[Guest]

A bag contains 4 balls.

Two balls are drawn at random and are found to be white.

What is the probability that all balls are white?

[superprismatic]

I can accept the answer being 0.6 as long as there is a good reason, like yours.

I don't appreciate explanations that are both inconsistent and condescending though.

Really this argument shouldn't be a hot point of debate after this long.

I realize that much of this dragging along is my fault for so eagerly debating.

I say everyone should walk away with whatever conclusion they like best.

Sorry if I sounded condescending. I really only meant to point out to marsupialsoup that his analysis was fine except for the fact that the three cases may not be of equal probability. I don't think this is a hot point. There is good reason to disagree with the 0.6 answer. Pointing out where the problem lies is good, though.

[Guest]

Sorry if I sounded condescending. I really only meant to point out to marsupialsoup that his analysis was fine except for the fact that the three cases may not be of equal probability. I don't think this is a hot point. There is good reason to disagree with the 0.6 answer. Pointing out where the problem lies is good, though.

No you did not sound condescending at all.

I was not talking about you at all.

Sorry for the misunderstanding.

I just felt that I might have overreacted to prior posts and I was babbling about why I was a little bit annoyed.

I realize it is silly of me to get so bothered by a simple puzzle.

My apologies for making it seem like I was accusing you.

I do not mean to come off so combative.

Edited March 28, 2010 by mmiguel1

[Guest]

Hi,

Firstly, apologies if it was me you were referring to for saying I was condescending - that was not the intention at all.

Also, if my comment about the assumption of infinite balls, etc. has confused matters, again I apologise. I did not intend this to mean that we can now calculate the probability of the bag initially having 4 white balls, I simply meant to illustrate that we know nothing about how the bag was initially made up and, for the purposes of answering the question, that is irrelevant.

In order to try to put this to rest:

The approach where people get to a "not answerable" solution is one where you want to first understand the probability the bag initially contained 2, 3 or 4 white balls and then understand how that probability changes based on an observation.

The approach that leads to an answer of 0.6 simply states that we know nothing about the initial probabilities. The OP starts with "A bag contains 4 balls." We have no idea what those 4 balls are. We have no understanding of how those balls may have been selected. But we do know it is a definitive bag - the balls have already been selected - they are fixed. Probabilities do not enter into it at this point. We are given a fixed bag with a set of 4 balls in it - that is all we know. We then pick out 2 balls and see they are both white. Now we can deduce that the bag either had 2, 3 or 4 white balls in it to start with. Again we do not know which, and we can not say what the probability of each was (it may be that each time this question is asked of anyone, they are always given a bag with 4 white balls - we simply do not know.) However, based on what we saw, we can determine what the likelihood of our result was in each of those scenarios. From that, we can then determine which scenario we are more likely to be in and what the probability of it is.

Note that the answer doesn't imply anything about the initial selection of the balls. It simply determines the probability that we are in 1 of 3 possible scenarios based on what we have observed.

This is a bit similar to the bonus part of the Coin Probability Puzzle that has resurfaced. In the bonus part of that puzzle, you are given no information as to whether your coin is fair or not, or what the probability is that you have a fair coin. You are simply told you have a coin, flip it 20 times and it keeps coming up heads. In that case, it is quite simple to observe that it is more likely that you have an unfair coin than not. You could (using a similar approach to I have used in this question) work out the exact probability, but that question doesn't ask for that as the answer is relatively obvious (although still caused some debate!) It is exactly the same case here - you know nothing about the initial selection of the bag but, based solely on your observations, you can determine which situation you are more likely to be in.

[Guest]

Hi,

Firstly, apologies if it was me you were referring to for saying I was condescending - that was not the intention at all.

Also, if my comment about the assumption of infinite balls, etc. has confused matters, again I apologise. I did not intend this to mean that we can now calculate the probability of the bag initially having 4 white balls, I simply meant to illustrate that we know nothing about how the bag was initially made up and, for the purposes of answering the question, that is irrelevant.

In order to try to put this to rest:

The approach where people get to a "not answerable" solution is one where you want to first understand the probability the bag initially contained 2, 3 or 4 white balls and then understand how that probability changes based on an observation.

The approach that leads to an answer of 0.6 simply states that we know nothing about the initial probabilities. The OP starts with "A bag contains 4 balls." We have no idea what those 4 balls are. We have no understanding of how those balls may have been selected. But we do know it is a definitive bag - the balls have already been selected - they are fixed. Probabilities do not enter into it at this point. We are given a fixed bag with a set of 4 balls in it - that is all we know. We then pick out 2 balls and see they are both white. Now we can deduce that the bag either had 2, 3 or 4 white balls in it to start with. Again we do not know which, and we can not say what the probability of each was (it may be that each time this question is asked of anyone, they are always given a bag with 4 white balls - we simply do not know.) However, based on what we saw, we can determine what the likelihood of our result was in each of those scenarios. From that, we can then determine which scenario we are more likely to be in and what the probability of it is.

