Thanks to reader Matt Lewis who sent me this link.
Martin Gardner, may he rest in peace, gave us a delicious probability problem:
Suppose that Mrs Smith has two children, at least one of whom is a son. What is the probability both children are boys?
Isn’t that lovely? Everybody immediately says, “One half!” But that answer is wrong because the counting is wrong. Discrete probability—which means all real-life probability—is all about counting, and counting isn’t easy. Counting is so difficult that once we depart from our standard ten-finger rules, our intuitions abandon us.
The first—and most important—rule of counting is this: What is everything that can happen? In the “Mrs Smith” problem, given the information provided, everything that can happen is this:
Boy, Girl
Girl, Boy
Boy, Boy.
This is every situation that gives old lady Smith at least one son out of two kids, as stipulated by the problem. Now, there is only one case which produces two boys, and three different possibilities. That means the probability—given only the knowledge provided to you in the problem; Mrs Smith obviously knows whether she has two boys or one—of having two boys is 1/3. Simple!
Given the information, the situation “Girl, Girl” is impossible (no jokes, please).
Since this kind of problem is always fun, a conference inspired by Gardner regularly convenes in which participants offer simple-sounding, but difficult, probability problems. Science News covered the most recent convocation, in which Gary Foshee offered this puzzler:
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
Intuition—the base of all our thoughts, but a sometime shaky base—practically screams at us, “Tuesday? What’s the day of the week have to do with it? It’s irrelevant! The probability must be one-half.”
In fact, I thought so too, upon my first reading. But knowing it was meant to be a puzzler, and noting that it had a probability smell to it—this faculty is one developed through time, exercised by being burned regularly by simple-sounding problems—caused me to pause.
The Science News article provides an explanation of why the probability is not 1/2. But that explanation is difficult to follow. So let’s do it ourselves slowly and see what happens.
The second rule in counting is this: count a simpler, but similar, situation first. Do not try to jump to the right answer; that is a sure way of making mistakes.
OK, what’s a simpler, similar situation? How about this: births can only happen on Tuesdays. Well, if that’s true, then we are back to Gardner’s original problem, right? Prove that to yourself before moving on.
So far, so good. Now let’s suppose that births can occur on Tuesdays or Wednesdays. What is everything that can happen given the information that a boy was born on a Tuesday and that there are two children? Just this:
(1) Tuesday: Boy, Girl; Wednesday: none
(2) Tuesday: Girl, Boy; Wednesday: none
(3) Tuesday: Boy, Boy; Wednesday: none
(4) Tuesday: none, Boy; Wednesday: Boy
(5) Tuesday: none, Boy; Wednesday: Girl
(6) Tuesday: Boy, none; Wednesday: Boy
(7) Tuesday: Boy, none; Wednesday: Girl.
(Incidentally, and this has nothing to do with nothing, the word “Girl” is one of the words the spelling of which looks weirder the longer you stare at it.)
There are seven possibilities—again, and I cannot emphasize this enough, given the premises—only three of which lead to two boys. The probability—conditional on your information—is 3/7, which is just under one half. (And to all those who have been following the past two weeks, this is a logical probability; no “randomness” needed.)
The way this reads is this: At least one boy must be born on a Tuesday. He may be the first born or second. If two kids are born on a Tuesday, then none can be born on Wednesday (1-3). The first born might have popped out on a Tuesday—which must be a boy—but the second on a Wednesday (6-7); or the stork might have shown up on Wednesday first, leaving a boy to be born on Tuesday (4-5).
Solving the general problem is now easy: (1-3) stay the same, even if we add in the remaining days of the week. For example, here’s the full (1):
(1) Tue: Boy, Girl; Wed: none; Thu: none; Fri: none; Sat: non; Sun: none; Mon: none.
Situations (4-7) just repeat, once for each other day of the week. For example, here’s (4′):
(4′) Tue: none, Girl; Wed: none; Thu: Boy; Fri: none; Sat: non; Sun: none; Mon: none.
The number of situations is 4*6 = 24 for the original (4-7) for each day of the week, plus 3 for (1-3), which stay the same. That’s 27 possibilities. The number of those with two boys is 2*6 = 12 for the original (4-7) for each day of the week, plus 1 for (1-3), which stays the same.
Thus, the probability is 13/27, which is 0.48, or nearly one-half.
Update: Challenge! If you’ve figured out the original, then you’ll surely get this:
I have two children, one of whom is a son born on 29 June. What is the probability that I have two boys?
Ignore leap years.
You are reading too much into the question. It does not state that the second child must be born on a specific day. The question is simply, “What is the probability that I have two boys?”. Since 1 out of the 2 is definitely a boy the only other available outcome is either boy/boy, or boy/girl. The answer is 1/2. The day is irrelevant in the question.
Chris,
If that is so, then you must also disagree with the solution to Gardner’s original problem; the “Tuesday” problem is just a generalization of it. But what is the flaw in the logic, whereby each possibility is spelt out?
Of course. The Gardiner problem is misstated. There is no distinction between “girl-boy” and “boy-girl” since the order of birth is not relevant. Thus there are not three countable solutions.
The sex of one child is already determined and all that remains a matter for probablity is the sex of the other child. 1/2. Don’t make more of the problem than is really there.
Here’s another way to look at it. The stipulated male child is either the eldest or the youngest. If he is the eldest the count choices are:
Boy-Girl
Boy-Boy
i.e. 1/2
If he is the youngest, the count choices are:
Girl-Boy
Boy-Boy
again, 1/2
Case closed.
Isn’t this simply a reframing of the “Monty Hall” problem?
http://www.stat.columbia.edu/~cook/movabletype/archives/2010/05/hype_about_cond.html
Always the problem is distinguishing cases. The version I recall is, “You visit a town where every household has just two children. You meet a girl. what is the probability that she has a sister at home?” Until you can convince yourself that ‘boy-girl’ and ‘girl-boy’ are distinct cases you get the wrong answer.
In the Tuesday case the problem (for me anyway) is seeing that “Tuesday: Boy, none” and “Tuesday: none, Boy” are distinct cases. I couldn’t explain why they are.
I agree that the probability is one-half. Another way to state the question (“I have two children, one of whom is a son born on 29 June. What is the probability that I have two boys?”) would be to say —
John (the son born on 6/29) has one sibling. What is the probability that John’s sibling is a boy?
Or simply, what is the probability that a child is a boy?
The problem with your original 1/3 answer is that you don’t handle the birth order issue properly. You said there are 3 outcomes (b/g, g/b, or b/b) but that splits the boy-girl possibilities by birth order and does not split the boy-boy possibilities by birth order. If you are going to use birth order, here are the four possibilities:
1. Boy (T), girl
2. Boy (T), boy
3. girl, Boy (T)
4. boy, Boy (T)
The reality is that birth order is not relevant to the question, but if you are going to use it, you have to be consistent. Can’t use it if the sibling is a girl, but not if the sibling is a boy.
Let me see if I’ve got this right.
The probability of the children being two boys in the first problem is one out of three.
The probability of the children being two boys in the second problem is roughly one out of two.
So if I change the statement of the first problem to:
“Suppose that Mrs Smith has two children, at least one of whom is a son who was born on a Tuesday. What is the probability both children are boys?”, then the probability of the there being two boys increases from one out of three to about one out of two?
