Probability puzzle
Radford Neal
extension of a well-known puzzle, but I think my modification makes it
harder, and more enlightening once you get the right answer...
A couple you've just met invite you over to dinner, saying "come by
around 5pm, and we can talk for a while before our three kids come
home from school at 6pm".
You arrive at the appointed time, and are invited into the house.
Walking down the hall, your host points to three closed doors and
says, "those are the kids' bedrooms". You stumble a bit when passing
one of these doors, and accidently push the door open. There you see
a dresser with a jewelry box, and a bed on which a dress has been laid
out. "Ah", you think to yourself, "I see that at least one of their
three kids is a girl".
Your hosts sit you down in the kitchen, and leave you there while they
go off to get goodies from the stores in the basement. While they're
away, you notice a letter from the principal of the local school
tacked up on the refrigerator. "Dear Parent", it begins, "Each year
at this time, I write to all parents, such as yourself, who have a boy
or boys in the school, asking you to volunteer your time to help the
boys' hockey team..." "Umm", you think, "I see that they have at
least one boy as well".
That, of course, leaves only two possibilities: Either they have two
boys and one girl, or two girls and one boy. What are the probabilities
of these two possibilities?
----------------------------------------------------------------------------
Radford M. Neal rad...@cs.utoronto.ca
Dept. of Statistics and Dept. of Computer Science rad...@utstat.utoronto.ca
University of Toronto http://www.cs.utoronto.ca/~radford
----------------------------------------------------------------------------
Henry
>Here's a probability puzzle to amuse and enlighten you. It's an
>extension of a well-known puzzle, but I think my modification makes it
>harder, and more enlightening once you get the right answer
I'm not convinced it is more enlightening, but I would say 2/3 chance
of two girls one boy and 1/3 of two boys one girl.
Andrew Harvey
if we accept the evidence as proof that there is in the family one boy and one
girl then the only unknown remains the gender of the third child. This child
could be male (p=0.5) or female (p=0.5) i.e. your two possibilities each have a
50% probability. (ignoring the complications of incompetent school secretariats
and cross-dressing adolescents).
Radford Neal wrote:
> Here's a probability puzzle to amuse and enlighten you. It's an
> extension of a well-known puzzle, but I think my modification makes it
> harder, and more enlightening once you get the right answer...
Radford Neal
that the odds are 2:1 in favour of there being two girls and one boy.
The old puzzle involves two kids, and information that at least one of
them is a girl. It can suffer from ambiguous phrasing, such as "you
know that they have two kids, and that one of them is a girl". The
extended puzzle is supposed to clarify this ambiguous phrase in two
ways, providing information that there is at least one boy and at
least one girl, but in different ways. To avoid falling into the trap
of thinking the situation is symmetrical with respect to boys and
girls, one has to realize that the statement "I see that at least one
of their three kids is a girl" is true, but that it is NOT a complete
statement of the information obtained from stumbling into the bedroom.
You know more: namely that one PARTICULAR child is a girl.
Avoiding all traces of ambiguity can be hard, though:
>3) Quibble: With 3 kids the possibility of two boys at different schools
>(eg, elementary & junior high) cannot be discounted. This would strentghen
>the evedince for multiple boys, but not to the extent that the bedrooms do
>for girls.
I tried to avoid this by the phrase "before our kids come home from
school at 6pm", which seems to imply that they all go to the same school.
I think I did avoid other problems. Eg, my initial phrasing had one
of the bedroom doors left open, but then I realized that that won't work,
since it could be that boys are more (or less) likely to leave their
bedroom doors open than girls.
Interestingly, one DOESN'T have to make any assumptions about identical
twins being impossible to get the answer.
Jack Tomsky
G B G
G G B
G B B
The probability of two girls and one boy is 2/3. The probability of
one girl and two boys is 1/3. So I would agree with Monte Hall, er,
Henry.
Bill Taylor
|> As several people have posted, the correct answer to the puzzle is
|> that the odds are 2:1 in favour of there being two girls and one boy.
Yep; nice puzzle! Congratulations on this new twist.
|> Avoiding all traces of ambiguity can be hard, though:
I was astounded and impressed! It was beautifully written.
If only that idiot vos Savant had had some of your nous before she
got into this territory... sigh... oh well, c'est la vie.