Note that the answer doesn't imply anything about the initial selection of the balls. It simply determines the probability that we are in 1 of 3 possible scenarios based on what we have observed.

This is a bit similar to the bonus part of the Coin Probability Puzzle that has resurfaced. In the bonus part of that puzzle, you are given no information as to whether your coin is fair or not, or what the probability is that you have a fair coin. You are simply told you have a coin, flip it 20 times and it keeps coming up heads. In that case, it is quite simple to observe that it is more likely that you have an unfair coin than not. You could (using a similar approach to I have used in this question) work out the exact probability, but that question doesn't ask for that as the answer is relatively obvious (although still caused some debate!) It is exactly the same case here - you know nothing about the initial selection of the bag but, based solely on your observations, you can determine which situation you are more likely to be in.

Sorry if I overreacted to perceived condescendence (if that is a word).

On theoretical grounds I disagree with what you said in your last post.

For one, I argue that it is impossible to calculate an answer to this problem without knowing the probability of the bag initially having 4 white balls.

You may not claim you have made any assumption about the initial probability, but from the pure mathematics, you have in fact made such an assumption or else your answer is not defined.

In short, it is extremely relevant what the initial probability of having 4 white balls in the bag is.

I will write equations and sentences to demonstrate this.

In general, for events A and B:

P[A and B] = P[A|B] P = P[b|A] P[A]

And is the set intersection operator and the pipe "|" is the conditional probability symbol for given.

Let's take the two left parts of the mathematical statement above and divide both sides by P.

P[A|B] = P[b|A] P[A] / P

Let A = having 4 white balls initially in the bag

B = drawing 2 white balls from the bag

Obviously P[b|A] = 1, because if you have only white balls, you will certainly draw 2 white balls in your first 2 draws.

So

P[A|B] = P[A]/P

Your claim is that P[A] is irrelevant to the calculation of P[A|B]

You can see from this equation:

P[A|B] = P[A]/P

that unless the P[A] is cancelled out somehow, P[A|B] will definitely depend on P[A].

To see if there is any cancellation, let us examine P.

What is P? It is the probability of drawing 2 white balls given no other assumptions.

Intuitively you can see that you must consider all possible cases for number of white balls in the bag initially, and use the probability of those cases as weights.

All possible cases for number of white balls are:

0 white balls

1 white ball

2 white balls

3 white balls

4 white balls

Each of these is mutually exclusive with all others.

P = P[b|0W]*P[0W] + P[b|1W]*P[1W] + P[b|2W]*P[2W] + P[b|3W]*P[3W] + P[b|4W]*P[4W]

This makes intuitive sense.

To get B (drawing two white balls), you must weight what the probability of getting two white balls in every possible (mutually exclusive) case, by the probability of that case, and add up all combinations.

Very quickly, P[b|0W] = P[b|1W] - 0

because if there are less than 2 balls initially, you cannot draw 2 balls.

Therefore,

P = P[b|2W]*P[2W] + P[b|3W]*P[3W] + P[b|4W]*P[4W]

Remember that event A is the same as having 4 white balls initially, and we showed earlier that P[b|A] = 1.

P = P[b|2W]*P[2W] + P[b|3W]*P[3W] + P[A]

Remember from earlier that

P[A|B] = P[A]/P

P[A|B] = P[A]/(P[b|2W]*P[2W] + P[b|3W]*P[3W] + P[A])

It is clear that there is no cancellation of A to get the answer you have argued for unless

P[A] = P[2W] = P[3W]

You cannot get this from the statement of the problem. You must assume it.

You are making an assumption about the initial distribution of balls in the bag.

As a result of your assumption, your equation simplifies to

P[A|B] = 1/(P[b|2W] + P[b|3W] + 1) = 1/( 1/6 + 1/2 + 1) = 0.6

The question is not answerable unless you make an assumption.

You made this assumption and got an answer.

You claim that you did not make this assumption, but I am confident that I demonstrated that you have though.

If you truly want to make no assumptions whatsoever, here is your answer:

P[A|B] = P[A]/(P[2W]/6 + P[3W]/2 + P[A])

which is undefined until P[A], P[2W] and P[3W] are defined.

It is IMPOSSIBLE to get an answer without making some sort of assumption/guess about these values.

[superprismatic]

In order to try to put this to rest:

Hey! Good luck with that!

[Guest]

Very quickly, P[b|0W] = P[b|1W] - 0

because if there are less than 2 balls initially, you cannot draw 2 balls.

I made a typo: supposed to be = 0 (both are)

Edited March 31, 2010 by mmiguel1

[Guest]

Just to clarify what we are debating - is it whether or not the answer is correct, or whether or not an assumption was made?

You may not claim you have made any assumption about the initial probability, but from the pure mathematics, you have in fact made such an assumption or else your answer is not defined.

I'm not sure where I claimed I didn't make an assumption. In my first post on this thread I gave an example of questions involving biased/unbiased coins/dice and said that there are fair and reasonable assumptions that are always made in order to answer questions. You demonstrated that yourself in a post where you said you made an assumption that the question was in English and not some other language that merely looked like English.

