Probability Of At Least One Chocolate In Dish
heich1
Nov. 18th, 2009. I and he have a difference in interpretation of this
puzzle and hence its solution. Here it is.
Diana walks by the candy dish in her home. She knows there are three
pieces of candy left in the dish and that those candies are either
chocolates or caramels, but she can't see them.
She also knows that there is an equal chance they could be a
combination of the two candies.
Her brother walks by and she tells him she'll buy him a pizza if he
can tell her what the probability is that at least
one chocolate candy is in the dish.
He answers "3/4"
No pizza for Diana's brother.
Can you tell Diana what the correct answer is?
Please give the answer(s) and your method to get the answer(s).
Eventually I will give both Terry Stickels' answer and mine.
Peter Heichelheim
Eric Sosman
S
P
O
I
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A
G
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If we view the dish as holding a set of three candies, there
are four possible states: A={3*Ch}, B=(2*Ch,Ca}, C={Ch,2*Ca}, and
D=(3*Ca}. Mixed and unmixed states are equiprobable, so we have
P(A)+P(D) = P(B)+P(C) = 0.5. The probability of "at least one
chocolate" is one minus the probability of "all caramels," or
1-P(D). But there we're stuck: We know that P(A) and P(D) sum
to 0.5, but we don't know their individual values (for example,
if Diana's mother likes caramels and would be unable to resist
the temptation of three, we could have P(D)=0 and P(A)=1/2.) We
can conclude that 1-P(D) is somewhere between 0.5 and 1.0, but
that's all.
If we view the dish as holding an arrangement of three candies,
there are eight possible states corresponding to the eight ordered
triples on the set {Ch,Ca}. Again, the mixed and unmixed states
are equiprobable, so P({Ch,Ch,Ch})+P({Ca,Ca,Ca}) = 0.5. Again, we
know the sum of the two terms but are unable to isolate the value
of P({Ca,Ca,Ca}) from that sum, and again we conclude that "at least
one chocolate" has a probability between 0.5 and 1.0.
So in both cases the brother's answer should have been
"Better than an even chance," nothing more definite.
(Mine's the one with the nougat center.)
--
Eric Sosman
eso...@ieee-dot-org.invalid
Mensanator
No, there are 8 states.
> A={3*Ch}, B=(2*Ch,Ca}, C={Ch,2*Ca}, and
> D=(3*Ca}.
That's combinations, you should be figuring permutations.
The 8 permutations are:
Ca, Ca, Ca
Ca, Ca, Ch
Ca, Ch, Ca
Ca, Ch, Ch
Ch, Ca, Ca
Ch, Ca, Ch
Ch, Ch, Ca
Ch, Ch, Ch
> Mixed and unmixed states are equiprobable, so we have
> P(A)+P(D) = P(B)+P(C) = 0.5.
This makes no sense.
> The probability of "at least one
> chocolate" is one minus the probability of "all caramels," or
> 1-P(D). But there we're stuck:
No, we're not, the probability of "all caramel" is 1/8.
So, in 7 out of 8 cases, there is at least one chocolate.
> We know that P(A) and P(D) sum
> to 0.5, but we don't know their individual values (for example,
> if Diana's mother likes caramels and would be unable to resist
> the temptation of three, we could have P(D)=0 and P(A)=1/2.) We
> can conclude that 1-P(D) is somewhere between 0.5 and 1.0, but
> that's all.
But is, in fact, 7/8.
>
> If we view the dish as holding an arrangement of three candies,
> there are eight possible states corresponding to the eight ordered
> triples on the set {Ch,Ca}. Again, the mixed and unmixed states
> are equiprobable, so P({Ch,Ch,Ch})+P({Ca,Ca,Ca}) = 0.5. Again, we
> know the sum of the two terms but are unable to isolate the value
> of P({Ca,Ca,Ca}) from that sum, and again we conclude that "at least
> one chocolate" has a probability between 0.5 and 1.0.
Maybe *you* can't isolate it, but *I* can.
>
> So in both cases the brother's answer should have been
> "Better than an even chance," nothing more definite.