Well done again, and thanks.
------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
------------------------------------------------------------------------
William Tell, missing the apple but hitting his son in the eye 3 times:-
YES! Look at that GROUPING!
------------------------------------------------------------------------
r.e.s.
[...]
: To avoid falling into the trap
: of thinking the situation is symmetrical with respect to boys and
: girls, one has to realize that the statement "I see that at least one
: of their three kids is a girl" is true, but that it is NOT a complete
: statement of the information obtained from stumbling into the bedroom.
: You know more: namely that one PARTICULAR child is a girl.
Hmm, I wonder...
That explanation seems not quite adequate to me,
since we can also generate the puzzle even if nothing
is learned about any *particular* child.
I think it's agreed that the heart of the puzzle is
an asymmetry expressed by the following equations,
where Event1, Event2 are the "bedroom", "kitchen"
events, respectively, in your version:
pr(Event1|1Boy,2Girls) = 2*pr(Event1|2Boys,1Girl)
pr(Event2|1Boy,2Girls) = 1*pr(Event2|2Boys,1Girl).
But we can design the puzzle so that these two
equations hold, with *neither* event conveying
information particular to any child.
As in your version, suppose that there are three
children's bedrooms, but now also suppose that the
following (admittedly unusual) message is conveyed:
"Between the two of them, the same-gender children
are sharing their bedrooms."
It seems to me that the same asymmetry exists as
before in the two probability equations -- but now
the "bedroom event", although still twice as likely
given 2Girls compared to 2Boys, seems not to convey
any info about a particular girl. (The information
is particular to a *bedroom*, but not to a person.)
--
r.e.s.
rs...@mindspring.com
Kazimierz Wiesak
The best way to solve such puzzles is to first clearly specify the
initial probability space. In our case it is
BBB BBG BGB GBB GGB GBG BGG GGG
all with equal probabilities.
First, we get information that there is a girl in one randomly
selected bedroom. Using Bayes theorem to employ this information I got
P(one girl) = 1/4
P(two girls) = 1/2
P(three girls) = 1/4.
If the additional information from the kitchen would say only that
that there is a boy in the house (i.e. the case of three girls is
excluded) then indeed the final probabilities would be P(one girl) =
1/3, P(two girls)=2/3. The problem is that that information says more
and when imposed onto what we already know (by using Bayes theorem
again) it returns us back to the original (no information at all).
Final probabilities are P(one girl) = 1/2, P(two girls) = 1/2 as
someone correctly suspected.
Kaz Wiesak
Will Dwinnell
"If the additional information from the kitchen would say only that that
there is a boy in the house (i.e. the case of three girls is excluded)
then indeed the final probabilities would be P(one girl) = 1/3, P(two
girls)=2/3. The problem is that that information says more and when
imposed onto what we already know (by using Bayes theorem
again) it returns us back to the original (no information at all).
Final probabilities are P(one girl) = 1/2, P(two girls) = 1/2 as
someone correctly suspected."
I agree. We are presented with evidence that there is at least one
girl, and at least oen boy. Given that there are three children and no
further evidence, it would seem that the probabilities are 0.5, 0.5.
Will Dwinnell
James Lymp
At a local bar there are two taps: one is Budweiser and the other is
Guinness. A customer walks in the door and is blindfolded. The bartender
pours three pints of beer. Each time, before pouring the beer, the bartender
flips a coin: if the coin is heads the pour is Budweiser and if tails it's
Guinness. The pints are then set on the bar at the customer's left, middle
and right according to the order poured. The customer then chooses any of
the three beers and drinks it - a Guinness. Now the customer is told that at
least one of the other two glasses is a Budweiser. What is the probability
that there is another Guinness?
--
James Lymp
http://staff.washington.edu/lymp
ly...@u.washington.edu
Mark W. Brown
ZERO! No one could resist a freshly poured Guinness sitting on the bar
in front of a blind-folded man! Someone would swipe it and drink it!
:-)
-Mark-
Kazimierz Wiesak
"the customer is told...". How does one probabilistically model it.
The first 2/3 of both puzzles are probabilistically identical.
Step 1. Both puzzles have the same initial probability space.
BBB BBG BGB GBB BGG GBG GGB GGG
where all outcomes are equally likely.