Maybe I’m totally dense, but it seems to me that Mrs. Smith’s kid has to be born on some specific day of the week, so for the life of me I can’t see how specifying that day somehow decreases the likelihood that his sibling is a girl.
Ok, my mind was just blown away.
I remember that tv game and understand its logic, but this just went over my head.
BTW, the answer to the final one is ~1/2. 😀 (yeah I’m very lazy today, but that’s a very precise answer, I’d venture!!)
1) Suppose that Mrs Smith has two children, at least one of whom is a son. What is the probability both children are boys? ANS: 1/3
2) I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys? ANS: 0.48
in (1), the known son had to have been born on day X. Let’ X=Tuesday. Essentially the same problem but two different answers. Goes against intuition.
Tricky indeed.
I think the problem (post coffee) I have with this explanation is that it treats “Tuesday: boy, boy” as a single case instead of two. I think it should be two cases. All other cases consider that we may have chosen either the elder or younger of the siblings.
Further, the distinguishing detail between this and Monty Hall is that while the doors are mutually exclusive, days of the week are not. So I would argue that the Tuesday problem hinges on whether you interpret it to say that “only one” is born on a Tuesday, which is, of course inconsistent with the way in which gender is described, and therefore, not a good reading of the puzzle.
Counting is hard.
Looking back at the comments, I’d venture they are right…. albeit the “tuesday” thing is even more outrageous.
I’d guess that the problem relies on the ungrounded usage of Monty Hall. In Monty Hall you had 3 doors and the choice you make in the first one conditions the showing of another open door. In this case there is no conditioning. The fact that the “first” son to be disclosed is a boy does not condition the sex of the other at all.
Another thing is the division of sex in 4 types, and that’s just wrong. There are not “4”, but actually “3” possible logical outcomes: either two boyes, one boy and one girl or two grils. 3, not 4.
To state that they are four, you are adding another element named “birth order”, and as stan explained, if you do that, you should also add the other boy boy possibility.
…. so my sanity came back.
If this were a fair coin & the question was:
Mrs. Smith tossed a head (H), what is the probability that the second — pending — toss will be a head (H)?
The answer would be 50 percent — the outcome of independent events are not conditional on prior outcomes (Monte Carlo/Gambler’s Fallacy).
Most people unconsciously latch onto that thought process & apply it to problems that appear fundamentally the same…and doing so subsequently blinds them to the real facts at hand:
With a fair coin the probablity of a specific sequence of outcomes (say, H-H) can be calculated and is found to be something other than 50% (Prob of any specific outcome is 25% in this case) as the possibilities –at the outset — are:
H – H
H – T
T – H
T – T
Implicit in the first question is that the Probability of a boy or a girl is equal (which in the real world is itself an overall average with most couples experiencing something more analogous to a biased coin). This implicit assumption makes the situation fundamentally similar, but not necesssarily precisely equal, to the conditions associated with the coin tossing problem…though the asserted solution is affected by partial information:
Boy – Boy
Boy – Girl
Girl – Boy
Girl – Girl (not possible given the conditions provided)
What is known in the problem is that one child is a boy–information is insufficient to conclude/assume that the first child born was a boy (i.e. the first child born may have been a girl)–thus the probability is one-third, not one-half, as there are three possible outcomes per the incomplete set of facts provided.
Had the question been worded to provide certain additional information, namely that the first born was a boy, with the query being what is the probability of the second born child, THEN the probabilty would be 50% (i.e. this would match exactly the physics one intuitively associates with the coin-toss problem illustrated by the Gambler’s Fallacy — that a subsequent child’s gender is independent of the prior child’s gender).
But that’s not the problem presented.
Which highlights just how easily one can reframe the facts & conditions of a situation (usually by oversimplifying them) to create an erroneous model….conclusions…etc.
Here’s another tricky problem: I have flipped this coin 9 times and have gotten heads 9 times (lucky me!). What is the probability the next toss will also be heads (100%, 50% or X%)?
I think the natural possibility (barring bizarre syndromes) of having a girl is 25% (requires XX chromosome pair I think). Presumably having a child of one sex does not imply a predisposition for the sex of following progeny so,
P(Boy|chance ) = 0.75
P(Boy|1rst=boy) = 0.75
P(2 boys) = 0.75 * 0.75 = 0.5625
It was a great relief to get to the comments to confirm that if I’m an idiot not to see that Mrs’ Smith other kid does not have a 50% chance of being a boy, when there is absolutely no hint int the question that birth sequence matters, at least I’m not the only one.
From the posing of the problem, choice one and two are therefore identical leaving the sex of the second child to either.
Briggs, I’m sure you are trying to get at some greater truth, but from one of the meanest understanding, I can’t follow you there.
For the question regarding the 29 June birthdate for a boy… I’d expect the answer to be between 0.500 & 0.48 (rounded down from 13/27) — i.e. that as more possible days/weeks are added to the scenario the calculated solution approaches 50 percent (probably asymptotically). Arguably, depending on how fussy one is about rounding, “50 percent” is probably “pretty darn precise” (though I’ll omit definitions for “pretty darn” in this situation for “later”). I’d guess 0.499, just based on crude trend fitting per the above observation.
John,
Let’s appeal. Now, the logical explanation suffices, but you can do a little experiment to prove the Gardner problem to yourself.
Get two coins and flip them (or flip one twice). Let “H” be a Boy and “T” be a girl. Here’s what to do.
(1) Flip both coins. Mark down that you made a flip ONLY IF you do not see two Ts. That would be two girls and is impossible by definition.
(2) Mark, in a separate column, if the two coins are both H.
(3) Do this at least 20 times.
(4) Divide the number of times it came up both H with the number of valid tosses (a pair of tosses counts as one, don’t forget; not as two).
(5) That frequency will soon approach the (logical) probability.
(6) Report back to me.
Boy-Girl and Girl-Boy are distinct for one important reason…. theoretical probabilities are calculated using equally-likely outcomes or by taking weights into account. The outcome of one boy and one girl is twice as likely to occur as two boys or twice as likely to occur as two girls. So we can either implicitly assume that birth order matters to make equally likely outcomes BB, BG, & GB, or we can or consider BG (with order not mattering) to be twice as probable as BB.
I saw this puzzle a week or so ago on another blog, and have had some time to think about why the result is so counterintuitive.
I think Ken does a good job explaining the confusion between “1/2” and “1/3” in Gardner’s original puzzle.
For this version, I think it comes down to the presumed mechanism for choosing the day-of-week “Tuesday”. Within the puzzle, this choice is made without reference to the actual dates of birth of the question-posing-parents’ children. People read it as if any random parent were telling you the date of birth of one of their boys. If that were the case, then any parent with at least one boy could play, and the question could be rewritten:
“I have two children, one of whom is a son born on [one of the days of the week]. What is the probability that I have two boys?”
…and then the probability of two boys would be 1/3.
The truth is, I don’t see anything that rules out that reading of the question.
But in the intent of this puzzle, only parents with Tuesday boys – specifically – can ask you the question – and that restriction limits the number of families that qualify for inclusion in the puzzle universe. And because Tuesday births are relatively rare, inclusion in this universe is a good deal more likely for those parents with two sons than for those with only one. If you have two sons, you have a much better chance of having one of them born on Tuesday.