>
> (Mine's the one with the nougat center.)
What do you suppose the probability of that is?
Hint: probability is the number of successes divided by total
number of possible outcomes.
>
> --
> Eric Sosman
> esos...@ieee-dot-org.invalid- Hide quoted text -
>
> - Show quoted text -
Sylvia Else
And in 6 out of 8 there is a combination. But we're told that the
probability of a combination is evens. So the candies are not simply a
random selection, and you have to take that into account.
Sylvia.
Mensanator
And yet, we are told that 3/4 is not the answer.
This will be good.
>
> Sylvia.- Hide quoted text -
Eric Sosman
That's the other analysis (also performed), not this one.
One views the candies as unordered sets, the other as ordered
triples. Both lead to the same conclusion (and maybe there's
a slick argument somewhere to show that the results are the
same; I've seen this called "The Law of the Unconscious
Statistician," but I don't recall a proof).
>> Mixed and unmixed states are equiprobable, so we have
>> P(A)+P(D) = P(B)+P(C) = 0.5.
>
> This makes no sense.
P(mixed) = P(unmixed); that's what Diana's second piece of
information says. Since the two P(*) add to 1.0 (because "mixed"
and "unmixed" exhaust the entire universe of possibilities), and
they are equal, P(*) = 0.5 is the only solution. What's nonsensical
about that?
>> If we view the dish as holding an arrangement of three candies,
>> there are eight possible states corresponding to the eight ordered
>> triples on the set {Ch,Ca}. Again, the mixed and unmixed states
>> are equiprobable, so P({Ch,Ch,Ch})+P({Ca,Ca,Ca}) = 0.5. Again, we
>> know the sum of the two terms but are unable to isolate the value
>> of P({Ca,Ca,Ca}) from that sum, and again we conclude that "at least
>> one chocolate" has a probability between 0.5 and 1.0.
>
> Maybe *you* can't isolate it, but *I* can.
How? I offer you the following set of probabilities:
P({Ch,Ch,Ch}) = 0.5
P({Ca,Ca,Ca}) = 0.0
P(each other) = 0.1666...
In this scheme, the probability of "at least one chocolate" is
unity. But I might instead have offered
P{(Ch,Ch,Ch}) = 0.0
P({Ca,Ca,Ca}) = 0.5
P(each other) = 0.1666...
... and now the probability of "at least one chocolate" is one-half.
I can set the first two probabilities to *any* pair of non-negative
values that sum to 0.5 and meet the conditions of the puzzle.
If you can find something *in the problem as stated* that
forbids either of these arrangements, or any other arrangement
in between these extremes, please point it out; I've missed it.
> Hint: probability is the number of successes divided by total
> number of possible outcomes.
Damn! I wish I'd known that! Thank you *so* much for
setting me straight.
--
Eric Sosman
eso...@ieee-dot-org.invalid
Mensanator
As Sylvia pointed out, I missed that bit.
> Since the two P(*) add to 1.0 (because "mixed"
> and "unmixed" exhaust the entire universe of possibilities), and
> they are equal, P(*) = 0.5 is the only solution. What's nonsensical
> about that?
Upon re-thinking, never mind.
>
> >> If we view the dish as holding an arrangement of three candies,
> >> there are eight possible states corresponding to the eight ordered
> >> triples on the set {Ch,Ca}. Again, the mixed and unmixed states
> >> are equiprobable, so P({Ch,Ch,Ch})+P({Ca,Ca,Ca}) = 0.5. Again, we
> >> know the sum of the two terms but are unable to isolate the value
> >> of P({Ca,Ca,Ca}) from that sum, and again we conclude that "at least
> >> one chocolate" has a probability between 0.5 and 1.0.
>
> > Maybe *you* can't isolate it, but *I* can.
>
> How?
Upon re-thinking, maybe I can't.
> I offer you the following set of probabilities:
>
> P({Ch,Ch,Ch}) = 0.5
> P({Ca,Ca,Ca}) = 0.0
> P(each other) = 0.1666...