The point I made in that post about the biased/unbiased coin was that, with no information about how the coin was initially selected, you can still come to an answer and it is reasonable to come to an answer. After all, it's probability we're dealing with, so you are just asked "based on what you know, what is the probability that...".

My argument is simply the same here. Given no information about how the balls were initially placed in the bag, all you can deduce from what you know is that you are in one of three possible circumstances. You have no information to suggest that any one of those circumstances is more favourable as a starting situation than another so you can't give any more weighting to one over another. (If you like, this is the assumption that each of the three scenarios is equally likely - as fair and reasonable an assumption as "we are using a fair coin". I haven't considered this necessary to state as an assumption, as you are given no information to suggest you should do anything else.)

I fully admit that the more information you are given, the more you can refine your answer. So if you are given further information to suggest, for example, that the person giving you the bag only has white balls to choose from, or that the 4 balls were initially randomly selected from a certain collection of balls, then that will change your answer. The point is, just because you don't have all the information up front, does not mean that you can't come to a solution.

Let me put it another way. Are you saying that, unless you have every piece of information about a situation to hand, there is no way you can determine the probability of something? Consider the coin tossing example in my last post - given no information about the likelihood of initially being given a biased or fake coin, are you saying that it is not possible to say, based solely on observation, which is more likely?

Of course, this type of approach is used all the time. For example, how do you know if a coin is fair in the first place? You can try tossing it a number of times - if it comes up heads far more often than tails then you'd naturally come to the conclusion that the coin is biased towards heads - and that would be an appropriate and correct conclusion. You couldn't possibly say that, because you didn't know the initial probability that you had a biased coin, that you couldn't reach that conclusion. However, if I later told you that only 1 in a billion coins was biased, you would then vary your conclusion and maybe determine that it was more likely you had a fair coin and witnessed an extraordinary event. That does not mean that your initial conclusion was incorrect - simply that it was different based on the information you had at the time.

The way probability questions usually work is that you start by giving information about the initial conditions and a probability is calculated. You then add information about subsequent observations and the probability is revised. Just because you didn't have the information about subsequent observations does not mean you can't work out a probability in the first instance - you just assume you don't know anything about future observations! What we have here is simply the normal situation reversed - exactly the same principle applies, but for some reason it is more difficult to accept that a solution is possible.

[Guest]

Anyone remember Superman flying so fast that he went back in time? Why is it that so many people get upset about this particular feat, saying that "It's impossible!", and yet mention nothing about Superman even being able to fly in the first place!? It's because going into that movie those people knew that in order to enjoy it they would need to be willing to accept some very improbable things. It's not until the movie exceeds what an individual had intended to accept that the Laws of Physics became so important.

One thing that this question has shown quite clearly is how, in a similar way, different people are willing to accept different levels of ambiguity in riddles. Obviously each person begins a riddle prepared to make a number of assumptions in order to fill in the gaps in a question that exist for the sake of brevity. However, when we find ourselves required to make more assumptions than we were initially prepared to make, it upsets us. The folk over in the Word Riddles forum, for example, are significantly more willing to leap tall buildings to arrive at conclusions for the sake of a riddle because that's what those riddles often take to solve. Here in the Math/Logic Puzzles section we tend to expect much more structure and rigidity in our riddles.

As for me, when I clicked on this riddle the first time I was quick to come to the same conclusion as many others have and discount the riddle as broken. Perhaps if the question had informed me that I was required to try to deduce how the balls were selected to go into the bag without any information I wouldn't have minded as much. But it didn't, and I was not interested in answering a question that wasn't even asked.

If I were to post the question here "What do I have in my pocket?", there may be a few who could entertain themselves speculating on what I'm likely to have based on what people ordinarily have in their pockets or whatnot. But I'm sure there would be many more who'd argue that my question makes for a very poor riddle indeed.

[Guest]

Just to clarify what we are debating - is it whether or not the answer is correct, or whether or not an assumption was made?

I'm not sure where I claimed I didn't make an assumption. In my first post on this thread I gave an example of questions involving biased/unbiased coins/dice and said that there are fair and reasonable assumptions that are always made in order to answer questions. You demonstrated that yourself in a post where you said you made an assumption that the question was in English and not some other language that merely looked like English.

The point I made in that post about the biased/unbiased coin was that, with no information about how the coin was initially selected, you can still come to an answer and it is reasonable to come to an answer. After all, it's probability we're dealing with, so you are just asked "based on what you know, what is the probability that...".

My argument is simply the same here. Given no information about how the balls were initially placed in the bag, all you can deduce from what you know is that you are in one of three possible circumstances. You have no information to suggest that any one of those circumstances is more favourable as a starting situation than another so you can't give any more weighting to one over another. (If you like, this is the assumption that each of the three scenarios is equally likely - as fair and reasonable an assumption as "we are using a fair coin". I haven't considered this necessary to state as an assumption, as you are given no information to suggest you should do anything else.)