So to answer the June 29th question, of the 1/4 of all two-child families who have two boys, 729/133225 have at least one born on Tuesday. Of the 1/2 of families with one boy, 1/365 have one born on Tuesday. The probability is [729/133225]/([729/133225]+2*[1/365]) = 729/1459, or 0.499657 that the family has two boys. Note: 729/133225 = 1 minus the probability that neither of the boys was born on June 29th, which is (364/365)^2 = 132496/133225.
If it’s exactly 1 boy born on Tuesday, then it’s 728/1458 = 364/729, I think.
Er, sorry about that answer to the June 29th question, where I lapsed into “born on Tuesday” rather than “born on June 29th”. Mentally substitute, please.
Is the answer to the siblings question different if it is asked ” What is probability that second child is boy?” from “what is probability that both are boys?”
I see a faint glimmer that I might be catching on.
The Gardner solution is also incorrect since order is irrelevant. Boy/Girl and Girl/Boy are the same thing. Therefore, since there are only 2 options: boy/boy or boy/girl, the probability is 1/2.
Tim,
Thanks for the link. Gelman and I differ on the nature of probability. He says there is such a thing as “conditional” probability, implying, of course, that there are “unconditional” probabilities, but these are never defined without reference to conditions (premises, evidence, etc.), which means, I say, that all probabilities are conditional (as are all other statements of logic).
His point about the possible ambiguity in the word “one” is relevant, however. Most misunderstandings in probability come from these sorts of ambiguities, particularly with the word “random.”
Chris,
Nope. See my experiment you can try.
The more I look at it, the more I think the question is posed in a way that is actually misleading.
“I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?”
This sounds like a parent who would have said “Wednesday”, “Thursday”, or whatever day of the week applied in his case. If so, then “Tuesday” is not really a condition. It’s only a description.
Another way of addressing the first “I have two children, one of whom is a son” is to simply restate it : “I have two children and they aren’t both girls”.
And thus the question “what is the other son?” translates simply to “Are they both sons or are they a son and a daughter?”.
We can see the possibilities here. The first statement tells us that these outcomes are possible:
Son1 – Son2
Son2 – Son1
Son1 – Daughter1
Daughter1 – Son1
If you are going to say that the first is equal to the second, what is exactly your logical argument to say that the third isn’t equal to the fourth?
I will make the experiment, although it makes the answer pretty clear… I am convinced.
Luis,
You are double counting. With 2 items there are only 4 possible unique arrangements: AA AB BA BB. In this case: GG GB BG BB and case GG is not possible from the givens. You have effectively counted an item from the first column (Boy1) as being in column 2 (or vice versa).
If one discounts the assertion that ordering is relevant, then how do you determine what the sex of the child is when they are born? You may as well say that your first born is a boygirl, your second born is also a boygirl, since it doesn’t matter in that case which one was born when. It makes no sense.
Luis,
OTOH, and after more thought, we aren’t drawing prejoined pairs out of a hat. You are correct in that the order or birth doesn’t seem relevant to the given problem therefore GB is the same as BG and we are left with only two possibilities (BB and BG). Briggs should have realized this (and perhaps did — he’s quite the tease) from the contradictory answers to the Tuesday and sans-Tuesday problems.
It’s that damn wording abiguity again.
Argument 1: The anwer depends on the method of determination of “at least one of whom is a son”. If we got the determination by probing (e.g., selecting Mrs. Smith’s two children at random) , then we must assume we drew BG or GB or BB but don’t know which. BB represents only 1/3 of the possibilities.
Argument 2: Mrs. Smith tells us this out of the clear blue (For whatever reason. She perhaps was struck in the head by a stat book and it warped the way she ineracts with the rest of humanity). In this case , the proportions of BG, GB abd BB in the rest of the population are irrelevant as there are only two remaining possibilities (G or B).
There was a toxic milk spill on the next farm when I was growing up. Must have affected my thought processes. I can be really slow at times as a result but I think I finally got this one right. Now if I can just learn to frame my thoughts more clearly.
—
“Beer. It’s the reason why I get up every afternoon.”
Q: I have two children. The firstborn is a boy. What is the probability that the other is a boy?
A: It depends on what process generated the evidence I provided above.
Situation 1: The evidence rule is “if only the firstborn is a boy, say my firstborn is a boy; if only the secondborn is a boy, say my secondborn is a boy; if both are boys, choose from ‘my firstborn is a boy’ and ‘my secondborn is a boy’ with equal probability.
Then, even though we could always eliminate from consideration one of B-G or G-B, we would not gain any information by doing so. ‘My firstborn is a boy’ occurs with no G-G, all B-G, and *half* of the B-B. ‘My secondborn is a boy’ occurs with no G-G, all G-B, and *the other half* of the B-B. So even though we’ve eliminated one of the family structures, we aren’t any better off. The probability is still 1/3.
Situation 2: The evidence rule is “truthfully tell the sex of the oldest child”, then ‘my firstborn is a boy’ occurs with no G-G, no G-B, all B-G, and all B-B. Now the probability is 1/2.
So there are assumptions regarding the evidence generation process and what evidence that process provides under various circumstances that are built into the answer.
I meant “‘My firstborn is a boy’ occurs with no G-G, no G-B, all B-G, and *half* of the B-B. ‘My secondborn is a boy’ occurs with no G-G, all G-B, no G-B, and *the other half* of the B-B.”
We can use a tree diagram and counting techniques to solve all the questions in this post, of course, assuming all possible outcomes are equally likely.
We need to first formulate the probability of interest carefully. If it’s a conditional probability, always solve it using its definition, i.e.,
P(A given B) = P( A and B) / P(B).
For simplicity, the premise “Smith has two children†is omitted in the following. The numbers in parentheses represent numbers of possibilities,.
Let A ={both are boys} and B ={at least one is a son}.
Q1 = P( both are boys given at least one is a son ) =P( both are boys and at least one is a son) / P(at least one is a son)=P(A)/P(B)
The 1st child is either a boy or a girl (2), so is the 2nd second child (2). There are 4=2*2 possible outcomes.
TREE DIAGRAN here… sorry, don’t know how to make one in html.
Of the four, three outcomes belong to A. So P(at least one is a son) = ¾.
Similarly, P( both are boys and at least one is a son) = P( both are boys) = ¼
Therefore Q1 = P(A)/P(B)= (1/4) / (3/4) = 1/3.
ooops, ” Of the four, three outcomes belong toB. So P(at least one is a son) = ¾.
In the spirit of bloody mindedness, I wrote a little simulation. Basically, I simulated 10,000,000 2 child families, and did some calculations, so that if you were to randomly sample the results of the simulation, you’d expect to get the following:
P( boy | family ) = 7498761 / 10000000 (74.99%)
P( 2 boys | family ) = 2498602 / 10000000 (24.99%)
P( 2 boys | at least one boy ) = 2498602 / 7498761 (33.32%)
P( am boy | family ) = 4372014 / 10000000 (43.72%)
P( am boy | at least one boy ) = 4372014 / 7498761 (58.30%)
P( 2 am boys | family ) = 623660 / 10000000 (6.24%)
P( 2 am boys | at least one am boy ) = 623660 / 4372014 (14.26%)
P( 2 boys | at least one am boy ) = 1873715 / 4372014 (42.86%)
And, indeed, 3/7 is roughly 42.86%. Note that I used the simplified two phase version, but rather than day of the week, looked for an AM or PM birth.