Step 2. (Same for both puzzles). After randomly selecting Guiness (G)
it is obvious that the case BBB is no longer possible. Since two G's
and one B makes the selection of G more likely than two B's and one G
would have, it is clear that the additional information of randomly
selecting G revises the initial estimates of probabilities. Applying
Bayes theorem one gets
P(one Guiness) = 1/4
P(two Guinesses) = 1/2
P(three Guinesses) = 1/4
Step 3. If the customer randomly selected Budwiser (from the remaining
two drinks) then this would restore the original equality of
probabilities P(two Guinesses) = P(two Budweisers). But the customer
is told! Being told excludes the case of "three Guinesses". But does
it change the relative probabilities of "one Guiness" and "two
Guinesses"? (Now, 1/3 and 2/3, respectively, after excluding the case
of "three Guinesses"). Here we are leaving the domain of Probability
and entering the domain of English. In other words, the English
statement is not precise enough, thus precluding any further
reasoning.
(In the first puzzle the assumption was that all three children had
equal chance of having a school message posted on refrigerator. This
restored the equality P(two girls) = P(two boys)).
Kaz Wiesak
James Lymp
> it is obvious that the case BBB is no longer possible. Since two G's
> and one B makes the selection of G more likely than two B's and one G
> would have, it is clear that the additional information of randomly
> selecting G revises the initial estimates of probabilities. Applying
> P(three Guinesses) = 1/4
I do not follow the need for Bayes Theorem. How about getting there this
way. Let's say I pick the middle pint and it's a Guinness. That leaves the
following possibilities (all equally likely) for the left and right pints
(since the pours were independent).
BB BG GB GG
This results in the same probabilities.
> Step 3. ... But the customer
> is told! Being told excludes the case of "three Guinesses". But does
> it change the relative probabilities of "one Guiness" and "two
> Guinesses"? (Now, 1/3 and 2/3, respectively, after excluding the case
> of "three Guinesses"). Here we are leaving the domain of Probability
> and entering the domain of English. In other words, the English
> statement is not precise enough, thus precluding any further
> reasoning.
Hmmm. You might have a point. What happens if we say the following. The
bartender will tell the customer "at least one is a Budweiser" if possible;
otherwise the bartender will say "at least one is a Guinness".
Radford Neal
Kazimierz Wiesak <kx...@psu.edu> wrote:
> I think I disagree with most of the people who sent their answers.
> The best way to solve such puzzles is to first clearly specify the
> initial probability space. In our case it is
>
> BBB BBG BGB GBB GGB GBG BGG GGG
>
> all with equal probabilities.
>
> First, we get information that there is a girl in one randomly
> selected bedroom. Using Bayes theorem to employ this information I got
>
> P(one girl) = 1/4
> P(two girls) = 1/2
> P(three girls) = 1/4.
>
> If the additional information from the kitchen would say only that
> that there is a boy in the house (i.e. the case of three girls is
> excluded) then indeed the final probabilities would be P(one girl) =
> 1/3, P(two girls)=2/3. The problem is that that information says more
> and when imposed onto what we already know (by using Bayes theorem
> again) it returns us back to the original (no information at all).
>
> Final probabilities are P(one girl) = 1/2, P(two girls) = 1/2 as
> someone correctly suspected.
This is obscure to me. What is this information that says more? Maybe
the following from another post by the same poster answer this question...
>... the assumption was that all three children had
>equal chance of having a school message posted on refrigerator. This
>restored the equality P(two girls) = P(two boys)).
This was not the intent of the story. The letter on the refrigerator
is supposed to convey only the information that they have at least one
boy. In particular, they would get exactly the same letter (and only
one letter) if they have one boy or if they have two boys. I tried to
make this clear.
It's always possible to imagine that the absence of something is
significant. Perhaps you know that the school has a special program
to help girls who are unfortunate enough to not have a sister to talk
things over with, and they sent out letters to all parents of girls
without sisters announcing this new program. Then the absence of this
letter from the refrigerator would be evidence in favour of two of the
children being girls.
However, if you consider possibilities like this that have no support
in the story, you will eliminate *all* probability puzzles. It
doesn't seem like much fun. Note that in the current puzzle nothing
changes even if you assume that there is a girls' hockey team, managed
in the same way as the boys' hockey team. You already know that they
have at least one girl, so apparently the letter regarding the girls'
hockey team just fell off the refrigerator.
