I have a bag that initially held 10 marbles.

In article <davefitz-0208940444410001@cts44.cc.utah.edu>,
davefitz@u.cc.utah.edu (Dave Fitzjarrell) writes:
> Wow!  I must take a stab at this.  First let's generalize:
>
> I will assume that there are C possible colors (one is black), all of
> equal probability of occuring.

That assumption does not generalize the problem at all; it results in
a very specific restatement of the problem.

I think the poster wanted to know the probability independent of any
additional assumptions.  In that case, I don't think it can be done.
You have to know something about the underlying population from which
the sample was drawn.  You'd suspect that after drawing nine straight
black marbles that the underlying population was entirely black, or at
least predominantly black, but without any additional information I
think it's impossible to calculate the probability.

Using Bayes' Theorem :

                                P(last black and first 9 black)
P(last black | first 9 black) = -------------------------------
                                        P(first 9 black)

Since we can't calculate the denominator without knowing something
about the distribution from which the sample was drawn, I don't think
the desired probability can be calculated (unless, like I say, you
make some additional assumptions).

-- Bob Marshall <marshall@lmsc.lockheed.com>

----------

Here is the way I resonate about the problem.  I do not state that
this is necessarily correct or that it is a solution, but it might set
you on the right track.

You have two possibilities:
1) All the ten marbles are black and that is why you pulled out 9
   black first.
2) The last marble is non-black and what has occured is an unlikely
   event.  But not very unlikely.  The probability is 1/10 that this
   would happen.  (One marble has to be the last to be pulled out and
   because there are ten of them the probability the the non-black one
   should be last is 1/10)

Consider you have a room full of bags, all containing ten marbles, P%
of them have 10 black marbles and (100-P)% of them has 9 black marbles
and one non-black marble.  You pick one bag and pull out 9 marbles at
random and discover 9 black marbles.

Let p = P/100 = probability of a bag containing ten black marbles.
The probability that you first picked a bag with ten black marbles is
p*1 and the probability that you picked a bag with ten black marbles
and one non-black marble and then had the "luck" to pick out the nine
black marbles first is (1-p)*1/10.  The probability B that the last
marble is black is equal to the probability that the first of the two
alternatives occured.  This probability is
B = p/(p+(1-p)*1/10) = 10p/(9p+1) = 10/(9+1/p).  If p = 1 we get B = 1
which seems reasonable and as p decreases towards zero B also
decreases towards zero wich shows that you can get any probability for
the last marble being black by choosing different values for p, so as
we know nothing of this p we also know nothing of B.

[ Editor: "a bag with ten black marbles and one non-black marble"
evidently means the bag with nine black marbles and one non-black. ]

-- Andreas Stabel <anst@vin.dolphin.no>

----------

In article <rloCtx5CE.Ltp@netcom.com> rlo@netcom.com (Robert
Orenstein) writes:
>Great puzzle!
>
>I don't think that this question is in any way answerable without
>more information.

The only other information you need, as other postings have pointed
out, is the number of possible colors for the balls.  Tack on an
assumption about total randomness and eliminate things like
multicolored balls and manufacturing tendencies, and the puzzle has a
solution.  The solution is not very exciting or counter-intuitive, but
it is a solution.

Mathematical derivations have shown that, given balls of C different
colors, the probability that the last ball is black is 1/C.  That is,
you cannot say that the last ball is more likely to be black just
because choosing the N-1 black balls from a bag of N balls with only 1
non-black ball is unlikely.

I think a mathematical derivation, though it does provide a solid
ground for an answer, is a bit much here.  Restating the problem makes
the answer more obvious:

"Given that N-1 of N balls are black, what is the probability that the
Nth ball is also black?"

It doesn't matter how they were chosen in the first place...once we
select N-1 black balls in a row, all we know is that N-1 of the N
balls are black.  That's it.  It doesn't matter how unlikely it is
that we could choose N-1 black balls without selecting the non-black
ball.  With C different colors, there are C combinations with N-1
black balls, only one of which has all black balls.  Thus, if we know
that N-1 of the balls are black, then the probability that the last
one is black is 1/C, as the equations showed.