The 1st child is either a boy or a girl (2) born on any of the seven days (7) with a possibility of 14=2*7 ways. So is the second child. There are 196 = 14*14 combinations.
SKETCH your tree diagram here.
Let BT={one of whom is a son born on a Tue} and BBT= {both are boys and one born on a Tue}.
Q2= P(both are boys given BT) = P( BBT )/ P(BT).
First, find P(BT) by selecting outcomes (branches) that belong to the set of BT from the tree diagram. Three cases.
1. If the 1st child is boy-Tue, the 2nd child can be any of the 14 sex(2)xday(7) combinations. So this case gives us 14 outcomes in {BT}.
2. If the 1st child is boy-notTue (6), the 2nd child has to be boy-Tue (1). Here we collect 6 outcomes in {BT} .
3. If the 1st child is girl-day (7 ), the 2nd child must be a boy-Tue(1). This case yields 7 outcomes belonging {BT}.
Therefore, a total of 27 = 14+6+7 outcomes in BT. So the denominator P(BT) = P(one of whom is a son born on a Tue) = 27/196.
Next, find the number of outcomes in the set BBT. Two cases.
1. 1st boy-notTue (6) and 2nd=boy-Tue… (6)
2. 1st = boy-Tue(1) and 2nd=boy-any of the 7days… (7).
(Sorry, got lazy in my writing.) Yeah.. with patience, we’ve found that there are 13=6+7 outcomes in BBT. So, P(BBT) = 13/196.
Therefore,
Q2 = P(BBT) /P(BT) = (13/196)/ (27/196) = 13/27.
One may argue that there are more Tuesdays in certain years (therefore not equally likely), which makes it a bit more complicated!
JH,
I’ve finally converted you! You are now a logical probabilist! What a happy day!
I think the wording of the statement is ambiguous when taken the context of normal language usage. It leaves the set size poorly defined.
Statement:
I have two children, one of whom is a son born on a Tuesday.
In normal conversation that is most likely to mean only one son born on a Tuesday with the possibility of a Tuesday daughter left undefined.
If the probability question isn’t asked, and you were to query about the possibility of another Tuesday child, do you realistically believe the response “My other child is a son born on Tuesday” would occur in equal proportion to the none or daughter response?
I would feel it necessary to give less weight to that element of the set. This leaves a set where all elements do not have equal probability.
This wording is less ambiguous.
I have two children. My son Joe was born on a Tuesday. What is the probability that I have two boys?
OK, here’s my take on it. Let’s say I am asked the original question – two children, one is a boy, what’s the probability there are two boys?
And then, instead of answering, I say “what’s the boy’s date of birth? and his name? and his favorite color? and his grade in arithmetic?” and after I get all the answers, I say – ah, it’s 0.5 minus epsilon.
The more data I have about a child, the more the probability approximates half.
Why is that? That is because, as you said, every probability in the real world is discrete probability. But in the real world, every boy has a name, a date of birth, a favorite color and a grade in arithmetic. But all that data is irrelevant, what is relevant is that IF I ASSUME THIS DATA EXISTS, then I know that the probability is 0.5 minus epsilon. I don’t really have to know any of it, just know that it’s there, and approximate the answer from that.
Intuition 1, math 0.
I think I understand why it works. Consider the original problem, where the answer is 1/3. Now consider changing it: “My first child is a boy…” In this case, the order of children is known, and we can throw out the GB case, leaving us with 1/2.
With the Tuesday problem, we only have partial information about the ordering. When the clarification has only 2 possibilities (AM vs PM), then we have a bit more information than we did with the days of the week. So we get closer to having knowledge of the order.
It’s like the logic puzzles where you take the clues and mark off boxes, looking for the answer. Very like that, when you start writing down the cases. I feel much better now.
“meanest understanding” still munching on first question. If time starts before woman has any kids, then the 1/3 works. if it starts when she has one, which I think was my mistake, and the questions only applies to the second, then i get the 50%. I can see now that the first is correct as you state it. Matt’s observation is the key.
it’s wonderful that such simple stuff can seem so tricky.
It’s really only about a definition of random variables . Where’s the problem ?
Let X and Y be 2 different random variables taking values in {B,G}
Knowing that X=B OR Y=B , what is the probability that X=B AND Y=B ?
(Please note the conjonction OR in the hypothesis while there is an AND in the question)
Well the answer is obviously 1/3 and it is quite clear that the states of the system
{X=B , Y = G} and {X=G , Y = B} are different because X and Y are different random variables .
Now I can call X = spin of an electron and Y = spin of a proton or X = Child1 and Y = Child2 or whatever .
As long as X and Y are different and 2 valued , the probability will always be 1/3 .
One can of course easily generalise for N variables and M values .
To find 1/2 one has to substitute to the original premise “knowing X=B OR Y=B” the premise “knowing X=B” or the premise “knowing Y=B” which are clearly different .
Even more compact 🙂
Knowing that “A or B” is true , what is the probability that “A and B” is true ? Answer 1/3 .
And just for fun the compactest solution for the Tuesday case I could find :
X and Y take values in {G,B} , Zx and Zy take values in {1,2…7}
Knowing that (X=B and Zx=2) OR (Y=B and Zy=2) what is the probability that
X=B AND Y=B ?
There are 28 (4×7) quadruplets {X,Y,Zx,Zy} that verify the premise .
But ONE is counted double namely {X=B , Zx=2 , Y = B , Zy =2} so there are in total 27 different quadruplets .
Half of the initial 28 namely 14 corresponded to X=B and Y=B . But the one counted in double belonged precisely among the X=B and Y=B . So there are 13 (14-1) quadruplets left with X=B and Y=B which answers the question .
The probability is therefore 13/27
Note that the 29 june case is the same thing with Zx and Zy taking values in {1,2 … 365}
TomVonk,
Are you sure it’s {1,2…365}? What about 366? Which happens nearly a quarter of the time. That seems like the trick behind his question, once we’ve mastered “simple” counting.
I’ve read all the comments and I still stand by my first comment. The probability is one-half. Why should birth order matter for the possibility of a girl, but not matter for the possibility of two boys? That’s nuts.
I have a son. What’s the probability the other child is a boy? 1/2
Doesn’t matter whether the known son came first or second. Let me repeat. Does Not Matter. I have a son John. What are the probabilities that his sibling is named Betty vs. Bob? We could have John then Betty or Betty/John or John/Bob or Bob/John. Four possibilities and half of them give us two boys.
1/2
Well, Stan, repeating what is false doesn’t make it true. Have you tried the experiment yet? That would convince you.
If I have time, I’ll come up with an experiment for the Tuesday/Wednesday problem.
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OK, so if someone approaches you and says “I’m Mrs. Smith. I have two children, at least one of whom is a son. Will you bet, at even odds, that you can predict my other child’s sex?” What do you reply and why?
What happens when Mr. Smith appears, puts a gun to your head and forces you to play the game?
If you guess randomly, with a 50/50 split between boys and girls, it seems to me that you’re minimizing your losses (i.e., if you play this game over and over, in the long run you’ll neither win nor lose money). If you do anything else, couldn’t Mrs. Smith exploit your choice?