>
> In this scheme, the probability of "at least one chocolate" is
> unity. But I might instead have offered
>
> P{(Ch,Ch,Ch}) = 0.0
> P({Ca,Ca,Ca}) = 0.5
> P(each other) = 0.1666...
>
> ... and now the probability of "at least one chocolate" is one-half.
> I can set the first two probabilities to *any* pair of non-negative
> values that sum to 0.5 and meet the conditions of the puzzle.
What's the justification from the puzzle "as stated" for any such
choice?
>
> If you can find something *in the problem as stated* that
> forbids either of these arrangements, or any other arrangement
> in between these extremes, please point it out; I've missed it.
>
> > Hint: probability is the number of successes divided by total
> > number of possible outcomes.
>
> Damn! I wish I'd known that! Thank you *so* much for
> setting me straight.
Can you explain why 3/4 is the wrong answer?
Eric Sosman
> [...]
>> I offer you the following set of probabilities:
>>
>> P({Ch,Ch,Ch}) = 0.5
>> P({Ca,Ca,Ca}) = 0.0
>> P(each other) = 0.1666...
>
> What's the justification from the puzzle "as stated" for any such
> choice?
Merely that it's not forbidden -- or equivalently, that
we're not told anything that would rule out such a set of values.
(I have a sneaking suspicion that the puzzle may have been
abridged at some point, omitting information about probabilities
of individual kinds of candies. If such information were at hand,
the probabilities I offer might be ruled out.)
> Can you explain why 3/4 is the wrong answer?
Hmmm. For the same reason that "four" can be the wrong
answer to "pick a number between one and ten." As I see it,
the information provided shows only that the probability is
somewhere in [0.5 .. 1.0]. Any value in that range is possible,
so naming just one value is wrong with um, er, probability one.
--
Eric Sosman
eso...@ieee-dot-org.invalid
Sylvia Else
Even asserting that it's the right answer requires some assumptions
about the kind of process that lead to the evens chance of a
combination. I imagined a partion of the permutations into two groups,
one that containing combinations, and one that doesn't, and then a
random choice between them. That seems a fair approach, and it appears
to give the 3/4 result, but it does rather go beyond the facts provided
in the puzzle.
Sylvia.
James Dow Allen
> heich1 wrote:
> ... [snip]
> > She also knows that there is an equal chance they could be a
> > combination of the two candies.
To me, this sentence is *so* unclear that EITHER
(a) the puzzle is about the interpretation of such
ambiguous seeming statements, OR
(b) Mr. Stickels's puzzle is very defective, OR
(c) Mr. Heichelheim has typed the puzzle in incorrectly.
I know Terry Stickels has a very high reputation.
At risk of offending Mr. Heichelheim, my vote is (c).
> ... Mixed and unmixed states are equiprobable, so ...
> there we're stuck:
My question for Eric is:
Am I correct that you're being "tongue in cheek"?
You interpreted the ambiguous sentence in a way that
seemed best and then demonstrated the puzzle's
defect.
> (Mine's the one with the nougat center.)
I'll take the pecan center, please.
My own solution to this puzzle?
The probability is 5/8 that Peter Heichelheim has
typed it in incorrectly. :-)
James Dow Allen
Mark Brader
> Here is a puzzle given Terry Stickels in the Toronto Daily Star of
> Nov. 18th, 2009.
No it wasn't. That hasn't been the name of the paper for 38 years.
> She also knows that there is an equal chance they could be a
> combination of the two candies.
Equal to what? This statement is meaningless in the given context,
and so the puzzle is insoluble.
--
Mark Brader, Toronto "C and C++ are two different languages.
m...@vex.net That's UK policy..." -- Clive Feather
Mensanator
But isn't it standard practice in probability problems to consider
that all outcomes have equal probability unless explicitly stated
otherwise? And it was stated that the set containing two outcomes
had a total probability of 1/2. It was not stated in the puzzle
that P(Ca,Ca,Ca)>P(Ch,Ch,Ch) nor that P(Ca,Ca,Ca)<P(Ch,Ch,Ch).
