4:31 pm |
OK, you’ve convinced me at last- I guess I need it spelled out
Thanks guys
May 26, 2009 at 4:58 pm |
The problem here is that depending on the information you have at the start, the question is different (even though most people don’t realise that). The details are quite complicated, but there’s a good explanation at: http://www.overcomingbias.com/2008/03/mind-probabilit.html
(which is a good site in general)
May 27, 2009 at 1:42 pm |
There is simply no way that knowing the child’s name changes the odds. He’s still an individual whether you know his name or not. That said, the odds are still 50-50, and here’s why:
If boy-girl is different from girl-boy, then boy-boy is different from boy-boy. In other words, you don’t need to know the boy child’s name, but there are still two combos that involve two boys. First is the boy child we know about and his little brother. Second is the boy child we know about and his big brother.
May 27, 2009 at 7:18 pm
Think of it this way – at least this makes it perfectly clear for me, and does not involve probabilities, combinations or other difficult stuff at all.
We have a big sample pool of randomly selected two-child families. We randomly select one family. If it happens to have two girls, we ignore it and move on (since we know the family must have at least one boy). Otherwise, we get either a family with two boys, or a family with one girl and one boy. Now, if we repeat this many times, do we get a roughly equal amount of single- and mixed-gender families (completely ignoring girl-girl families)?
It’s easy to see that the answer is no – we get (roughly) twice as many mixed than boy-boy families, because there’s twice as many of them in the pool to begin with. And to return to the original problem, in a mixed family a boy has a sister, but in a boy-boy -family, a boy has a brother, that’s where the probability 1/3 comes from.
It might be tempting to count a boy-boy family twice (there are two boys there that both have a brother), but the original question only reveals the gender of one (in case of boy-boy, randomly selected) child.
May 27, 2009 at 2:18 pm |
@jeniferm:
>He’s still an individual whether you know his name or not.
That’s actually the core of the issue. If you just know “at least one boy”, then there is not a particular individual in mind – we only know something about the family as a whole, rather than a particular individual member. As for why Boy1-Boy2 / Boy2-Boy1 is different from Boy-Girl / Girl-Boy, looking at the probability tree may help:
1st 2nd __ Boy = Boy-Boy : 25% / Boy--< / \__ Girl = Boy-Girl : 25% / < __ Boy = Girl-Boy : 25% \ / \Girl-< \__ Girl = Girl-Girl : 25%Each branch is 50% likely. It doesn’t actually matter how you order them – you could make 1st the first born, 2nd the second born, or order them by height or shoe size – it doens’t really matter except that each sex has a 50% chance of being in each slot.
As you can see, before we discard Girl-Girl, there is a 75% chance that a family will have at least one girl in it (or at least one boy), and a 50% chance that it will have one of each. Since we know
Girl-Girl didn’t happen, we are left with just 25% 2 boys, 50% Mix. You can subdivide Boy-Boy into older/younger possibilities, but each of these is only half as likely as the general case of Boy-Boy. Rescaling the probabilities to the new range, Boy/Boy is now 25/75, or 33%, while Mixed is 50/75, or 66%. ie. the probability of a girl being present has reduced from 75% to 66%
May 27, 2009 at 4:19 pm
Doh – just realised that “height or shoe” size are really stupid examples to pick if I’m trying to make the point that they have 50% probability to be in either slot. Please substitute some non-sex linked trait, such as alphabetical order by name, day of the week they were born, or how far they were from their house at exactly 4:17pm yesterday – ie arbitrary but independant.
May 27, 2009 at 2:45 pm |
The original question is: “Mrs Brown has two children. One of them is a boy called Albert. What is the probability of her other child also being a boy?”
There is no mention of order or age in this question.
Suppose Mrs B is 9 months pregnant with her second child and due any moment. In this scenario the probability is 1/2, surely?
However, if you read it as Mrs B is chosen from a pool of 2-child families and has at least one child who is male then 1/3 is the answer, yes?
May 27, 2009 at 8:23 pm |
I think a few people seem to be treating this question the same as “what’s the probability her next child will be a boy” – but that’s not the same question.
The chance of next child being a boy is 50%, no matter what gender any existing children are. They’re independent events, so it’s 50% every time.
However, if instead you ask “What’s the chance of her child next being boy and the one after that being a boy” then it’s 25%. Each child has a 50% chance of being a boy, so that gives a 25% of them both being a boy. If you forget about excluding girl-girl combinations, then this is effectively the same question, we’re just asking it after she’s had two children instead of before (in both cases, we don’t know the genders so we have the same information).
If you then disregard any girl-girl combinations (because we’ve effectively been told “it’s not two girls”) and then re-normalizes we get 33% for boy followed by boy. The other 66% being girl-then-boy and boy-then-girl.
It’s the same as flipping a coin and getting two heads in a row. 25% chance. Even though each individual flip of the coin is 50/50.