Radford Neal
Will Dwinnell
"This was not the intent of the story. The letter on the refrigerator
is supposed to convey only the information that they have at least one
boy. In particular, they would get exactly the same letter (and only
one letter) if they have one boy or if they have two boys. I tried to
make this clear."
That much seems clear to me, but I find myself back at:
what is known:
1. there are 3 children
2. at least one is a girl (room)
3. at least one is a boy (letter)
With no further information, I don't see why we wouldn't settle on 0.5,
0.5 probabilities. The analysis which led us to 2/3, 1/3 would seem to
be something which could be performed in either sense, leading to the
opposite (2/3, 1/3) conclusion.
Will Dwinnell
pred...@compuserve.com
Radford Neal
Ah, but the point is that although (1), (2), and (3) are all true,
there is indeed some further information. You know that the child in
the bedroom you stumbled into is a girl. This is definitely further
information. You can't deduce it from (2). On the other hand, you
don't know any further information from the letter. Suppose you
stumbled into the first bedroom in the hall. We can list the eight
possibilities for boys and girls in these bedrooms as follows:
BBB BBG BGB BGG GBB GBG GGB GGG
When you stumble into the first bedroom, the first four of these
possibilities are eliminated. When you find out that there is at
least one girl, the last is eliminated. Of the remaining
possibilities, GBB, GBG, and GGB, two out of three have two girls.
Radford Neal
Robert Froese
rad...@cs.toronto.edu (Radford Neal) wrote:
> Suppose you
> stumbled into the first bedroom in the hall. We can list the eight
> possibilities for boys and girls in these bedrooms as follows:
>
> BBB BBG BGB BGG GBB GBG GGB GGG
>
> When you stumble into the first bedroom, the first four of these
> possibilities are eliminated. When you find out that there is at
> least one girl, the last is eliminated. Of the remaining
> possibilities, GBB, GBG, and GGB, two out of three have two girls.
Order of discovery is a common element in most examples of basic
probability. I reviewed my basic probability text (by J.A. Rice) just
now, and each example refers to an explicit *series* of events. The
difference between Rice and Mr. Neal is that Rice states this in his
question, for example,
"Driving to work, a commuter passes through a sequence of three
intersections with traffic lights. At each light, she either stops, s,
or continues, c. The sample space is the set of all possible outcomes
omega = {ccc,ccs,css,csc,sss,ssc,scc,scs}"
Furthermore, "the event that the commuter stops at the first light is
the subset of omega denoted by A={sss,ssc,scc,scs}." However, many
events are possible. For example, the event that the commuter stops at
least once and continues at least once is denoted by
A={ssc,scs,css,ccs,csc,scc}.
Now our task as your audience is to decide exactly what the experiment
is, what the two events are, and then determine their probabilities.
Can we can agree that the experiment may be defined as "a guest
discovers the gender of each of the three children of his hosts, each
born with probability of one gender or the other equal to 0.5?" The
sample space is the set of all possible outcomes
omega = {ggg,ggb,gbb,gbg,bbb,bbg,bgg,bgb}.
Now to your question: "either they have two boys and one girl, or two
girls and one boy. What are the probabilities of these two
possibilities?" Here comes the interpretation! Is your question the
probability associated with the events:
A= the first is a girl, and both of the remaining two are boys, and
B= the first is a girl, and one each of the remaining two are a boy and
a girl;
Or,
C= at least one is a girl and one is a boy?
Alternatively, may I provide a different definition for the experiment?
Rice notes that the probability of an event may be calculated as the
number of ways the event may occur divided by the total number of
outcomes, *only* if all outcomes are equally likely. In your example,
the first iteration of the "experiment" has but one outcome - a girl -
with a probability of 1. The other two children have probability 0.5 of
each gender, leading to the sample space:
omega = {bg,gb,gg,bb}
and your question becomes the probability associated with the event that
at least one child is a boy.
Rice defines probability as "a mathematical model for chance phenomena."
Shall we be a bit more clear on what model you are basing your answer
upon?