This argument works because, given total randomness, the color of one
ball is independent of the color of any other ball.  Therefore, the
color of the Nth ball in no way depends on the colors of the other
N-1, and we're back to the probability of one ball being black, given
C colors.

-- Michael Wolfe <wolfe@ils.nwu.edu>

----------

In article <31npng$1s3a@unixfe.rl.ac.uk> hm@rlssp1..bnsc.rl.ac.uk (Huw
Morris) writes:
>As another poster pointed out, I think all you can do is to start
>with absolutely no assumptions about the bag, ie all possibilities
>have equal probabilities.

And how many possibilities are there?  Doesn't this depend on how many
distinct colors are possible?

>As you take a marble out of the bag, you can look at it and refine
>the probabilities.  The first marble is black.  This means that the
>probability that all the marbles are red is 0.  Also, the probability
>that 9 marbles are red and 1 black has decreased, while the
>probability that 1 marble is red and 9 black has incresed.

Two points:

The probabilities "increase" or "decrease" only if they are not zero
to begin with, i.e., you have assumed that red marbles exist and might
have gotten mixed with black marbles in the bag.

Under reasonable assumptions about how the bag of marbles was
assembled, i.e., ten marbles were drawn at random from a large supply
of marbles of many colors, then after drawing one black marble the
chance of (9 red, 1 black) has *increased*, not decreased as you
claimed, simply because all the (none black) combinations have been
eliminated.

>This is a nice puzzle.  I guess that the probability that the last
>marble is black is pretty high, but whether it can actually be
>calculated without extra assumptions is beyond me.

You have to make extra assumptions.  If your assumptions are that the
marbles in the bag were initially selected at random from a very large
supply of marbles somewhere (e.g., all the marbles in the world), then
there's a very simple way to see that the odds of the tenth being
black are exactly the proportion of black marbles in the large supply
(a fact that several people have gone to great lengths of calculation
to show):

Drawing ten marbles at random from a large supply and putting them in
a bag, then drawing them out of the bag again at random, gives you
exactly the same probability distribution for the colors of the first
through tenth marbles as if you had simply drawn them from the large
supply in the first place without using a bag.  The whole business of
putting the marbles in the bag and shaking them up is just stirring a
subset of the original supply a bit more.  All possible outcomes
(assigning a unique ID to each marble in the large supply) are still
equally likely.

-- David A. Karr <karr@cs.cornell.edu>

----------

In article <1994Aug3.022353.15581@cs.cornell.edu>, karr@cs.cornell.edu
(David Karr) writes:
> If you make fewer assumptions, you could treat this as a problem in
> statistics rather than probability, and try to guess the color of the
> next marble and compute the amount of "confidence" you have in your
> answer.

Right.  Let's state that the bag of marbles was drawn randomly from
some large population of marbles, a proportion p of which are black
(we don't know p, of course).  We have a sample of size 9 from this
population, all of which are black.  Therefore, our point estimate of
p is 1.0.  And, as David mentions here, we can calculate a confidence
interval for our estimate.  In this case, invoking the usual
assumptions about random, independent drawings, "large" populations,
etc, etc, we can state with 90% confidence that p is at least .774
(this one is easy to calculate; it's just (0.1)**(1/9)).  So, while we
cannot compute the probability that the tenth marble is black, we
*can* state with 90% confidence that there is at least a 77.4% chance
that it's black (I'm taking liberties with the wording here; you
statisticians out there, don't hang me!).

-- Bob Marshall <marshall@lmsc.lockheed.com>

----------

OK, let me try this again.  I posted earlier with a calculation of a
confidence limit, but I don't think I did it quite right.