I think “Morgan 29 June 2010 at 9:42 am” is right. There is an ambiguity about how Mrs Jones was “chosen” to talk to you. Does the following conversation make sense?
Mrs Jones: I have exactly two kids, at least one being a boy. What’s the probability that I have two boys?
Briggs: The probability is 1/3.
Mrs Jones: The boy I just referred to was born on a Tuesday.
Briggs: In that case the probability is closer to 1/2.
Mrs Jones: Wait, on second thought, I really have no idea what day he
was born on.
Briggs: Well then the probability of two boys is 1/3 after all.
Mrs Jones: I do remember he was born on April 7th though.
Briggs: Around 1/2 then.
Mrs Jones: Silly me! It’s my husband who was born on April 7th!
Briggs: Around 1/3.
This doens’t make much sense to me. The problem is that you are not interacting with a randomly chosen person from populations that change with each statement above. Unlike Brigg’s suggested coin experiment, there is not a mechanism by which a parent is pulled from the population, and rejected if he or she doesn’t satisfy some criterion. It’s the same Mrs Jones each time, even though the selection criterion is changing.
I think a lot of the confusion can be avoided by stating the problem in terms of portions of an underlying population, and avoiding the mentioning of probability:
Q: Of the two-kid families having at least one boy, what portion have
two boys? A: 1/3.
Q: Of the two-kid families having at least one tuesday-born boy, what
portion have two boys? A: 13/27.
First, i want to say i have the utmost respect to Mr. Gardner. That said, i think the confusion starts with the semantics: when the mother says ‘I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?’ – does she mean AT LEAST ONE OF WHOM… or EXACTLY ONE OF WHOM… well, the question in the end indicates she means AT LEAST. in this case the answer is clearly 13/27. Let me show this:
Let us look at 210,000 cases: 70,000 BG, 70,000 GB and 70,000 BB. in how many would the statement be correct?
in 10,000 of the Bg, 10,000 of the BG and… almost 20,000 (app. 19,572) cases of the BB!
(Note, please, that if youy sum up all the days: sunday, monday etc. you do NOT get 70,000 but much more).
The solution is 50%. There are 27 possible outcomes, but one of the outcomes (Tuesday Boy, Boy) is twice as likely to happen as any other possibility given the information in the problem. So you really have 14/28 ie 50% chance. A fun 3-legged dog problem. Just like saying I have flipped a coin twice and one was a head, what are the chances I have flipped 2 heads, still 50 even though there are only 3 possible outcome (HT, TH, HH), the HH has twice the probility as the HT/TH given the info that one must be heads.
The probability that someone with two children has two boys, given that you know they have one boy born June 29, and assuming a boy or girl child is equally likely, and that there are only 364 days possible birthdays (each equally likely) is (2*364-1)/(4*364-1) = 0.49966.
Cool problem, Matt. Thanks.
The trick here is that you have been told that there’s a *boy* AND that he was born on *Tuesday*. When you start eliminating cases things don’t add up evenly to 1/3 or 1/2 anymore. Why? If the other child happened to be a girl, she could have been born any day of the week. But there is now a linkage between boy and Tuesday.
Mathematically:
B-tue x B-tue: 1/3 * 1/7 * 1/7
B-tue x B-not-tue: 1/3 * 1/7 * 6/7 * 2
B-not-tue x B-not-tue: 1/3 * 6/7 * 6/7
B-tue x G-tue: 2/3 * 1/7 * 1/7
B-tue x G-not-tue: 2/3 * 1/7 * 6/7
B-not-tue x G-tue: 2/3 * 6/7 * 1/7
B-not-tue x G-not-tue: 2/3 * 6/7 * 6/7
Now let’s eliminate cases w/out a boy on Tuesday leaving:
B-tue x B-tue: 1/3 * 1/7 * 1/7
B-tue x B-not-tue: 1/3 * 1/7 * 6/7 * 2
B-tue x G-tue: 2/3 * 1/7 * 1/7
B-tue x G-not-tue: 2/3 * 1/7 * 6/7
If we assume “at least 1 boy on Tuesday”, then the number of cases is: 27/(3*7*7).
And the two boy outcome is 13/(3*7*7). Ratio is 13/27.
If we assume “only 1 boy on Tuesday”, then we have to eliminate the first case so we get total of: 26/(3*7*7). Two boy outcome (different days) of 12/(3*7*7). Ratio is 12/26 = 6/13.
Bet on the girl! 🙂
Matt
William said Ignore leap years
That’s why the birth dates (Zx and Zy) take values in {1,2,….365}
SteveBrookline
Briggs: Around 1/2 then.
Mrs Jones: Silly me! It’s my husband who was born on April 7th!
Briggs: Around 1/3.
This doens’t make much sense to me.
On the contrary it makes complete sense and is in essence the heart of this rather simple problem .
The probability of a certain outcome DEPENDS on the premises . Is here somebody who would say no to that ? Probabilities independent of premises ? Anybody ?
What your Mrs Jones is doing is that she keeps changing premises . And your Briggs is doing what is the only rationnal reaction – he correctly computes the result consistent with the premises . As the latter change , the former change too .
Stan
I have a son. What’s the probability the other child is a boy? 1/2
Doesn’t matter whether the known son came first or second. Let me repeat. Does Not Matter.
You are right . The fact that it is 1/3 actually vitally depends on the fact that it doesn’t matter . On the fact that you don’t know which one of the children is the son .
It is a logical “one OR the other” 🙂
But the point where you horribly err is that your question is completely different from William’s question .
William asks says what is the probability to have {Child1 = Boy AND Child2=boy} knowing that {Child1 = boy OR Child2 = boy}
You ask what is the probability to have Child1 = Boy knowing that Child2 = Boy (or having Child2 = Boy knowing that Child1 = Boy) . It is 1/2 in both cases .
Your problem is not that you give a wrong answer , your problem is that you answer a completely different question and don’t realize it .
If you had read my previous post , you’d have seen that I already adressed this misconception by saying :
To find 1/2 one has to substitute to the original premise “knowing X=B OR Y=B†the premise “knowing X=B†or the premise “knowing Y=B†which are clearly different .
William :
You were a bit unfair . Actually Stan keeps repeating a truth and his logics is good .
His problem is that he can’t see that he started with premises different from yours .
That’s why one could anticipate that he won’t be convinced by the experiment because he won’t see where the relavance is . He could say something along the line “But all this has nothing to do with the problem .”
Again a correct and true statement because the problem as he states it , has indeed nothing to do with the problem as you state it 🙂
Tom,
Ok. Stan, I’m sorry. But try the experiment!
Incidentally, I do have a son born on 29 June. And I do know the probability of whether I have another son. My information is different than the information in the problem. Remember: all probability (all logic) are conditional on certain stated evidence—and only on that evidence.
This is becoming something of an obsession.
Briggs said: “Remember: all probability (all logic) are conditional on certain stated evidence—and only on that evidence.â€
But it seems to me that there are times (and this is one) where the evidence itself is ambiguous, in this case because the model that generated the evidence is unknown.
My default presumed evidence generation model indicates that statements of the form “I have two children, one of whom is a son born on a Tuesday†are at least as likely to be generated by a person who was simply providing an additional detail (and would have said “on a Wednesday [Thursday, Friday, Saturday, Sunday, or Monday]†had Tuesday not been available) as they are to be generated by a person randomly selected from a population consisting entirely of two-child parents with a son born on a Tuesday.