I fully admit that the more information you are given, the more you can refine your answer. So if you are given further information to suggest, for example, that the person giving you the bag only has white balls to choose from, or that the 4 balls were initially randomly selected from a certain collection of balls, then that will change your answer. The point is, just because you don't have all the information up front, does not mean that you can't come to a solution.

Let me put it another way. Are you saying that, unless you have every piece of information about a situation to hand, there is no way you can determine the probability of something? Consider the coin tossing example in my last post - given no information about the likelihood of initially being given a biased or fake coin, are you saying that it is not possible to say, based solely on observation, which is more likely?

Of course, this type of approach is used all the time. For example, how do you know if a coin is fair in the first place? You can try tossing it a number of times - if it comes up heads far more often than tails then you'd naturally come to the conclusion that the coin is biased towards heads - and that would be an appropriate and correct conclusion. You couldn't possibly say that, because you didn't know the initial probability that you had a biased coin, that you couldn't reach that conclusion. However, if I later told you that only 1 in a billion coins was biased, you would then vary your conclusion and maybe determine that it was more likely you had a fair coin and witnessed an extraordinary event. That does not mean that your initial conclusion was incorrect - simply that it was different based on the information you had at the time.

The way probability questions usually work is that you start by giving information about the initial conditions and a probability is calculated. You then add information about subsequent observations and the probability is revised. Just because you didn't have the information about subsequent observations does not mean you can't work out a probability in the first instance - you just assume you don't know anything about future observations! What we have here is simply the normal situation reversed - exactly the same principle applies, but for some reason it is more difficult to accept that a solution is possible.

You make many good points.

I would first like to address one thing you said:

"Given no information about how the balls were initially placed in the bag, all you can deduce from what you know is that you are in one of three possible circumstances. You have no information to suggest that any one of those circumstances is more favourable as a starting situation than another so you can't give any more weighting to one over another. (If you like, this is the assumption that each of the three scenarios is equally likely - as fair and reasonable an assumption as "we are using a fair coin". I haven't considered this necessary to state as an assumption, as you are given no information to suggest you should do anything else.)"

I do not believe that assuming each of these three possible circumstances is equally likely is as fair and reasonable as assuming a fair coin.

In both examples you provided, the coin, and the dice, the fair and natural assumptions that you propose assume that each physical state is equally likely. These fair and natural assumptions do not assume that each event of interest is equally likely.

Suppose we toss two die and add the face-up numbers on each to get a random variable.

How many possible circumstances do we have? Well the sum can take any value from 2 to 12.

That gives 12-2+1 = 11 possible circumstances.

The fair and reasonable assumption here is not that each of these circumstances has probability 1/11.

Given no information about the die, it is not fair and reasonable to assume that if you keep rolling the die, about every 1 in 11 times you will get a sum of 2, and with about equal frequency you will get a sum of 7.

The fair and reasonable assumption is that each physical state of the dice is equally likely as any other.

There is only 1 physical state that sums to 2, when both dice have 1 face up.

For a sum of 7, you may have 1+6,6+1,2+5,5+2,3+4,4+3 or 6 different physical states.

Because of my assumption each one of these 6 states is worth the same from a probability standpoint as the single state that sums to 2.

Thus it is fair and reasonable to assume

P[2] = 1/36

P[7] = 6/36

and similarly for all the other numbers.

This is a very similar situation with the bags.

I am saying that you are doing something akin to saying that the sum of dice rolls can take any value from 2 to 12 each with probability 1/11, which is not in my opinion the most fair and reasonable assumption to make.

Your argument for doing so by analogy is saying we have no information about the fairness of the dice, so we have no reason to weight the outcome of a sum of 2 any different than an outcome of a sum of 7.

I have argued for what I believe to be the most reasonable assumption about 10 times (maybe not).

But I can't even get to that point unless we agree on what colors of balls are in the bag.

If just black and white, then I say the most fair and reasonable assumption is to say that any physical state of black and white balls is equally as likely as another.

This provides the distribution of the balls initially in the bag, and from that distribution, you may calculate that the probability is 0.25.

I showed in a previous post that if you assume that a ball may take one of N colors, and you make what I call the most fair and reasonable assumption, then the probability of getting two more white balls will be 1/N^2.

The reason that I worked so hard to argue that you made an assumption is because in previous posts you seemed to me to simply dismiss the need for the solver of the puzzle to make an assumption. When you say there is plenty of information to solve the problem, you act as if the puzzle leaves no arbitrary decision to the solver.

You also deny that there is any importance to the initial distribution of white balls in the bag.

Your equation shows no signs of the initial distribution, and you seem to use that to argue that the distribution is irrelevant.

I will explain what I feel about this by analogy:

f = x + y

Assume y=5:

Write down and then interpret:

f = x + 5

Conclusion: f does not depend on y, but only on x and 5.

My point: If you make an assumption about a parameter, your result will not show signs the parameter as a variable influenced the result.

I hope this conveys why I disagree with what you were saying.