Assume that the bag of ten marbles was randomly drawn from a large
population of marbles, a proportion p of which are black.  We don't
have any idea what the value of p is, of course.  Then we draw nine
marbles randomly from this sample of ten marbles.  The first nine are
black.  The probability of such an occurrence is :

P(9 straight blacks) = P(all were black)P(9 straight blacks | all were black)
                     + P(one non-black)P(9 straight blacks | one non-black)

                     = (p^10 * 1.0) + [10*p^9*(1-p)]*0.9

                     = p^10 + 9*p^9 - 9*p^10 = 9*p^9 - 8*p^10

Now, to compute a 90% lower confidence limit on p we need to find a
value of p such that P = .10.  This turns out to be about .671.  So,
we can state with 90% confidence that the original population was at
least 67.1% black.

[ Editor: with one non-black marble the first nine drawn are all black
only if it is drawn last, probability 1/10 (as Andreas Stabel says
above), not 0.9; with 1/10 the sum is p^10 + p^9*(1-p) = p^9, the
expression of the earlier posting. ]

-- Bob Marshall <marshall@lmsc.lockheed.com>

----------

Firstly, thank you all of you who responded to this question.
Although your analyses are mostly mathematically correct, you still
have not convinced me (don't go jump in the lake) BECAUSE your
fundamental assumptions are different.

Let me explain.

Most of you insist that there has to be a underlying distribution,
which, simply put, is the probability that ANY marble will be black in
its natural state.  You assume this probability is p, and so the
probability of it being non-black, 1-p.

Some of you assume that there are c possible colours that marbles can
have, and since black is one of them, the probability that ANY marble
is black is 1/c.  NOTE: This is an unfounded assumption!

Some of you say (rightly) that there are c^10 possible combinations,
but you also assume that the chance of getting each combination is
1/c^10.  NOTE: Again, this is an unfounded assumption!

So we know nothing of external distributions or natural tendancies.
Throw away all that.  Assumptions that whoever put the marbles in the
bag would probably have put 10 black marbles or any other combination
do not help.  This is an isolated experiment independent of
distribution.  p is unknown.  (This problem, I suspect, is not purely
mathematical, but involves the perpetual protest that inductionists
(philosophers) hold against all who practise science and mathematics
using the scientific method as they define it.)

What is important, however, is that we took RANDOM selections from a
FINITE group of items.  The items can be anything (no guarantee this
time that they're all marbles); each and every item can be absolutely
anything in the world.  All that we're told is that they are randomly
selected, so that each item still remaining in the group has a EQUAL
chance of being selected the next time you pick.  Now, given that nine
are exactly the same, what is the chance that the last too is the
same?

As some of you say, that chance is anything from 0% - 100%, depending
on your initial/external distributions/assumptions.  But here there
ARE NO external distributions.  So given only the limited info, what
is the best estimated probability?  I cannot give an answer, but I
still say that, logically, it is at least more likely than not that
the last item is the same.  This probability will INCREASE the larger
the SIZE of the group, eg. if having picked 99 out of 100 items, and
finding that they're ALL, say, baseballs bats.  (Or 999 out of 1000,
or n-1 out of n, where n is very large, in which case the probability,
I venture, is close to 1.)  Therein lies a possible answer to the
critique of induction.

Yet if the first 9 out of 10 were all DIFFERENT, then there is
absolutely no way of knowing what no. 10 would be!  So why does the
fact that the first 9 are the same make a difference?  Take, for
example, stars.  If there are n stars in the known universe and we
have managed to observe haphazzardly about m stars, where m is very
close to n and n is very large, and all of the m stars we observed
have characteristic @, then astronomers or astrophysicists will say
that most likely all stars have characteristic @.  (NOTE: This is a
general statement.)  Or they might say most likely the NEXT star we
observe will have characteristic @.  (NOTE: This is a particular
prediction.)  That is the natural, or intuitive impulse we all have as
inquiring minds in an environment.  If something keeps happening, we
all assume it'll will continue to happen next time, especially if (a)
it has happened for MANY MANY times and (b) we know of no external
information that would indicate otherwise (as would be the case with
lotteries, which we know must be independent events.)