Therefore, P(Ca,Ca,Ca)=P(Ch,Ch,Ch), so the probability of
P(Ca,Ca,Ca) is 1/4.
I still await a valid explanation of why the answer isn't 3/4.
>
> Sylvia.
Mensanator
> Mensanator wrote:
> > [...]
> >> I offer you the following set of probabilities:
>
> >> P({Ch,Ch,Ch}) = 0.5
> >> P({Ca,Ca,Ca}) = 0.0
> >> P(each other) = 0.1666...
Oh, by the way, that last one sums to 1. I trust you didn't mean that.
> >> [...]
>
> > What's the justification from the puzzle "as stated" for any such
> > choice?
>
> Merely that it's not forbidden -- or equivalently, that
> we're not told anything that would rule out such a set of values.
> (I have a sneaking suspicion that the puzzle may have been
> abridged at some point, omitting information about probabilities
> of individual kinds of candies. If such information were at hand,
> the probabilities I offer might be ruled out.)
>
> > Can you explain why 3/4 is the wrong answer?
>
> Hmmm. For the same reason that "four" can be the wrong
> answer to "pick a number between one and ten." As I see it,
> the information provided shows only that the probability is
> somewhere in [0.5 .. 1.0]. Any value in that range is possible,
> so naming just one value is wrong with um, er, probability one.
>
> --
> Eric Sosman
> esos...@ieee-dot-org.invalid
Sylvia Else
The fact that there's an evens chance of a combination already conflicts
with the idea that all outcomes have an equal probability. The
implication is the the outcome probabilities have been skewed in some
way. But we aren't given any idea of the mechanism.
Sylvia.
Mensanator
And when the outcomes are divided into sets, it then
extends to the elements within a set, again, unless
specifically stated otherwise. Unlike Eric's example,
the sums of all the sets still has to equal 1.
> The
> implication is the the outcome probabilities have been skewed in some
> way. But we aren't given any idea of the mechanism.
Sure we are. We are told the set of non-combinations
constitutes 50%. Unless explicitly stated otherwise,
we can assume each component of that set is 50%/n where
n is the number of elements within that set. For the set
of non-combinations that's 50%/2, so each element is 25%.
Likewise, the six elements of the set of combinations also
sums to 50% making each element 50%/6 or 8.333%.
Now, if that's not what was intended, then it needs to be
so stated.
>
> Sylvia.
Sylvia Else
>> The
>> implication is the the outcome probabilities have been skewed in some
>> way. But we aren't given any idea of the mechanism.
>
> Sure we are. We are told the set of non-combinations
> constitutes 50%. Unless explicitly stated otherwise,
> we can assume each component of that set is 50%/n where
> n is the number of elements within that set. For the set
> of non-combinations that's 50%/2, so each element is 25%.
>
> Likewise, the six elements of the set of combinations also
> sums to 50% making each element 50%/6 or 8.333%.
>
> Now, if that's not what was intended, then it needs to be
> so stated.
>
>> Sylvia.
>
Well, we'll have to differ on that.
However, note that the puzzle does NOT ask us to explain why the
probability is not 3/4. It TELLS us that it's not 3/4. That is, it's
providing an additional piece of information. If one could show that the
probability, without that constraint, had to be either 3/4 or X, then
with the constraint, one could show that it is X.
Sylvia.
Willem
) Mensanator wrote:
)> Therefore, P(Ca,Ca,Ca)=P(Ch,Ch,Ch), so the probability of
)> P(Ca,Ca,Ca) is 1/4.
)
) The fact that there's an evens chance of a combination already conflicts
) with the idea that all outcomes have an equal probability. The
) implication is the the outcome probabilities have been skewed in some
) way. But we aren't given any idea of the mechanism.
Perhaps we can get away with the (imo small) extra assumption that
the probabilities of the three candies are independent ?
I.E.: P(Ca,X,X) = P(X,Ca,X) = P(X,X,Ca) ( = P(Ca) ?)