May 29, 2009 at 10:58 am |
The actual probability of having another boy after the first is slightly higher than 50% (about 55%), since there are factors which influence sex determination (for example differences in the mobility of X or Y sperm under different vaginal and uterine conditions). If people have a boy once, they are more likely to have another boy, since they probably do not physiologicaly select against male offspring in the same way that other mothers might. There is also a slightly higher proportion of boys born than girls – they’re just more likely to die as they grow up, which is what gives an approximate 50:50 sex ratio in the population.
As to the logical solution, it makes sense at the level of possible outcomes (possibility), but it does not account for the probable outcomes (probability) of Mrs Brown having another boy. The same rationale could be applied to the lottery. The possible outcomes are:
Win
Lose
This gives a 50% possibility of winning the lottery, but the probability is far lower.
For the solution given, the question should have been phrased:
Mrs Brown has two children. One of them is a boy called Albert. What is the POSSIBILITY of her other child also being a boy?
June 1, 2009 at 6:18 pm |
The probability that any child will be a boy is about 52%, so the presumptions made there are wrong.
–JRM
June 1, 2009 at 8:43 pm |
“The probability that any child will be a boy is about 52%, so the presumptions made there are wrong.”
thats correct.. (baby boys got more chance tho to die cos of SIDS (sudden infant death syndrom)
and with every time the women is pregnant her chances are the same. (52%)
June 2, 2009 at 7:03 am |
Your first group of possibilities is correct giving rise to a 50% chance. Your mistake arose when using the name Albert to substitute for the word “Boy” in the first group so that the “Boy Boy” combination gave you “Albert Boy” but not also “Boy Albert”. The error was probably along the lines of “OK, I have made my substitution of Boy for Albert in this combination, move on to the next”.
@msh you are probably correct, but for which part of the Earth’s population? Infant mortality is not uniform all over the globe.
June 2, 2009 at 3:51 pm |
To really work it out you need to know the age of the mother and the age of the father; this influences sperm motility, amongst other things which changes the ratio of male:female. So too does ethnicity. So the actual likelihood cannot really be worked out at all as the puzzle has too little information to go on. Still fun though.
June 5, 2009 at 4:20 am |
If A=2 for B=1 or C=1 then
B+C=A
C+B=A
C+C=A
B+B=A
June 5, 2009 at 7:03 pm |
If Albert gets to get swapped in (Boy, Girl) and (Girl, Boy), then he sure as heck gets to get swapped in (Boy A, Boy B), and (Boy B, Boy A). Probability is 50%.
June 7, 2009 at 5:55 am |
The reason this does not give the same counter-intuitive result as the Monty Hall problem is that there are two parameters of that problem that are not replicated: In the Monty Hall problem, everyone knows there are prizes behind all but one of the curtains, that one having the goat. But *Monty* knows which one has the goat, and will never reveal what’s behind that curtain.
There is no privileged participant in your version of the problem, who knows the entire state of the possibility space and will only reveal certain information about it, nor is there any restriction on the balance of possibilities, so only biology-based estimates affect the chances that Mrs. Brown’s other child is a boy. Having one boy increases the probability that her reproductive system is biased to produce boys, or she might have had identical twins, and so on.
Okay, imagine I flip two coins, covering both with my palms before either of us see them. I then expose one coin and ask you if the other coin is the same or different. This is the same as asking you “heads or tails” for the first coin (when you had no information, and neither did I). It doesn’t matter what side came up on that coin, the other coin is a completely independent result (except insofar as they may have shared the same ballistics and might have started out facing the same way). The possibility space for the second coin is not meaningfully restricted by having collapsed the space for the first.
June 12, 2009 at 6:03 pm |
I can see why this question causes the same debate as the Monte Hall problem as it is equally prone to interpretation.
Answer 1 – a random mother of 2 is selected and a random child is selected and the sex revealed to be a boy. In this case, the odds of the 2nd child being male is an independant 50/50 event.
Answer 2 – the set of all mothers of 2 kids are examined and from those with at least one male child, this mother is selected. In this interpretation, all cases of girl/girl mothers are explicitly excluded at the knowlege of the questioner – so 25% of possible options for 2 kids are excluded and the answer is 25%/50%
So both answers are correct and depend on if mrs brown were a random choice or selected to have at least 1 son.
June 12, 2009 at 7:15 pm
Sorry, this is quite unrelated, but are you by any chance Nick Sharratt, the amazing children’s illustrator??
June 12, 2009 at 8:09 pm |
To @clementine – no, not that Nick Sharratt – I do get asked that faily often though by e-mail. (did meet his mum once
June 26, 2009 at 9:04 pm |
of course i didn’t read all comments.
in theory we have 4 different combinations, practically only 3 when dropping birth-order out of calculations.
therewore answer is 50%, it can be only boy or girl, not boy or girl or boy.
mathematics is cool and useful thing, but need specific care when dealing with material examples.
this comment has little bit of troll in it.