You also say:

"Note that the answer doesn't imply anything about the initial selection of the balls. It simply determines the probability that we are in 1 of 3 possible scenarios based on what we have observed."

Your claim here implies that your answer makes no assumption and needs no assumption.

As a matter of fact, your answer does imply something about the initial selection of balls. It implies that you assumed that the probability of 4 white balls initially is the same as having 3 is the same as having 2.

I have explained in many posts, but I can do so again, why I think this is not the most fair and reasonable distribution.

You made some other points.

You said:

"I fully admit that the more information you are given, the more you can refine your answer. So if you are given further information to suggest, for example, that the person giving you the bag only has white balls to choose from, or that the 4 balls were initially randomly selected from a certain collection of balls, then that will change your answer. The point is, just because you don't have all the information up front, does not mean that you can't come to a solution."

I still assert that you cannot come to a mathematical answer unless you have the distribution of balls in the bag. It is not all information, just probabilistic information. If you come to a solution, you are not doing so by the rules of probability, you are making a guess. Questions like these are not asking for guesses, they are asking for you to use probability theory.

You said:

"Consider the coin tossing example in my last post - given no information about the likelihood of initially being given a biased or fake coin, are you saying that it is not possible to say, based solely on observation, which is more likely? "

This is more complicated of a topic.

Here is what you were referring to:

"You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails? "

From what is written here, not enough information is given to make a confident assumption about what types of coins we may consider. Probability theory as a tool requires us to make these assumptions, to specify the sample space, to assume distributions in order to give us a useful answer, etc.

To make an assumption about this problem, you cannot use probability theory. This question cannot therefore be answered by probability theory unless you are provided with the information required to make the necessary assumptions.

This question is instead answered by some other human decision making tool (gut feeling, superstition, positive reinforcement, etc), but not probability theory.

Not to say that your answer cannot incorporate probability theory, but the assumption you make in order to use probability determines the answer to the question.

Your comments on the validity of using observations and empirical data to determine whether or not something is true also extend beyond the realm of probability theory. Most mathematics, including probability, is logically deductive except for axioms and assumptions. The scientific method you are talking about is logically inductive. I do not say the scientific method is invalid, because it is obviously useful and I appreciate it. However it is not relevant as an argument on how to calculate a probability (which was the original question). The scientific method is not a sub-theory of probability.

Edited April 1, 2010 by mmiguel1

[Guest]

OK, this may be one we'll have to agree to disagree on, but one last point (I'll try to keep it brief)...

Given no information about the die, it is not fair and reasonable to assume that if you keep rolling the die, about every 1 in 11 times you will get a sum of 2, and with about equal frequency you will get a sum of 7.

The difference between your reasoning and mine with the bag, is that you are looking for some way of determining the probability of the initial state of the bag. Here for example, you refer to "keep rolling the die", as that is a way you can determine the probability of any particular state. Similarly, in this discussion, you look for a way to determine a probabilistic way of initially selecting the balls that went into the bag. Given that information you are on absolutely solid ground when determining the conditional probability.

However, my point is that there is no probability calculation of the initial state of the bag. It is a bag with four balls. They have already been selected. As to how they were selected, we have no clue. It could be that the person filling the bag always fills it with white balls, or always fills it with exactly two white balls and two others. We haven't been given any information to suggest that there was a rule for how the balls were initially selected for the bag, so how can we give any one situation more weight than any other (or in other words, how can we determine the probability of each initial state)?

We are probably actually in agreement up to this point. The disagreement is that you assert that without that information there is no answer, whereas I (and the OP and the exam writers where the question was posted) assert there is. The approach to get to it is different to traditional probability theory, but that does not mean it is not probability (others in the thread have credited the approach to Bayes Theory which certainly rings a bell).

Finally, the logic part (or the gut instinct you referred to): If I knew a bag contained either 2, 3 or 4 white balls and I just picked out 2 white balls in a row, I'd be thinking it's more likely I've got the bag with 4 white balls than the other two. This contradicts the answer of 0.25 you gave. Although "gut instinct" or non mathematical methods may sometimes help understand a method, they usually do not prove the maths wrong. But it can suggest the approach is wrong, or maybe there's a mistake. If I was asked to divide 543 by 973, my "instinct" would tell me the answer is greater than 0.5. If I did some long division and got an answer of 0.3, I would question where I went wrong in my calculation, not argue that the maths must be correct. Indeed it is often this very approach that has led to new theories historically when it has been clear that the traditional approach isn't working or doesn't allow for certain scenarios, and may even by what initially prompted Bayes theory...

If you are still in disagreement then I call a truce! I hadn't intended to spend this long on this topic - but I do enjoy a passionate debate, which you have certainly provided and so got me sucked in!

[Guest]

If I were to post the question here "What do I have in my pocket?", there may be a few who could entertain themselves speculating on what I'm likely to have based on what people ordinarily have in their pockets or whatnot. But I'm sure there would be many more who'd argue that my question makes for a very poor riddle indeed.

After promising a truce, this made me think of a way of hopefully getting some agreement!