And from there find the probabilities that satisfy
P(Ca,Ca,Ca)+P(Ch,Ch,Ch)=0.5
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
Nick Wedd
<m...@vex.net> writes
>Peter Heichelheim:
>> Here is a puzzle given Terry Stickels in the Toronto Daily Star of
>> Nov. 18th, 2009.
>
>No it wasn't. That hasn't been the name of the paper for 38 years.
>
>> She also knows that there is an equal chance they could be a
>> combination of the two candies.
>
>Equal to what? This statement is meaningless in the given context,
>and so the puzzle is insoluble.
But that's what the puzzle is about.
>> She also knows that there is an equal chance they could be a
>> combination of the two candies.
is obviously meaningless as it stands. We are required to rewrite it
into something which
1.) makes sense
2.) leads to a unique answer
3.) that answer is not 3/4.
I suggest "She knows that each candy is either a chocolate or a caramel,
with independent probabilities of 50%". Ok, it's quite a stretch, does
anyone have a better suggestion?
Nick
--
Nick Wedd ni...@maproom.co.uk
Eric Sosman
> On Nov 20, 8:49 pm, Eric Sosman <esos...@ieee-dot-org.invalid> wrote:
>> Mensanator wrote:
>>> [...]
>>>> I offer you the following set of probabilities:
>>>> P({Ch,Ch,Ch}) = 0.5
>>>> P({Ca,Ca,Ca}) = 0.0
>>>> P(each other) = 0.1666...
>
> Oh, by the way, that last one sums to 1. I trust you didn't mean that.
Your trust is well-placed: I didn't mean that, or I
only half-meant it. Thanks for the correction.
--
Eric Sosman
eso...@ieee-dot-org.invalid
The Qurqirish Dragon
> In message <P7adnVWBSOHN65rWnZ2dnUVZ_uydn...@vex.net>, Mark Brader
Here is my lunatic idea (lunatic because I cannot see why anyone would
have meant THIS to be the question, but it was a cute idea, IMO), but
at least it gives a different answer than 3/4 or 7/8 (I personally
believe 7/8 is the answer desired, for reasons others have stated).
Here it goes:
There is a 50% probability of a set of 3 like candies, and 50% of a
mixed set. Thus, we need to find the probability of each type.
Let x = P(one candy is chocolate), and so 1-x = P (one candy is
caramel)
Then, we have have these probabilities:
x^3 = P (3 chocolate)
3 * x^2 * (1-x) = P (2 chocolate, 1 caramel)
3 * x * (1-x)^2 = P(1 chocolate, 3 caramel)
(1-x)^3 = P(3 caramel)
using the given that 50% of the time they are all the same, we get
(after simplification):
6x^2 - 6x +1 = 0
As a quick check, you can figure this equation using the homogeneous
or mixed cases.
Solving this gives x = 0.5 +/- sqrt(3)/6
Using a 3-decimal roundoff, this gives x = 0.789 or x = 0.211
thus, x ^ 3 = 0.491 or 0.009 (and, symmetrically, (1-x)^3 = 0.009 or
0.491, respectively
and so the probability of at least one chocolate is either 0.991 or
0.509
Unfortunately, we cannot determine any further which is the correct
answer, since the symmetry in the problem will always give two
results. The only time it is not ambiguous is if x = 0.5 (which gives
the 7/8 answer), but then there is a 75% chance of a mixed set of
candies.
heich1
You are wrong. I did type in the sentence correctly. In fact this
sentence is the reason I put in this puzzle into
rec.puzzles. While Terry Stickels has a very high reputation in this
case he for some reason refused to understand
my reasoning (maybe he was too busy answering other people) and
insisted on his answer with an irrelevant explanation.
So below on my post after "The Quqirish Dragon".
Peter Heichelheim
Phil Carmody
More literally - there's a 50% chance that the sweets are mixed
combination of the two types, and a 50% chance that they're all
of one type. Which sounds like a very bizarre statistic to have.