While we're at it, I would like opinion on induction.  The theory of
induction says that induction is completely unfounded, illogical and
unjustified.

Eg. since Einstein, thousands of experiments have confirmed that
e=mc2.  But how do we know the next experiment will do so?  In other
words, how to we formulate a general, predictive rule on nature based
on a FINITE set of experiments?  Since the number of occurrences to
which e=mc2 is potentially applicable in nature is INFINITE (or very
nearly so), what information does doing a FINITE number of experiments
give us, even if the experiments SO FAR are consistent with e=mc2?
How are we justified in formulating a UNIVERSAL rule from INDEPENDENT,
INDIVIDUAL events?  (Doing so is the procedure called induction, on
which, it is claimed, most of science is based.)

That is the problem, and the critique of induction.

(If I am wrong, any philosophers out there, please correct me, but
this is the general idea I get about induction.)

Any answers or rebuttals to that from scientists, engineers or
mathematicians out there?

(Sorry if this bores anyone.  I would just like to share my thoughts
with someone, and I find this a fascinating subject.)

-- KOH KEW SOON ("Baboo") <law20065@leonis.nus.sg>

----------

In article <31pq5o$pa9@nuscc.nus.sg>, law20065@leonis.nus.sg (KOH KEW
SOON) writes:

[an interesting musing on induction deleted]

Supposing we have a theory which we wish to substantiate via
induction; for the purposes of example, how about the theory that "all
crows are black".  Obviously, if we see a non-black crow, we have
disproven the theory.  Now, the interesting question Baboo is asking
is,

"Does seeing a black crow increase the probability that all crows are
black?"

I don't have an answer to this, although the human tendency is, is
think, to believe that it does increase the probability.  But the
nifty fact is, that if you do believe in the inductive increase in
probability, then you also must believe that "seeing a white moose
increases the probability that all crows are black," and that "seeing
a brown coffee-cup increases the probability that all crows are
black," and more generally that "seeing anything that's both *not*
black and *not* a crow increases the probability that all crows are
black."  This is because the statement "all crows are black" is
equivalent to "no non-black things are crows."

Instinctively, induction makes sense to me, but since it seems to me
that seeing a white moose shouldn't have anything to do with the truth
or falsity of the assertion that all crows are black, I guess I don't
really trust in its logical validity all that much.

-- JoXn S Costello <costellj@carleton.edu>

----------

Great discussion on this one...  Let me add a few more comments
pushing a Bayesian rather than frequentist interpretation of
probability for this problem: (I'll delete a number of thoughtful
comments to emphasize my points, which may mangle the logical flow of
previous arguments.  Sorry...)

David Karr (karr@cs.cornell.edu) wrote:
: In article <31q3eo$sr2@nuscc.nus.sg> law20065@leonis.nus.sg (KOH KEW
: SOON) writes:
: >If we assume that since no information is given, we'll take (at out
: >best estimate), the ratio of P(10B) to  P(9B) be 1 : 1.

: There's absolutely no basis for such an assumption.

: Yes, it's common to give even odds in probability when there are two
: events, but in those cases there's some obvious symmetry between the
: events that makes it "natural"---e.g. the toss of a fair coin, or an
: adversary who has hidded an egg under one of two cups.  There's no
: such symmetry here.

I agree that frequentist probalities in this context have little
meaning.  The Bayesian approach is that we don't know the distribution
of marbles in the bag before we start drawing them (uninformative
priors).  They might be all different colors, they might be all the
same, they might be nine of one color and one of another, etc.
Conventionally, when there is no information about the prior
probabilities, they are ASSUMED to be the same.  But that is just a
convention of convenience.  You can make any other ASSUMPTIONS about
the initial probabilities that you see fit.

Each marble drawn serve as a "test" which can be used to convert those
prior probabilities into "posterior" probabilities of the likelihood
of various initial distributions.  After the first marble is drawn,
Some possibilities will be rejected (e.g., that there were no black
marbles), some will have a lower posterior probability than prior
probability (e.g., that there is only black marble in the bag), and
some will have a higher probability (e.g., that all of the marbles are
black).