July 6, 2009 at 3:46 am |
Alright, let me restate the question here….
“Mrs Brown has two children. One of them is a boy called Albert. What is the probability of her other child also being a boy?”
The probability that the other child is a boy is 50%. The probability that it’s a girl is 50%. Why? Because Albert has nothing to do with this, it’s useless information. Each and every child-birth is independent of one another, and has nothing to do with the other child or child-birth.
October 23, 2009 at 5:29 am |
I’ve been trying to find which category of probability problems this kind of puzzle falls under. We are dealing here with a system with a “memory”. For this reason the rules of probability for tossing coins don’t necessarily apply. We can’t be certain that the ratio of XY chromosomes to XX chromosomes is 50%. The ratio varies depending upon the sperm sample. Some men produce an overwhelming majority of XY chromosomes with a minute percentage of XX chromosomes whilst others produce an overwhelming majority of XX chromosomes with very few XY chromosomes. I doubt the ratio is ever exactly 50%.
I’ll suggest a hypothetical problem which is very similar in nature. Consider a bag which contains blue and red counters. We don’t know the ratio of the blue to red counters in the bag. We know that the ratio may have a bias but we don’t know in advance which way. Sometimes we may pull out a blue counter and at other times a red counter. It is only by pulling out a large number of counters from the bag that we can estimate the ratio of blue to red counters in the bag. We can not automatically assume it is 50%. Likewise we can’t automatically assume that the probability of the second child has a 50% probability of being the same sex as the first child.
November 1, 2009 at 8:42 am |
I think I solved it. Here’s how:
–Notation–
^ AND
~ NOT
P(A) the probability of event A occurring
–Events–
B The child is a boy.
G The child is a girl.
A The child is named Albert.
X At least one of the two children is a boy named Albert.
Y At least one of the two children is a boy named Albert and the other child is a boy.
–Assignments–
P(B) = .5
P(G) = .5
P(A|B) = R
–Calculations–
P(B ^ A) = P(B) * P(A|B) = .5R
P(B ^ ~A) = P(B) * [1 - P(A|B)] = .5 – .5R
–Table of all possible child combinations–
Child 1 Child 2 Event X Event Y
P(B ^ A) * P(B ^ A) = .25RR TRUE TRUE
P(B ^ A) * P(B ^ ~A) = .25R – .25RR TRUE TRUE
P(B ^ A) * P(G) = .25R TRUE FALSE
P(B ^~A) * P(B ^ A) = .25R – .25RR TRUE TRUE
P(B ^ ~A) * P(B ^ ~A) = .25 – .5R + .25RR FALSE FALSE
P(B ^ ~A) * P(G) = .25 – .25R FALSE FALSE
P(G) * P(B ^ A) = .25R TRUE FALSE
P(G) * P(B ^ ~A) = .25 – .25R FALSE FALSE
P(G) * P(G) = .25 FALSE FALSE
–More calculations–
P(Y|X) =
(.25RR + .25R – .25RR + .25R – .25RR) / (.25RR + .25R – .25RR + .25R + .25R – .25RR + .25R) =
(.5R – .25RR) / (R – .25RR) =
(.5 – .25R) / (1 – .25R)
–Final answer–
If we know that at least one of the two children is a boy named Albert, then the probability of the other child being a boy is (.5 – .25R) / (1 – .25R). Again, R is the probability of naming a boy Albert.
Notice that if R = 1, then the probability is 1/3. That is exactly what it would have been if we hadn’t chosen to name the boy. What’s weird is that if R = 0, we get a probability of 1/2. If we aren’t naming any boys Albert, we would expect that probability to be 0, but its not. Try figuring that one out.
November 1, 2009 at 8:49 am
Oops. The table didn’t come out right. Let me try that again.
-–Table of all possible child combinations–-
Child 1 Child 2 Event X Event Y
P(B ^ A) * P(B ^ A) = .25RR TRUE TRUE
P(B ^ A) * P(B ^ ~A) = .25R – .25RR TRUE TRUE
P(B ^ A) * P(G) = .25R TRUE FALSE
P(B ^ ~A) * P(B ^ A) = .25R – .25RR TRUE TRUE
P(B ^ ~A) * P(B ^ ~A) = .25 – .5R + .25RR FALSE FALSE
P(B ^ ~A) * P(G) = .25 – .25R FALSE FALSE
P(G) * P(B ^ A) = .25R TRUE FALSE
P(G) * P(B ^ ~A) = .25 – .25R FALSE FALSE
P(G) * P(G) = .25 FALSE FALSE
November 1, 2009 at 9:03 am
Ok, it still didn’t work. There were supposed to be five columns. The first and second columns were for the first child and the second child respectively. The third column was for the probability associated with each row. “Event X” and “Event Y” were the headers for the fourth and fifth columns respectively. The first two columns were multiplied together to get the third column and the last two columns contain truth values. I guess you’re just going to have to decipher the table the way it is. Sorry.