I agree your question about the pocket in it's current state would be a very poor riddle. But what if I changed it slightly to the following:

"I have 100 things in my pocket. What is the probability that they are all the same?"

Again, this isn't a good riddle, as you don't know how the "things" were selected and I would be asking you for the probability of the initial state. However, if I let you pick something randomly out of the pocket one at a time (assuming that you couldn't tell the difference between the items by touch) and I let you place a bet each time on whether they were all the same or not, what would you now do? At the start, I'd say it's unlikely you place the bet. But after you've picked 99 things and you see that all 99 things were the same, would you place the bet now? You still have absolutely no idea about how I initially picked those 100 things, but is it not more likely that all 100 are the same, rather than you somehow have managed to avoid the one different item? Is there surely no form of probability that would allow you to determine this, or does maths let us down entirely here and say "no answer" so we have to rely on our gut?

The bag and balls question is exactly the same. Extend it out to 1 million balls in the bag. You have no idea how those balls were selected. When you randomly pick one out, you can't tell anything about the other balls as they all feel the same and you pick them "blindly". Before you start picking you have no idea if they are all the same colour. But, one by one, you pick 999,999 balls out of the bag and notice they are all white. Do you stand by your assertion that you do not have any information to determine whether that last ball is more likely to be white or not? Even though that last ball could be one of a million different colours, does that really change the probability that you have managed to avoid it all this time?

For anyone who now says "Yes, but you can't give a mathematically definitive probability - it's just guesswork, or gut feel" - really? Does maths let us down that much? If I was to say I'd double your money on whatever you bet, would maths really not allow us to work out at what point your odds make that bet worthwhile?

Typically, where current mathematical models let us down and say "you can't answer that question" - even though the answer is pretty clear - that's when a new model is created. This is what I was getting at with my last post in suggesting that Bayes Theory may have come from a similar recognition that the traditional approach just wasn't good enough.

[Guest]

OK, this may be one we'll have to agree to disagree on, but one last point (I'll try to keep it brief)...

The difference between your reasoning and mine with the bag, is that you are looking for some way of determining the probability of the initial state of the bag. Here for example, you refer to "keep rolling the die", as that is a way you can determine the probability of any particular state. Similarly, in this discussion, you look for a way to determine a probabilistic way of initially selecting the balls that went into the bag. Given that information you are on absolutely solid ground when determining the conditional probability.

However, my point is that there is no probability calculation of the initial state of the bag. It is a bag with four balls. They have already been selected. As to how they were selected, we have no clue. It could be that the person filling the bag always fills it with white balls, or always fills it with exactly two white balls and two others. We haven't been given any information to suggest that there was a rule for how the balls were initially selected for the bag, so how can we give any one situation more weight than any other (or in other words, how can we determine the probability of each initial state)?

We are probably actually in agreement up to this point. The disagreement is that you assert that without that information there is no answer, whereas I (and the OP and the exam writers where the question was posted) assert there is. The approach to get to it is different to traditional probability theory, but that does not mean it is not probability (others in the thread have credited the approach to Bayes Theory which certainly rings a bell).

Finally, the logic part (or the gut instinct you referred to): If I knew a bag contained either 2, 3 or 4 white balls and I just picked out 2 white balls in a row, I'd be thinking it's more likely I've got the bag with 4 white balls than the other two. This contradicts the answer of 0.25 you gave. Although "gut instinct" or non mathematical methods may sometimes help understand a method, they usually do not prove the maths wrong. But it can suggest the approach is wrong, or maybe there's a mistake. If I was asked to divide 543 by 973, my "instinct" would tell me the answer is greater than 0.5. If I did some long division and got an answer of 0.3, I would question where I went wrong in my calculation, not argue that the maths must be correct. Indeed it is often this very approach that has led to new theories historically when it has been clear that the traditional approach isn't working or doesn't allow for certain scenarios, and may even by what initially prompted Bayes theory...

If you are still in disagreement then I call a truce! I hadn't intended to spend this long on this topic - but I do enjoy a passionate debate, which you have certainly provided and so got me sucked in!

I do still disagree but I can accept a truce.

I would like to clarify my ending position though.

My approach also used Bayes' theorem.

I'm a little hesitant to part with saying that my claim is there is no answer.

One can come to an answer, but it is only as good as their interpretation of the problem and the assumptions they make.

In my opinion, there is no definite correct answer that everyone can agree on because the question does not explicitly tell the solvers what assumptions to make. You can say A is the best assumption, and I can say B is the best assumption, but there is nothing in the original statement of the problem which we can go back to and resolve our conflict. Kewal can come afterwards and say the assumption he wanted us to make is A because in his mind A is much more reasonable than B. It is not so obvious which is the best assumption: A or B, and so a person's answer comes down to their opinion. If it were so obvious, this debate would not have lasted so long. If you are marked wrong on a math problem because of your opinion is that a fair math problem? A math problem should test your mathematical ability, not whether or not you have the same opinion as the author of the problem.