These posterior probabilities from the first test are then used as the
prior probabilities for the second test, i.e., drawing the second
marble.

After drawing 9 black marbles, all but two of the initial
probabilities have been rejected: that all the marbles are black, or
that 9 are black and one is some other color.  The probability (in the
Bayesian sense) is much higher that all are black (otherwise it is
likely that the one of the other color would have been drawn by now.
[Unless of course the initial prior probability of nine black marbles
and one of another color was ASSUMED to be much higher than the
probability of all ten being black.]  So, I'd bet the tenth will be
black.

: >Then, ASSUMING that chance of every combination is the same as every
: >other,

: Again, this is an unfounded assumption.  Much worse than that,
: there's a much more plausible assumption generally used in
: probabilistic analysis of such a situation: namely, that every
: *permutation* is equally likely.  This is in fact a good
: approximation of what happens if you take a million marbles of each
: color, stir them up, and shake ten at random into your bag.  (In
: contrast, you'd have to go to quite a bit more trouble to make all
: combinations equiprobable.)

See above thoughts on necessary ASSUMPTIONS.

: > Of course, to clarify, the probability I am talking about is
: >mathematical probability (as if there were any other type).

: Usually when people draw statistical inferences on the basis of some
: number of observations, what they get is not a mathematical
: probability at all, but rather some statistical "level of
: confidence" that a certain fact is true.  Level of confidence is
: defined in part using probabilities, but it is not a probability that
: the fact in question is true.

What you are describing are the Bayesian approach to confidence
intervals, known as fiducial confidence intervals.  They were
originally developed by R.A. Fisher himself.  That is not the
frequentist interpretation of confidence intervals, which is defined
strictly in terms of repeated sampling from an underlying population.
It has nothing to do with "confidence" but it strictly a measure of
statistical precision.

Let me repeat my disclaimer that I'm not a statistician, so take with
a grain of salt.  I'm also not a die-hard Bayesian; this just seems
like a situation where that approach seems more fruitful than the
traditional frequentist methods.  If there are any facile Bayesians
out there, please chime in!

-- Peter M. Layde <playde@post.its.mcw.edu>

----------

In article <31rtlo$i8q@post.its.mcw.edu> playde@post.its.mcw.edu
(Peter M. Layde) writes:
>Great discussion on this one...  Let me add a few more comments
>pushing a Bayesian rather than frequentist interpretation of
>probability for this problem:

Let me just comment that almost all the analyses (including one of
mine) *have* been Bayesian rather than frequentist, and we all come up
with it being *unlikely* that the next marble is black.

>I agree that frequentist probalities in this context have little
>meaning.  The Bayesian approach is that we don't know the
>distribution of marbles in the bag before we start drawing them
>(uninformative priors).  They might be all different colors, they
>might be all the same, they might be nine of one color and one of
>another, etc.  Conventionally, when there is no information about the
>prior probabilities, they are ASSUMED to be the same.

Fine.  Assume this.  Most of us have.  Now please enumerate all the
possibilities---not necessarily one by one, but in enough detail that
you can tell me, according to your "uninformative" prior, what is the
exact probability that the marbles in the bag are all black?  Until
you do this, I don't believe you're making a Bayesian analysis.

Now what's the probability of 9 black and 1 white?  Remember, if there
is just one white marble in the bag, then (laying the bag on a table
and patting it down so I can feel where each marble is) it might be
the marble *here* or the marble *here* or here or here or here or here
or here or here or here or here---that's ten possiblities.

Now, the probability of 9 black and one red?  One yellow?  One orange?
How many colors of marbles are there, anyway?

>After drawing 9 black marbles, all but two of the initial
>probabilities have been rejected: that all the marbles are black, or
>that 9 are black and one is some other color.  The probability (in
>the Bayesian sense) is much higher that all are black (otherwise it
>is likely that the one of the other color would have been drawn by
>now.  [Unless of course the initial prior probability of nine black
>marbles and one of another color was ASSUMED to be much higher than
>the probability of all ten being black.]