On the other hand, the presumed evidence generation model that operates for such statements appearing in the context of probability puzzlers favors the second interpretation.
And clearly, the presumed evidence generation model makes a difference to the probability of the other child being a boy.
We have both kinds of evidence available to us – the statement “I have two children, one of whom is a son born on a Tuesdayâ€, but also our knowledge of how such statements generally come to be made – don’t we? Shouldn’t we take both kinds of evidence into account?
As I tried to state earlier there are actually 4 possibilities on Tuesday (not 3 as Briggs solution shows). There are 3 types of children Boy(Known), this is the boy that we know exist, Boy(potential), and girl(potential), so for both children born on Tuesday there are 4 possibilities:
Boy(Known), Boy(potnetial)
Boy(potnetial), Boy(Known)
Boy(Known), girl(potential)
Girl(potential), Boy(known)
just like every other day there is a 50% chance of having a boy.
William Briggs,
Your coin toss analogy and the other examples are wrong. For example, for the correct coin toss analogy you need to have two coins, one of which is a regular coin with a heads and a tails side. The other has heads on both sides. You mix them up so you can’t see which is which. Now flip them both at the same time. By initially forcing a response that at least one heads has to come up, THIS DIFFERENT COIN IS NECESSARY. Now you look at all possible results: one head is assured and the other coin has a 50-50 chance of head. It matters not where the coins land, so the head-tail or tail-head issue is not distinct. When you allowed the tail-tail possibility by using normal coins, but then rejected it AFTER the toss, you were looking at a separate issue than the stated one. I am sorry to disagree, but the probability is 1/2 for all cases.
Mr. TomVonk,
Right. Let’s not forget that we also assume that boys and girls are equally likely, and that 7 days are equally likely, and that gender of the second child is independent of the first one (and vice versa). Oh my, so nerdy.
____
My actuarial students have asked the same questions posted here every year for…many… years. Students can translate the probability of interest differently due to the ambiguity in the wording. This kind of problems and those involving Bayes Theorem tend to show up in the probability exam (now called Exam P) issued by the Society of Actuary.
Try the urn problem. E.g., an urn contains 10 balls: 5 blue and 5 green. Two balls are drawn one after another from the urn without replacement.
1. Knowing that I have one blue ball, what is the probability that I have two blue balls?
2. If one of the balls is blue, what is the probability that I have two blue balls?
Students would argue whether “I have one blue ball†means “one of the balls is blue.†However, they all seem to understand the outcome of each draw is relevant. My guess is that they all know the probability of 2nd draw being blue changes with (is not independent of) the knowledge of the the 1st draw.
My previous post reminds me of one funny incident. I once conducted the urn experiment to compare the theoretical and empirical results in an intro-stat class. Sarah came in late and didn’t get a container with balls in them. When other students were busy drawing, she was sitting pretty in the chair and doing nothing. I asked her why she wasn’t doing anything. She said, “Professor, I don’t have balls.â€
Leonard,
No, sir. I am right. That is to say, Gardner and every other probabilist and I am are.
I’ll try—if I can find the time—to hack out some R code so that you can prove it to yourself.
Leonard
For example, for the correct coin toss analogy you need to have two coins, one of which is a regular coin with a heads and a tails side. The other has heads on both sides. You mix them up so you can’t see which is which. Now flip them both at the same time. By initially forcing a response that at least one heads has to come up, THIS DIFFERENT COIN IS NECESSARY. Now you look at all possible results: one head is assured and the other coin has a 50-50 chance of head.
You make exactly the same 2 mistakes like everybody who thinks that the probability is 1/2 .
Mistake 1 : You don’t read the thread because this “problem” had already been explained and rigorously demonstrated above .
Mistake 2 : You substitute to the correct premise that William uses another and different one like Stan did .
In your example this substitution is strikingly obvious .
If we call the head-head coin variable X and the head-tails coin variable Y , you made sure that in every toss you will have X = heads with probability 1 . In other words you arbitrarily excluded the state of the system in which {X=tails , Y = head} .
In your example probability {X=tails , Y = head} = 0 .
Yet this state is not only perfectly allowed by William’s premise but its probability is non 0 !!
Hence your conclusion is in contradiction with the (William’s) premise and therefore answers a very different problem .
QED
.
May I remind for the third time already what the problem with the 1/2 “answer” is ?
To find 1/2 one has to substitute to the original premise “knowing X=B OR Y=B†the premise “knowing X=B†or the premise “knowing Y=B†which are clearly different .
JH
You are perfectly right . When reading this article I immediately thought at the traditional exercice “draw a ball and keep it” versus “draw a ball and put it back” .
I am sure that people who have problems to see that the answer to the William’s teaser is 1/3 would have problems to distinguish between the 2 cases in the ball exercice .
I have almost written a post about it but you just did it 🙂
Morgan
We have both kinds of evidence available to us – the statement “I have two children, one of whom is a son born on a Tuesdayâ€, but also our knowledge of how such statements generally come to be made – don’t we? Shouldn’t we take both kinds of evidence into account?
No .
I am not sure what was William’s purpose about this example .
Personnaly I have read it as an advanced introduction to formal logics and random variables .
And it is clear that “evidence” is meant here the formal transcription of a statement .
Williams statement is transported from general language which may be vague , confused , ambiguous to the formal language of logics where the is no ambiguity .
Can’t be per definition .
That’s why what he meant transcribes :
X and Y take values in {G,B} , Zx and Zy take values in {1,2…7}
Knowing that (X=B and Zx=2) OR (Y=B and Zy=2) what is the probability that
X=B AND Y=B ?
This statement is perfectly clear with no ambiguity .
And the answer which is clear and easy is : 13/27 .
It doesn’t matter at all “how” such statements are made !
William could have made the opposite transcription .
Make the statement above and then decide (arbitrarily) that X and Y are Mars moons and Zx and Zy are codes for colors .
He would then say a natural language sentence about moons and their colors .
He could decide that X and Y are cats and dogs and Zx , Zy is the number of squirells they caught .
Etc .
An infinity of natural language sentences yet always the same answer : 13/27 .
Good morning Mr. Briggs,
I see that there’s a lot of discussion on this problem. My take is, when you state that Mrs. Smith has two children and one is a boy, that reduces the problem. I’m a physicist, we like to do that. So, when you ask what the probabily of both children being a boy, what you are really asking is what the probability of the other child being a boy is. Since one child is know to be a boy, regardless of what day he’s born on, or on what planet, he’s 100 % a boy. that only leaves the probability of what another child’s being a boy is. If we assume the boy/girl probability is 50/50, which is a little off, then the probability is 1/2. Birth order, day of the year, or any other issue regarding the known child doesn’t matter, since we already know his gender.,
Here’s an easy way to analyze Gardner’s original problem. It may or may not produce an answer, but it does show that the dispute over BG vs. GB is not easy to resolve.
Since the two children already exist, the situation from the standpoint of the person being asked the question is exactly the same as if Mrs. Smith has a bag containing tokens each labeled with a possibility, such as BB or BG, but no token labeled GG; she tells you that the “prize token” has B as part of its label; and she asks you what the probability that its label is BB.