To clarify my position then, I say there is no fair answer from the original statement of the problem.

Gut instinct topic:

We are talking about completely different realms of ideas here.

"If I knew a bag contained either 2, 3 or 4 white balls and I just picked out 2 white balls in a row, I'd be thinking it's more likely I've got the bag with 4 white balls than the other two."

This conclusion is not related to probability whatsoever. This follows from a habit of human nature to assume that what they observe now or in the past will repeat in the future --- pattern recognition.

You do not know if the colors of the balls are dependent on one another or independent, you make a guess from experience. This guess is arbitrary from a mathematical point of view and is based on how our brains process data I'm assuming.

(In statistics, this guess is called the null hypothesis).

The only time that probability comes in to the picture is to test your guess.

From the point you start using probability, as far the math is concerned, your guess is absolute truth.

The answer to any question: "what is the probability of ..." is therefore only as true as your initial guess.

This is how probability is used in science.

You tell me when you see 2 white balls in a bag of four that you guess that there are more because you recognize the pattern.

That is not enough to calculate a probability. What is required is a guess on the initial distribution in the bag.

If you do all the math correctly, then you must accept the probability you calculate if you think your guess is right. The math does not lie. If it goes against your gut feeling, then change your guess.

0.25 is in fact a reasonable probability for black and white balls where each ball can be either black or white with equal probability.

Intuition 1:

You are very likely to get 2 white balls in 2 draws if there are lots of white balls in the bag.

Intuition 2:

You are more likely to get mixes with near equal numbers of black and white balls then mixes dominated by one color.

Your pattern recognition in this case is based off of intuition 1, and ignores intuition 2.

Yet these two intuitions compete. Probability tells us how to merge these intuitions.

It is not that the math failed, it is that human nature failed.

Of course intuition 2 is more deeply rooted in an assumption on the initial distribution of the bag than intuition 1.

That doesn't mean you can ignore it. Most distributions will give intuition 2 effects that are significant. The answer to the conundrum is to make a guess and test it.

Questioning your own assumptions is just as, if not more important than questioning your mathematics.

With that, I can stop talking about this stuff I think.

[superprismatic]

All I have to add is that the reverend Bayes would be proud of this debate!

But, ultimately, he agreed with me. That's why the CBSE board exam accepts

my answer as correct. I do, however, understand other opinions. After all,

Bayes was just an infallible human as all of us are. Basically, prior

probabilities are ALWAYS subject to debate. So, be calm......

[superprismatic]

All I have to add is that the reverend Bayes would be proud of this debate!

But, ultimately, he agreed with me. That's why the CBSE board exam accepts

my answer as correct. I do, however, understand other opinions. After all,

Bayes was just an infallible human as all of us are. Basically, prior

probabilities are ALWAYS subject to debate. So, be calm......

I meant "fallible".

[Guest]

Wow, I didn't even that half the posts on this last page existed.

I posted my response and didn't even check.

I only read you first message mentioning truce, neida.

Unfortunately, I have to go right now, but I will read the other posts later.

Edited April 4, 2010 by mmiguel1

[Guest]

All I have to add is that the reverend Bayes would be proud of this debate!

But, ultimately, he agreed with me. That's why the CBSE board exam accepts

my answer as correct. I do, however, understand other opinions. After all,

Bayes was just an infallible human as all of us are. Basically, prior

probabilities are ALWAYS subject to debate. So, be calm......

?

Is Bayes' Theory different from Bayes' Theorem relating a conditional probability to it's conditional converse?

If it isn't, then I wasn't aware of what it meant.

If it is the same, then I don't see how it agrees with you anymore than it agrees with me, as I also used Bayes' theorem in my derivation.

[superprismatic]

?

Is Bayes' Theory different from Bayes' Theorem relating a conditional probability to it's conditional converse?

If it isn't, then I wasn't aware of what it meant.

If it is the same, then I don't see how it agrees with you anymore than it agrees with me, as I also used Bayes' theorem in my derivation.

I was talking about Bayes' treatment of prior probabilities especially when nothing is known about them. My point was that these are quite subjective.

[bushindo]

I was talking about Bayes' treatment of prior probabilities especially when nothing is known about them. My point was that these are quite subjective.

I'd like to submit a point of view on this problem. I'd like to review the problem first. The crux of the debate, according to my skimming of the last 10 pages, boils down to this.

Let A(n), similar to superprismatic's definition, be the chance of initially having n white balls in the bag of 4.