Of course the latter assumption is the only honest one---the contrary
assumption is essentially saying that we already know that marbles in
bags are far more likely to be black than to be any other color, or
that there is some very *unequal* process assembling the bag.

In other words, the "Bayesian" argument in favor of high likelihood of
black has three steps:

(1) Divide all marbles into "black" and "non-black", and give each
    set equal weight in further probability estimates.
(2) Consider only the two cases (10 black) and (9 black, 1 not) as
    worth assigning probabilities to.  Moreover, ignore the fact that
    the non-black marble might be in 10 different places in the bag.
(3) Assign P(10B) = P(9B) in the prior, and then *do* consider the
    9 separate cases where the non-black ball is drawn early.

I put "Bayesian" in quotes because I don't believe the above is a
Bayesian analysis at all.  It's a highly contrived way of raising the
probability of a tenth black marble in the posterior distribution.

-- David A. Karr <karr@cs.cornell.edu>

----------

A week or two ago I posted this puzzle here.  I think some of you may
have missed the solution I posted so here it is again:

To reiterate the problem in terms as clear as possible:

Given:

There are 10 items in a bag.
I take out one at a time randomly.
The chance of my taking out any one item is always = 1/n, where n is
the number of items still remaining in the bag.
I have already taken out 9 items, leaving 1 in the bag.  All 9 were
identical and they were black marbles.

Question:

At this stage, what is my best possible estimate of the chance that
the last item will also be a black marble?

Possible Solution
-----------------

Define E as the event that I took out 9 black marbles in a row.
Let P(10B) be the "initial" or "seed" probability that the bag
initially contained 10 black marbles and nothing else.
Let P(9B) be the "initial" or "seed" probability that the bag
initially contained 9 black marbles and one other item that is NOT a
black marble.
Let P(B) be the probability that the last item is also a black marble.

Then P(E) = P(10B) x 1 + P(9B) x 1/10.

Now, there are two ways to look at it.

First we may assume that since no information is given, we'll take
(at out best estimate) the ratio of P(10B) : P(9B) to be 1 : 1.

If so, we will have:

P(B) = P(10B) / P(E)
     = ...
     = 10/11.

However, if we assume that there are only c different types of items
that could be possibly be inside the bag, then:

The total number of different combinations of 10 items in the bag

= c^10/10!

Among these, the number of combinations of 10 black marbles

= 1

and the number of combinations of 9 [Editor: "10" in the original]
black marbles + 1 anything

= c-1.

Then, ASSUMING that the initial chance of having every combination is
the same as every other,

P(10B) = 10!/c^10 ;    and
P(9B) = 10!/(c-1)c^10.

In which case,

P(B) = P(10B) / P(E)
     = ...
     = 10/(c+9).

Check for yourself.

[ Editor: two slips.  The number of unordered combinations of 10 items
of c types is C(c+9,10), not c^10/10! (for c = 2 it is 11, while
2^10/10! is less than 1).  And the answer 10/(c+9) follows from
P(9B) = (c-1) x P(10B), i.e. (c-1)10!/c^10, not 10!/(c-1)c^10. ]

This would be a good answer that induction DOES work for finite groups
of items each with a finite number of possible states.  This is so
since 10/(c+9) > 1/c.

The induction I am talking about here is not mathematical induction
but what philosophers call reasoning by induction as opposed to
reasoning by deduction.  They claim there is no logical basis for
induction.  Of course, to clarify, the probability I am talking about
is simply mathematical probability (as if there were any other type
:)).

..........

Later, I posted this article to qualify my solutions:

..........

Firstly, I'am glad my deceptively simple marble problem has elicited
responses and contributions from so many people.  (I posted this
problem initially in both rec.puzzles and rec.org.mensa, though
rec.puzzles had more responses.)