The probability depends on whether the bag contains 3 tokens {BB,BG,GB} or only 2 tokens {BB,GB}. I hope we can all agree that birth order doesn’t matter. This may or may not exclude the possibility that there are 3 tokens, but if it does not exclude it, this does not mean that birth order or anything else that might be described by having both a BG and a GB token matters.
So which it is, 3 tokens or 2 tokens? Only Mrs. Smith knows, and she’s not telling. One might object that I’m allowing her to have any number of tokens in the bag, in any combination, giving any probability. Is that a valid objection? Not if she has implied that there is one token for each possibility. Mr. Briggs would think she had more than implied it, i.e., that she had implied 3 tokens, and not 2 or any other number.
Suppose we take the 2 and the 3 to be equiprobable. In this case the probability of 2 boys is (1/2)(1/3)+(1/2)(1/2)=5/12.
At this point I’m puzzled.
“she tells you that the “prize token†has B as part of its label; and she asks you what the probability that its label is BB.”
Sorry, I should have said, “she tells you that every token has B as part of its label…” I don’t think this affects the rest of what I said.
@john galt, you are misinterpreting the problem that’s posed here; let me try to rephrase it:
Say you have a large number of families with exactly two children. Assuming boy and girl children are equally likely, and that the sex of the second child in a family is uncorrelated to the sex of the first child, then there are exactly four different cases for the sex of the two children:
1. The first child is a boy, and the second is a boy as well. (B,B)
2. The first child is a boy, but the second is a girl. (B,G)
3. The first child is a girl, but the second is a boy. (G,B)
4. The first child is a girl, and the second is a girl as well. (G,G)
These are each distinct cases, and each is equally likely. Overall, then, the chance of a two-boy family is 1/4. The chance that a family will have one boy and one girl is 1/2. That makes sense, right?
Now, you pick one family among these and discover that one of the two children is a boy. What is the probability that the second child is a boy as well?
If the family you picked has a least one boy, then it’s one of the first three cases listed above: (B,B), (B,G), or (G,B). Because there are more boy-girl families than there are two-boy families, it’s likely that the family you picked is a boy-girl family. Because 1/3 of the families with a boy are two-boy families, the probability that the second child in the family you picked is also a boy is 1/3.
The other problems posed in Matt’s article can be analyzed the same way.
Get two coins and flip them (or flip one now and the other later). Let H be a Boy and T be a girl. Here’s what to do.
(1) Flip both coins, but only look at the first.
(2) Say the following sentence, but substitute “son”/”boy” for #{coin} if the first coin is H, and “daughter”/”girl” if the first coin is T: “Suppose that Mrs Smith has two children, at least one of whom is a #{coin}. What is the probability both children are #{coin}s?”
(2) Mark, in a separate column, your prediction: the value of coin.
(3) Do this at least 20 times.
(4) Divide the number of times your prediction was right by twenty.
(5) That frequency will soon approach the (logical) probability.
(6) Report back to me.
Does the stated original question map to Briggs’s experiment or mine?
It’s ambiguous.
It’s ambiguous even though my experiment doesn’t require us to throw out inconvenient runs (flip the coins 20 times and you get 20 valid results with mine).
It’s easy enough to state a problem without the ambiguity, but the original problem fails that test. That’s why you can be made to lose money if someone uses the original wording, and nothing else, and then asks you to take less than 50/50 odds on a bet that you can predict the other sex. With just the information stated, you can’t do better than a 50/50 guess.
The missing prerequisite, that Mrs. Smith’s children were picked at random and that the decision to announce the sex of at least one of them was also done at random, is indeed *missing*. “Remember: all probability (all logic) are conditional on certain stated evidence–and only that evidence.”
BTW, above I tried to create an experiment that is close to the wording of Briggs’s experiment. One objection could be that my experiment requires use of the word “daughter” and “girls”. If you flip two coins and decide that whatever the first coin lands on is a boy, you will never need to use “daughter” or “girl”, so if the first coin is “heads”, then the second coin will need to be heads for the second child to be a boy. If the first coin is “tails” then the second coin will need to be tails for the second child to be a boy. Again, you’ll get 50/50.
This question is very similar to the Monty Hall question where one flavor of answer depends on whether or not Monty will *always* reveal a goat. If the problem doesn’t state that Monty will *always* reveal a goat, you can’t assume that he will just because he does it in the one instance that you’re asked to play the game. For all you know, he could be an “Evil Monty” who will only reveal a goat when he knows you’ve correctly picked the car. It’s easy enough to state the Monty Hall question and explicitly state that he will always reveal a goat, but typically people omit that required statement and then get into an argument about an incomplete problem.
TomVonk:
I agree that the intended answer is 13/27 (or 12/26 if we are to read “one” as meaning “exactly one”), and that within the context of brain teasers we can presume (as you elegantly put it) that “by ‘evidence’ is meant…the formal transcription of a statement”. No problem there at all, and I computed the probability of a June 29th boy above assuming that reading.
But the interesting thing (to me) about this problem is that it is counterintuitive and spurs debate, and I think part of the reason is that not everyone translates it as a formal proposition. What’s more, the formal logical proposition is not a particularly natural reading of the natural language statement (in my opinion), so the implications of the alternative reading can be discussed.
More broadly, we have to interpret evidence in real life. Having some experience with life, we both know that treating every statement as though its elements can be validly transcribed on a one-to-one basis into a formal logical proposition is clearly wrongheaded (i.e. sometimes our understanding of how a piece of evidence came to be implies that the preferred translation into a formal proposition is *not* the one-to-one reading). But if we have a way to identify the correct evidence-generating model (sorry about the made-up lingo, but I don’t know what else to call it) we can still perform a valid transcription, even if not one-to-one.
And even if we don’t know with certainty what the operative evidence-generating model is, if we can assign probabilities to the various possible models, we can (I think) make valid inferences within a Bayesian framework.
“Suppose that Mrs Smith has two children, at least one of whom is a son. What is the probability that I chose to mention the son (or daughter) by merely revealing the sex of the eldest child (who happened to be a son)?”
For those who say that the answer is unambiguously 0%, please show your work.
Deadhead:
We’re talking the same language, I think. You’re positing an evidence-generating model that includes the possibility that Mrs. Smith might have said “girl” and just happened to say “boy” because of her specific family structure (either she had no choice because of two boys or she happened to pick “boy”, but might equally likely have picked “girl”, because she has one of each). So you would restate the problem “I have two children, one of whom is a [name the sex of a randomly chosen child]. What is the probability both children are [sex of the randomly chosen child]?”
And under your assumptions:
P(H|E) = P(E|H)*P(H)/P(E)
The universe is all parents of two children.
P(E|H) = 1 (if the hypothesis that there are two children of [named sex] is true, then the evidence will always be “I have a at least one [child of named sex]”)
P(H) = 1/2 (because the probability of 2 boys or 2 girls (inclusive or) is the same as that of 1 of each)
P(E) = 1 (every parent in the two-child universe can play this game and will provide evidence of the form “I have two children, one of whom is a [boy or girl])
So P(H|E) = 1/2
I have to admit, even when I was arguing that “Tuesday” might be an uninformative descriptor, it didn’t occur to me that “boy” might be as well. But of course that reading isn’t ruled out either.
Morgan: We cross posted. My short post @ 10:31 was not aimed at you, but at people who may still claim that the question is not ambiguous.