Cheers,
...Robert
--
Robert Froese, RPF
Student and Research Assistant, Forest Biometrics, University of Idaho
MF Candidate, Quantitative Silviculture, University of BC
Of course, all opinions are mine and mine alone, and do not represent
anyone else's silly thoughts.
Sent via Deja.com http://www.deja.com/
Before you buy.
Jerry Dallal
b for boy, g for girl--without regard to order, Thus,
P(bbg) is the probability of two boys and a girl.
To keep the notation clean, g-s will written before b-s.
If we make the rash assumption that, overall, the probability of
having a boy is equal to the probability of having a girl and, within
family, the sexes of children are independent,
P(ggg) = P(bbb) = 1/8
P(ggb) = P(gbb) = 3/8
Assume that when a bedroom is examined, the sex of the occupant
can always be determined, without error.
Let G be the event that a bedroom is chosen *at random*, its
contents are examined, and the occupant is found to be a girl.
P(G|ggg) = 1
P(G|ggb) = 2/3
P(G|gbb) = 1/3
P(G|bbb) = 0
Let x = P(G|ggg)P(ggg)+P(G|ggb)P(ggb)+P(G|gbb)P(gbb)+P(G|bbb)P(bbb)
= 1 1/8 + 2/3 3/8 + 1/3 3/8 + 0 1/8
= 1/2
Then,
P(ggg|G) = P(G|ggg)P(ggg)/ x
= (1 1/8) / (1/2)
= 1/4
P(ggb|G) = P(G|ggb)P(ggb)/ x
= (2/3 3/8) / (1/2)
= 1/2
P(gbb|G) = P(G|gbb)P(gbb)/ x
= (1/3 3/8) / (1/2)
= 1/4
P(bbb|G) = P(G|bbb)P(bbb)/ x
= (0 1/8) / (1/2)
= 0
r.e.s.
[snip derivation of:]
: P(G|ggg) = 1
: P(G|ggb) = 2/3
: P(G|gbb) = 1/3
: P(G|bbb) = 0
So you've shown that P(G|ggb)=2*P(G|gbb).
We can also show that P(B|ggb)=1*P(B|gbb),
where B is the event that occurs in the kitchen.
(This is the essential asymmetry in this puzzle.)
The solution, due to this asymmetry, is then
P(ggb|GB)=2/3 and P(gbb|GB) = 1/3,
because by Bayes theorem we have
P(ggb|GB) = 1/(1+1/r)
where r is the odds ratio:
r = P(GB|ggb)/P(GB|gbb), since P(gbb)=P(ggb)
= P(G|ggb)P(B|ggb) / (P(G|gbb)P(B|gbb))
= (P(G|ggb)/P(G|gbb)) * (P(B|ggb)/P(B|gbb))
= ( 2 ) * ( 1 )
= 2
(Writing Bayes theorem in terms of an odds ratio
makes it clearer that the the above-stated
asymmetry is the core of this puzzle.)
Then P(ggb|GB)=1/(1+1/r)=2/3, and P(gbb|GB)=1/3.
--
r.e.s.
rs...@mindspring.com
William D. Allen Sr.
mail was received from PhDs and math profs with incorrect solutions.
Lack of skill in statistical analysis is apparently a common problem in the
technical professions.
WDA
end
r.e.s. wrote in message <7u7qfm$fjs$1...@nntp4.atl.mindspring.net>...
r.e.s.
be sure whether you agree or disagree with what
I wrote. But if you believe there is a mistake,
then please, if you have the skill, point it out,
so that all may learn by the discussion.
--
r.e.s.
rs...@mindspring.com
William D. Allen Sr. <ball...@home.com> wrote ...
: "Ask Marilyn" posed this question in Parade magazine
: some years ago. Lots of mail was received from PhDs
: and math profs with incorrect solutions.
:
: Lack of skill in statistical analysis is apparently
: a common problem in the technical professions.
:
: WDA
:
: end
:
: r.e.s. wrote ...
Radford Neal
Robert Froese <rfr...@my-deja.com> wrote:
>Now to your question: "either they have two boys and one girl, or two
>girls and one boy. What are the probabilities of these two
>possibilities?" Here comes the interpretation! Is your question the
>probability associated with the events:
>
>A= the first is a girl, and both of the remaining two are boys, and
>B= the first is a girl, and one each of the remaining two are a boy and
>a girl;
>
>Or,
>
>C= at least one is a girl and one is a boy?