Phil
--
Any true emperor never needs to wear clothes. -- Devany on r.a.s.f1
heich1
>
> - Show quoted text -
You got the same answer as I did so at least one person agrees with
me. Mensantor gave the
same answer as Terry Stickels got, i.e. 7/8. However there is nothing
in the text of the puzzle,
in the way I understand it, that states that the probability of the
Diana getting either of the two
candies is equal and all we are given is that probability of two
different candies in the dish is equal
to not having two candies in the dish.
By the way if anyone wishes to see Terry Stickels' emails to me please
send your email to
hei...@istar.ca. Maybe you can make better sense of them than I can.
If they know some way to
get these emails into this thread of rec.puzzles that would be even
better because I dont.
Peter Heichelheim
Mark Brader
>> Equal to what? This statement is meaningless in the given context,
>> and so the puzzle is insoluble.
Nick Wedd:
> But that's what the puzzle is about.
No, that wouldn't be Terry Stickels's style of puzzle.
--
Mark Brader "It's okay to have our own language if we feel
Toronto we need it, but why does it have to be used
m...@vex.net as a nose to look down?" -- Becky Slocombe
dgates
wrote:
>Here is a puzzle given Terry Stickels in the Toronto Daily Star of
>Nov. 18th, 2009. I and he have a difference in interpretation of this
>puzzle and hence its solution. Here it is.
>
> Diana walks by the candy dish in her home. She knows there are three
>pieces of candy left in the dish and that those candies are either
>chocolates or caramels, but she can't see them.
> She also knows that there is an equal chance they could be a
>combination of the two candies.
I don't like that last sentence. If that's literally the wording in
the original puzzle, then no wonder it causes confusion.
> Her brother walks by and she tells him she'll buy him a pizza if he
>can tell her what the probability is that at least
>one chocolate candy is in the dish.
> He answers "3/4"
> No pizza for Diana's brother.
> Can you tell Diana what the correct answer is?
If there are 3 candies, and each has an independent 50-50 chance of
being chocolate, then there's only one of the eight combinations that
has no chocoloate: Caramel-Caramel-Caramel. The other 7/8 combos
contain chocolate.
Gerry Quinn
@s31g2000yqs.googlegroups.com>, hei...@istar.ca says...
It sounds like a mistranslation of a simple puzzle to which the answer
would be the obvious 7/8.
If we do get hung up on the "equal chance of a combination" theory, it
seems to me that the 3/4 answer is defensible, i.e. there is a 50%
chance that both candies are present, and if there is only one type
present, there is a 50% chance that they are all chocolate.
On the other hand, in idiomatic English, a combination of the two
candies would be a chocolate-caramel candy, and thus neither a chocolat
nor a candy. On this basis the answer is 1/4, as there is a 50% chance
of chocolat-caramels, while if they are not chocolat-caramels, there is
a 50% chance they are chocolate.
Then again, the answer 1/2 can be defended by assuming there is a 50%
chance of a caramel-chocolate quantum superposition, which will
collapse to a chocolat or candy state when she reaches for it. Again,
if it is not in such a state, the chances of chocolate are 50%.
All this is so incredible that I am forced to conclude that Diana and
her candies do not exist, and thus the probability is 0.
But somewhere, in some universe, they must exist. Hence the
probability is 1.
Anyway, I'm pretty sure that even if my answers are wrong, it's
somewhere in the range I suggest.
- Gerry Quinn
Sylvia Else
> Then again, the answer 1/2 can be defended by assuming there is a 50%
> chance of a caramel-chocolate quantum superposition, which will
> collapse to a chocolat or candy state when she reaches for it. Again,
> if it is not in such a state, the chances of chocolate are 50%.
Well, unless the system is phase entangled with another somewhere in the
Universe where a selection has already (I use the word "already"
loosely) been made.
Sylvia.
Mark Tilford
Perhaps the brother reasoned as follows:
All chocolate, 2 chocolate, 1 chocolate, no chocolate.
In the first three cases, there is at least one chocolate; in the
fourth, there isn't. Therefore, he answered 3/4.
His error was assuming all four cases were equally probable.
Sylvia Else
No, his error was assuming that he'd been given enough information to
provide an answer at all.
Sylvia.