Like the Monty Hall problem, it’s a shame that the question is worded ambiguously, because even when each question is stated in a way that it requires no additional qualifications, it’s still a fun problem that many people get the wrong answer to. It doesn’t help when people then dig in their heels and insist that it’s not ambiguous, well, unless it’s fun to tug at people with their heels dug in.
The fact that this is worded ambiguously ought to be the point. Treated as an excercise in probability, we would want a precise statement that we could then compute probabilities from. However as an excercise in statistics, it is crtical to remind us that we cannot simply compute probabilities from some statement someone gives. We need to know the basis upon which they arrived at that statement.
That this is critical is most pertinently observable in the court. If a suspect was found using a trawl of a DNA database then the statement that the subject’s DNA matches that at the crime scene counts for nothing.
If on the other hand the suspect was found at the scene of the crime, and a DNA match test was done then the fact that there was a DNA match counts for a large amount.
In this case is the reason the “Tuesday” was given was just to provide information about the boy, then this is no information from the point of view of this Q, and the probability is 1/3. If on the other hand someone asked “hands up if you have a son born on Tues, and Mrs Smith (
whoops… continuing.
…put up her hand, then the alternative values given here are the probabilities.
Lets never forget that how you come by information is as statistically relevant as the information you come by. That is why we really really need to worry about bias in any information collecting excercise.
Amos Storkey:
Nicely put.
As to Tuesday:
I have two children, a boy and a girl. Only one of them was born in June. What are the chances it was the girl, given only this information?
50% — the number of months is irrelevant. The alternatives here are the children, not the months.
I have two children, one a son born on Tuesday. Probability of two boys?
50% . The probability of at least one boy is 1. So only the probability of the other child matters. There are exactly two alternatives: a boy or a girl. The alternatives are the sexes, not the days of the week.
Think about a reductio ad absurdum: There are only 25 boy’s names allowed in my country. I have two children, one of whom is a son named Fred. Probability of two boys?
— we clearly don’t need a list of all the possible names to answer this.
Matt —
Prove it by experiment.
Take two coins. Set one on the table, heads up. That’s the son born on Tuesday. Don’t touch it; that’s the information you’re given.
Flip the other coin at least 500 times and see how often it comes up heads.
@Tom Pollard
2 July 2010 at 9:02 am
But birth order is irrelevant. The operative alternatives are:
2 boys
2 girls
One of each
“2 girls” is eliminated. So once again, there are two equal probabilities.
William —
Consider the following logic, based on the red-blue block diagram in the article, which is like this:
1. BB
2. BG
3. GB
4. GG
Now, we know that Mrs. Smith belongs to EITHER the case where her eldest is a boy (1,2) OR the case where her youngest is a boy (1,3). At least of those alternatives must be true.
If case (1,2), then in half (50%) of the alternatives, she has two boys.
If case (1,3), then in half (50%) of the alternatives, she has two boys.
Therefore, in every possible case, the probability is 50%.
What’s wrong with this logic? Let’s ask de Morgan or Quine or somebody like that.
Putting it into good old symbolic logic, the statistician points out that case 1 factors out:
(BB v BG) v (BB v GB) >> BB v (GB V BG) — i.e. there are more ways to roll a 7 than any other number …
BUT if your firstborn is a son, there’s a 50% probability that you will have two sons. (And if your firstborn is not a son, and two is the limit, then there is a zero % probability that you will have two sons.) The more I think about this, the less sense it all makes…
It is extremely counterintuitive that if Mrs. Smith told you whether the son she was telling you about is the oldest or youngest, it would change the probability of the sex of his sibling, but apparently it would. Genuinely weird, since you know in advance that the kid has to be either the oldest or youngest.
Craig Goodrich:
“It is extremely counterintuitive that if Mrs. Smith told you whether the son she was telling you about is the oldest or youngest, it would change the probability of the sex of his sibling…”
It would not change the probability of the sex of his sibling unless we take it to be a condition that restricts the universe of parents from which Mrs. Smith was drawn.
If “my oldest son is a boy” means that only someone whose oldest son is a boy can pose the “what’s the chance of two boys?” question to you, then only parents from the B-G and B-B quadrants can play the game, and yes, it changes the probability to 1/2. But if it simply describes the boy (i.e. it can take values “older” or “younger” depending on the circumstances) then all parents of at least one boy (parents from the G-B, B-G, and B-B quadrants) can still play, and it doesn’t change the probability – it’s still 1/3.
Same thing with Tuesday, same thing with “boy”. If “boy” simply describes the child, rather than acts as a condition that restricts the universe of parents (i.e. if we could have heard “at least one is a girl, what is the probability of two girls” instead), then all four quadrants can play.
@Craig Goodrich
You say that
> The operative alternatives are:
> 2 boys
> 2 girls
> One of each
You can describe the alternatives this way, but it’s misleading, because these three alternatives are not equally likely. A two-child family is twice as likely to have “one of each” as it is to have two boys. When you evaluate probabilities by enumerating alternatives, you need to be careful that the alternatives you list are equally probable.
The reason you need to enumerate the alternatives as
first child BOY, second child BOY
first child BOY, second child GIRL
first child GIRL, second child BOY
first child GIRL, second child GIRL
is because each of these cases is equally likely. If this isn’t obvious to you, look at it this way: what’s the probability of having two children who are both boys? The probability that your first child is a boy is 1/2, and the probability that the second child is a boy is also 1/2. These two events are independent of one another, so the chance that both the first and second children are boys is 1/4. Counting things the way you proposed would lead you to believe that the chance of two boys was 1/3.
Hi,
Given such an information, regarding an event that involves at least one male child, Can the probability of having two boys ever be exactly 0.5?
You are all wrong – including the author of the question. No one seems to have noticed the “at least”. The original question says “at least”, so the answer is 1/3. The second question does not say “at least” so it is not the same type of question and the answer is 1/2. If the “at least” had been left out of the first question it would also be 1/2.
“Suppose that Mrs Smith has two children, at least one of whom is a son. What is the probability both children are boys?”
To me this is like saying “Mrs Smith, do you have at least one son?” and she says yes, meaning that there are now three equally likely outcomes – BB , BG , GB. Hence 1/3.
But if the question had said:
“Suppose that Mrs Smith has two children, one of whom is a son. What is the probability both children are boys?”
This is different. This says to me that Mrs Smith has randomly chosen one of her children, and he happened to be a son. The other child then can be either a boy or girl – i.e. probability is 1/2.
The author probably meant to put “at least” into the second question
“I have two children, at least one of whom is a son born on a Tuesday. What is the probability that I have two boys?”
If he had written it this way his answer would have been correct.
But without the “at least” he is randomly choosing a child and he happened to be a boy who was born on a Tuesday. The other child is simply a boy or a girl i.e. the probability is 1/2.
It’s really disappointing to me to see so many people denying the actual solution in favor of their own simplification. It’s not 1/2 no matter how many of you insist it is….including one who claims to be a physicist. The fact that one son is born on Tuesday is relevant.
One common mistake here is the assumption that when we say “One is a boy” that we know which one that is. That does not reduce the problem down to what gender the other one is, because we don’t know which one is the son…the first or second….and that matters. There is a sad but clear lack of understanding here in the basics of probability. Some of you would do well to open your minds and ask “What am I missing here?”.