We need to have the vector of probabilities [ P( A(2) ), P( A(3) ), P( A(4) ) ] to solve this problem. We are not given information about these states, and this is where mmiguel1 and superprismatic diverges.

mmiguel1 argues that in the lack of information about the balls-bag generation process, he will assume that the bag was generated by a binomial process with 4 trials and chance of black ball = .5.

superprismatic and the OP, and the CBSE board as well, argue that the uniform distribution is most representative of the lack of information, otherwise known as an un-informative prior in Bayesian statistics, and thus use the vector of probabilities [ P( A(2) ), P( A(3) ), P( A(4) ) ] = [ 1/3, 1/3, 1/3 ]. The uniform distribution is the oldest and simplest un-informative prior, so I can understand where this is coming from. The issue is that the uniform distribution is not the only un-informative prior there is (see http://en.wikipedia.org/wiki/Prior_probability#Uninformative_priors). There are a wide number of un-informative priors, and they probably would yield different answers. One example, for instance, would be to assume that each ball in the bag has a chance of being black p, where p belongs to the Beta distribution with the improper prior Beta(0,0).

superprismatic made the point that prior distribution, especially in case with no information, is subjective, which I think is where the discussion should have ended. However, my beef with the problem is that the CBSE puts this problem on a board exam, and presumably expects only 1 right answer, which is derived from assuming a uniform prior distribution as the non-informative prior. If the CBSE taught all their students to always assume the uniform as a non-informative prior, then this is a fair problem as a board exam. It is not, however, valid to post this problem to the brainden and board and expect .6 as the answer, just because the CBSE said so.

Edited April 4, 2010 by bushindo

[Guest]

I'd like to submit a point of view on this problem. I'd like to review the problem first. The crux of the debate, according to my skimming of the last 10 pages, boils down to this.

Let A(n), similar to superprismatic's definition, be the chance of initially having n white balls in the bag of 4.

We need to have the vector of probabilities [ P( A(2) ), P( A(3) ), P( A(4) ) ] to solve this problem. We are not given information about these states, and this is where mmiguel1 and superprismatic diverges.

mmiguel1 argues that in the lack of information about the balls-bag generation process, he will assume that the bag was generated by a binomial process with 4 trials and chance of black ball = .5.

superprismatic and the OP, and the CBSE board as well, argue that the uniform distribution is most representative of the lack of information, otherwise known as an un-informative prior in Bayesian statistics, and thus use the vector of probabilities [ P( A(2) ), P( A(3) ), P( A(4) ) ] = [ 1/3, 1/3, 1/3 ]. The uniform distribution is the oldest and simplest un-informative prior, so I can understand where this is coming from. The issue is that the uniform distribution is not the only un-informative prior there is (see http://en.wikipedia.org/wiki/Prior_probability#Uninformative_priors). There are a wide number of un-informative priors, and they probably would yield different answers. One example, for instance, would be to assume that each ball in the bag has a chance of being black p, where p belongs to the Beta distribution with the improper prior Beta(0,0).

superprismatic made the point that prior distribution, especially in case with no information, is subjective, which I think is where the discussion should have ended. However, my beef with the problem is that the CBSE puts this problem on a board exam, and presumably expects only 1 right answer, which is derived from assuming a uniform prior distribution as the non-informative prior. If the CBSE taught all their students to always assume the uniform as a non-informative prior, then this is a fair problem as a board exam. It is not, however, valid to post this problem to the brainden and board and expect .6 as the answer, just because the CBSE said so.

I can agree with that.

I think continuing this debate is trying everyone's nerves.

Great arguments all around, but I think enough is enough.

I will really try hard to resist any urge to post on this thread again

[Guest]

since 2 balls are already white, the rest of the two balls can be either:

1. both non-white,

2. One white and one non-white

3. both whites.

Hence the probability of all 4 whites is 1 out of 3 i.e 1/3

[Guest]

1/3.

The possible collection of balls is

(1) two white and two non-white,

(2) three white and one non-white, or

(3) four white

Therefore, there is a 1/3 chance that all four balls are white.

Edited July 17, 2010 by Dej Mar

[Guest]

Incredibly easy:

There are only three possibilities:

WWOO

WWWO

WWWW

where O is any other colour

Because there are 3 possibilities and only one is correct, the probability is 1/3

Simple

[Guest]

Dear lord, not this again.

[Guest]

I believe those last three posts are wrong. Even if you assume that each ball has a probability of 50% to be white (which is not what I understand from the riddle) the chances of two white balls left are 1/4, not 1/3.

The possibilities are actually

WWOO

WWOW

WWWO

WWWW

However, I don't think we have enough information to answer it. To begin with, I don't think 50% of all balls are white. I think they are much more likely to be more colourfull. (Wait! Maybe they are golf or ping-pong balls, and we're all wrong?) Also, we do not know who and why put them in the bag to begin with. Maybe it's supposed to be a bag of identical balls. If it's in a store or warehouse or something like that.

Incredibly easy:

There are only three possibilities:

WWOO

WWWO

WWWW

where O is any other colour

Because there are 3 possibilities and only one is correct, the probability is 1/3

Simple

[Guest]

Incredibly easy:

There are only three possibilities:

WWOO

WWWO

WWWW

where O is any other colour

Because there are 3 possibilities and only one is correct, the probability is 1/3

Simple

And regarding your statement "Because there are 3 possibilities and only one is correct, the probability is 1/3", what if I told you that you have two possibilities: winning the lottery or not winning the lottery. Would that make the posibbility of winning 1/2? The fact that there are three options doesn't mean that they all have equal probabilities.