Thank you all:

Dave Budd
David Wagner (who wrote twice)
HuaiLuu
Bob Marshall
Dave Fitzjarrell
Andreas
Thomas Hin Shun Tsang
Karl H
Seth
Harri
...

(Anyone I missed out, please forgive me)

Needless to say, almost every opinion was different from every other.
(If you wish, I could post them all, just tell me and I'll do it.)

I hope you've read the article Probability Problem (SOLUTION at last?)
in which I detailed two possible solutions, resulting from two
different assumptions.  Now there may be a need to add a third one,
which was the answer posted by Dave Fitzjarrell (which I will append
below).

All of you who said we need more info are correct, in a sense.  If we
knew the initial distribution, or set, from which the sample bag was
drawn, or from which the sample marbles were drawn, we would have an
ACCURATE answer.  However, given this problem, from a scientist's
point of view, one has to do with whatever info one has.  The essence
of giving an answer in terms of a probability is that one can always
give a best possible estimate (I mean this in the lay sense, not as a
statistical term :-)) no matter how little info one is given.  If
necessary, assumptions have to be made.

To clarify, it is given (not assumed) that of any n items remaining in
the bag, the chance of drawing out any of them is always 1/n.  (So
forget whether this is founded or unfounded.)

Now, it seems we can all agree that

P(event) = P(10B) + 1/10  x P(9B1NB)

and

P(B) = P(10B)/P(event),

where

P(event) = probability that the event defined as my drawing out 9
black marbles in a row (which did happen) could have happened, given
what we know now.  In other word conditional probability is involved.
P(10B) = probability that the bag initially contained 10 black marbles
P(9B1NB) = probability that the bag initially contained 9 black
marbles and one something that is NOT a black marble.
P(B) = probability that the last marble we draw is black.

* Notice that P(B) is not necessarily equal to P(10B)

Now where do we differ?  I submit it is in the assumptions as to the
likelihood the initial contents of the bag being 10B as opposed to the
likelihood that it contained 9B1NB.  This depends of course on what
kind of distribution or set or group the bag was drawn from,
hypothetically.  This requires external info.  But given no such
info, we can only guess which distributions are most likely.

The first answer I gave is based on the assumption that, knowing
nothing else, we take the chance of the bag being 10B to be equal to
that of the bag being 9B1NB.  Assuming this, then P(B) = 10/11.  In
general, this is independent of the number of possible different TYPES
of items that could be inside the bag (let's call it c).  Also, in
general, for a bag of n items out of which we have drawn out (n-1)
which are the same or which all possess a certain characteristic @,
P(B) = n/n+1.  So the larger the value of n, the greater our
confidence that next item will be the same.

The second answer I gave is based on the following assumption:

If there are c possible types of items, then there are c^10/10!
possible different COMBINATIONS of items.  This says nothing about the
natural tendancies of these combinations occuring, so we shall take
those tendancies to be the same.  In other words, the probability of
any one UNIQUE combination appearing would be 10!/c^10.  Notice that
this approach does not emphasize any sequence in which the items are
placed into the bag, hence the division by 10!.  The end result, as I
have shown, is that P(B) = 10/(c+9).  In general, P(B) = n/(c+n-1).
Hence the larger the value of n, the higher is P(B); and the smaller
the value of c, the higher is P(B).  This makes sense (ie. it is
intuitively "correct").

Now for approach number 3, which depends on the assumption that
somehow the bags were filled in a particular sequence.  This focuses
on the individual probability of each item appearing in nature, as
opposed to the probability of each different combination of items
appearing.  This makes a whole world of difference.  By assuming each
item has a 1/c chance of appearing in nature, it follows that some
combinations are more likely than others!

I shall now bring in Dave's reply (abbreviated)

.........

I will assume that there are C possible colors (one is black), all of
equal probability of occuring.  Let's say there are N marbles in the
bag, and we have taken out (N-1) of them and they were all black.  Let
us imagine that we have picked our bag of marbles randomly from a
barrel in which are stored all possible bags of N marbles.