Oh come on! Perhaps one can dispute whether some assumed facts
earlier on in the puzzle are made absolutely clear or not, but I just
don't believe that there's anything at all ambiguous about the
phrasing of the final question. It is quite clear that the question
is neither asking for the probabilities of your events A and B, nor
for the probability of your event C, but rather for the probabilities
of D and E, where
D = one child is a girl and two children are boys
E = one child is a boy and two children are girls
It has been previously established that the probability of the event
(D or E) is one.
Here's the puzzle again from my original posting:
A couple you've just met invite you over to dinner, saying "come by
around 5pm, and we can talk for a while before our three kids come
home from school at 6pm".
You arrive at the appointed time, and are invited into the house.
Walking down the hall, your host points to three closed doors and
says, "those are the kids' bedrooms". You stumble a bit when passing
one of these doors, and accidently push the door open. There you see
a dresser with a jewelry box, and a bed on which a dress has been laid
out. "Ah", you think to yourself, "I see that at least one of their
three kids is a girl".
Your hosts sit you down in the kitchen, and leave you there while they
go off to get goodies from the stores in the basement. While they're
away, you notice a letter from the principal of the local school
tacked up on the refrigerator. "Dear Parent", it begins, "Each year
at this time, I write to all parents, such as yourself, who have a boy
or boys in the school, asking you to volunteer your time to help the
boys' hockey team..." "Umm", you think, "I see that they have at
least one boy as well".
That, of course, leaves only two possibilities: Either they have two
Kazimierz Wiesak
>What is the chance that we'll ever reach consensus about it?
I thought we did reach the consensus!
As you know, to get an answer, any answer, one MUST start from some
data and assumptions. One cannot get an answer starting from nothing!
Here is the difference:
(A) I started from the assumption (in the second part of the puzzle)
that each child has an EQUAL chance of getting a message from the
school posted on the refrigerator and obtained (correctly!)
P(two girls) = P(two boys) = 1/2.
(B) Jerry Dallal started from the assumption that the probability that
girl(s) could get a message from the school posted on the refrigerator
is ZERO ("P(GL | ggg) = 0") and obtained the correct (!) answer P(two
girls) = 2/3.
The difference between us is on the level of building the model, not
on the level of reasoning within the model. Which model is more
"correct"?
Read the relevant part of the original puzzle:
"a letter from the school tacked up on the refrigerator (...) "Each
year at this time, I write to all parents, such as yourslf, who have a
boy or boys in the school, asking you to volunteer your time to help
the boys' hockey team..."
and pick your own probability that girl(s) can get any message posted
on the refrigerator. All what a probabilist can say is
1. IF you pick P(girl(s) get a message on the fridge) = 0, then P(two
girls in the house) = 2/3
2. IF you pick P(girls can get a message on the fridge) > 0, ie. there
was a chance that the school could write any kind of message to girls
that could be posted on the refrigerator, then
P(two girls in the house) is less than 2/3.
The bigger P(girl(s) get a message on the fridge), the lower P(two
girls in the house) becomes.
3. IF you pick P(girl(s) get a message on the fridge) = 1 (say, the
principal writes to all girls asking for help on girls' volleyball
team, by analogy to boys' hockey team), then P(any girl in the house)
= 0, which, of course, makes the puzzle contradictory.
Kaz Wiesak
Ivan Verdenik
James Lymp wrote:
> I heard this version once and I think I like it even better.
>
> At a local bar there are two taps: one is Budweiser and the other is
> Guinness. A customer walks in the door and is blindfolded. The bartender
> pours three pints of beer. Each time, before pouring the beer, the bartender
> flips a coin: if the coin is heads the pour is Budweiser and if tails it's
> Guinness. The pints are then set on the bar at the customer's left, middle
> and right according to the order poured. The customer then chooses any of
> the three beers and drinks it - a Guinness. Now the customer is told that at
> least one of the other two glasses is a Budweiser. What is the probability
> that there is another Guinness?
>
>
As the first is Guiness and one of others Budweiser, there are three
possibilities left:
G GB
G BG
G BB
Therefore the probability that there is another Guinness is 2/3.