P(event) = P(N) + 1/N * P(N-1)  where:

P(N) = the chance of getting a bag with all black marbles.
P(N-1)  = the chance of getting a bag with all marbles black except
one.

So, let's now evaluate the terms in the above formula.

P(N):
-----
The chance of any one marble being black is, of course, 1/C.  (If we
have 10 colors, chance of one marble being black is 1/10).  So, the
chance of two random marbles being black is 1/C * 1/C = 1/(C^2).  One
can see that the chance of N marbles being all black is 1/(C^N).

So, P(N) = 1/(C^N)

P(N-1):
------
With C colors, N marbles, there are C^N possible bags.  [If we have 2
marbles per bag, three possible colors (w,b,g), you have 3^2 = 9
possible bags - (w,w), (w,b), (w,g), (b,w), (b,b), (b,g), (g,w),
(g,b), (g,g).]

Now, how many of these C^N bags have (N-1) black marbles?  The easiest
way to see this is to ask instead how many bags have only one marble
not black.  Since for each of the N marbles per bag, there are C - 1
ways to not be black, there should be (C-1) * N possible bags that
have exactly (N-1) black marbles.  So, the probablilty of one of these
bags being chosen is (number of these bags) / (total number of bags)
= [(C-1) * N] / C^N.

So, P(N-1) = [(C-1) * N] / C^N [Editor: "C^" in the original]

... the probability of the remaining marble being black [P(B)] is
P(N)/P(event).

Our solution, then, is:

P(B) = P(N) / [P(N) + 1/N * P(N-1)]

Where:
P(N) = 1/(C^N)
P(N-1) = [(C-1) * N] / C^N

P(B) = (1/C^N) / [1/C^N + N(C-1)/NC^N] = 1 / [1 + N(C-1)/N] = 1/C

Which shows that the color of the final marble is completely
independent of the colors of the first nine.  (Which makes sense.)

..........

That was Dave's reply.  Now you see that using this approach, the
probability of the last item being anything is unaffected by my having
drawn out 9 black marbles in a row.  It has always been, and will
always remain 1/c.

If you asked me which solution to follow, I have no answer.
Intuitively, though I prefer the first or second one.

Any comments?

-- KOH KEW SOON ("Baboo") <law20065@cobra.nus.sg>, quoting Dave
   Fitzjarrell <davefitz@u.cc.utah.edu>

----------

[ Editor: a postscript to the thread, under the subject "Black Marble
Bags and Bar Bets": ]

I have no idea what this implies, and leave it to anyone who wants to
to debate probabilities, etc. without debating it myself...

After reading the threads about the black marbles, I realized that
there was a neat bar bet here.  So I got a bag and put in nine black
marbles and one green marble (green was the only color I could find
that was exactly the same size and texture as the black marbles).

For a week, I played the following trick on my friends, neighbors, and
co-workers.  I gave each of them the bag and asked them to pull out
one marble, then another, then another etc. until there was only one
left.  If at any point during the trick they pulled out the green
marble, I called the trick off and told them what was going on, asking
them not to tell anyone else what I was doing.

If the first nine marbles they pulled out of the bag were black, I
asked them how certain they were that the next marble was black.  Out
of 15 people that I did this to, three of them (!) got to this point.
Interestingly, although they all said something like "99% sure",
no-one would bet me (of course, why would I be betting if I knew the
marble was black?)

So I changed the trick a bit.  This time, I gave the trickee a 3x5
card with instructions not to read it until they had pulled nine
marbles out of the bag.  The card read "I will bet you $10.00 that the
last marble is not black."  If they drew the green marble in the first
nine (as of course most did), I showed them the card and told them
what was going on.  Of the two people who DID draw nine black marbles
first, each bet the $10.00, each lost.

I'm not drawing any conclusions, except to say that in THIS instance,
of course, the chances of them drawing a black marble at the end was
precisely 0.

-- Robert Orenstein <rlo@netcom.com>