This doesn’t work on a couple of levels, Frylock. The big problem is that you are ASSUMING that the probability of there being one or zero marbles is equally likely, and then using that assumption to show that the probability is 0.5. What if I’m the person putting the marble in the bag and I like to put a marble in 90% of the time?
Even if the probability of the two outcomes was equal (say I determine it with a fair coin toss), what would happen if it was 2 and 3 marbles instead of 0 and 1? The probability can’t be (2+3)/2 = 2.5!
Without knowing how likely it is that a particular number of marbles is in the bag you cannot make any calculations, period.
I think you have to look at the big picture. Let’s say there is a machine that is filling up bags of marbles. It randomly inserts a number of blue and red marbles into each bag. It fills up a warehouse full of these bags. What are the odds of picking out a blue (or red) marrble from a randomly selected bag in the warehouse?
Unless there was a bias for one color in the manufacturing process - the chance of picking a specified colored marble should be 1/2.
If there are few marbles the chance would be closer to 3/5 than 1/2. Starting out with the case of one marble of each color in an average bag the chances are 4/6, 5/7, 6/8, …103/105, etc. In a large enough sample the chance will approach 1/2.
Ah, but that’s a big “unless”. Sure, if you equate “unknown” with “equally likely”, or as the OP has apparently morphed “unspecified number” into “equal number”, you can get away with your argument. But the answer to the originally stated OP question is that there is no exact numerical answer. You can come up with reasonable answers if you make additional assumptions beyond the original assumptions.
Frylock, the ‘probability’ of any event is a simple calculation based on two things:
the number of possible states or results
the number of possible states or results which satisfy a given condition
Example #1. Coin toss. Possible results = heads or tails = 2. Probability of ‘heads’ result = 1 / 2.
Example #2. Man has two children of different ages. What is the probability that the eldest one is female? Possible states, listing children from eldest to youngest:
Boy, boy
Boy, girl
Girl, boy
Girl, girl
Probability that the eldest is female = 2 / 4 or 1 / 2.
Example #3. Man has two children of different ages. They are not both boys. What is the probability that the eldest one is female? Possible states, listing children from eldest to youngest:
Boy, girl
Girl, boy
Girl, girl
Probability that the eldest is female = 2 / 3.
Returning to your OP, we don’t know anything about how many marbles of either colour are involved, so we don’t know anything about the range of possible states or results, so we can’t calculate the probability.
For all Frylock knows, you may equally likely be the kind of person who puts in a marble only 10% of the time. If Frylock has no way of preferring one alternative over the other, from his point of view, both alternatives are equally likely. Another example: Frylock tosses a coin, and you, Valgard, look at it. Seeing that it’s tails, from your point of view the likelihood of tails and heads are 1 and 0, respectively. From Frylock’s piont if view it is still 50/50.
If you have no way of knowing which way the process is biased, both ways are equally likely (from your point of view), and the likelihood of red/blue is 50/50.
There seems to be some misunderstanding going on that there is some real, underlying probability distribution that you have to know in order to make probability calculations, but the whole point of these calculations is to infer from incomplete knowledge. If you look at my example of the coin toss above, it is clear that, given Frylock’s information, the probability is 50/50, and from Valgard’s it is 100/0.
To put it into other words: If you have no information about the probability distribution, then every distribution is equally likely, leading to a flat (uniform) distribution.
0.5 that it’s 2, 0.5 that it’s 3, leading to an expected value of 2.5 (really.)
To address the OP:
If you have no way of knowing the number or red/blue distribution of marbles, then it is equally likely that the red outnumber the the blue, as the other way around, so you would expect there’s an equal number of each. As to the total number: if every number between 0 and infinity is equally likely, then there will, with probability 1 be an infinite number of marbles in the bag, leading to a 50/50 probability of drawing a red versus blue marble.
This is, however, misleading. You do have some information about the number of marbles–using common sense. There isn’t room in the universe for an infinite number of marbles–much less in a bag. You have to make an educated guess of a reasonable upper limit of marbles. You can call this upper limit L, and frame your answer in terms of L, using methods described by other posters, or, alternatively, you can posit some probability distribution on L, and include this in your calculation.
The reason this problem becomes complicated is the number L. If you have no information about a quantity, assuming a flat probability distribution is the way to go, but with L you clearly know that the PD must decrease and go to zero. The problem needs to be more clearly stated as to what kind of information you get. Are you allowed to see the bag, etc…
Unless I missed some clarification, “unspecified” means just that: unspecified.
It doesn’t matter how many other bags there are in the universe-- of if each “bag” were a universe itself. What matters is "how many marbles of each color are in this bag. It would be just as impossible to predict the statistics of a large number of bags, as it would be to predict a single bag absent more information than is provided in the problem.
There is no physical law that dictates that the number of red and blue marbles must be equal in the long run – and in fact, none of us have any evidence that this is actually the case on Earth today. Saying “absent any imbalance in colors” may sound "reasonable’ but it’s a huge bit of information we simply don’t have. (When marbles were a popular toy, the makers deliberately made varying numbers of various colors/styles, to encourage trading and accomodate varying polularity)
You can’t make a correct prediction about the final probablility from that bag. Numerous counterexamples have been provided.
You can only say that as T (= R+B) increases, the effect of any finite assymmetry (3R + 2B) being added to the bag decreases to zero, and the probability after the marbles are added is increasingly close to the original (unknown) probability. In fact, even with only a literal handful of marbles - a dozen or two, suitable for almost any bag - the addition of 3R +2B would already have a tiny effect on the existing population.
If R-B=1 is a small percentage of the total number of marbles (as it must be for all but the tiniest numbers of marbles), how can it DETERMINE the statistical outcome of a larger unknown population? Answer: it can’t.
It’s be like saying “if a spaceship carrying two men and three women lands on Planet X, what are the chances that the next person to die (or have a birthday) on Planet X will be female?” The number of people on that ship would have little relevance; whether the existing population of Planet X is billions or a mere handful of people, it would swamp out the newcomers. If the existing population were small (say, five people) the answer could vary from 30% (if the existing colony is all male) to 80% (if the existing colony is all female). Even if the existing population consisted of a single individual, the probability could range from 50% (if the existing inhabitant was male) to 66% (if the sole prior inhabitant was female)
I don’t find this convincing. If you have no information, it’s true that you have no reason to conclude that any particular distribution is more likely than another. But your ignorance has no effect on the actual distribution, which may follow any rule at all.
It’s hard to accept that the state of knowledge of an observer can control the probability in a specific case.
Choosing the uniform distribution as your prior uninformed distribution is subject to the classic conundrum of the choice of variables. If I tell you that some quantity x will be between 0 and 1, and I tell you that the distribution is unspecified, you might be tempted to treat x as uniformly distributed between 0 and 1. But an exactly equivalent statement of the problem is: some quantity x>=0 will be such that x[sup]2[/sup] is between 0 and 1. Would you choose x or x[sup]2[/sup] as uniformly distributed? Or how about arcsin(x)? These priors give quantitatively different answers, and I have given you no information with which to choose among them.
The only way to do inference when information is missing is to come up with a distribution for the unknown stuff using whatever knowledge is available. A mostly-uniformed (flat in some variable) prior is often used – sometimes solely for calculational convenience – but the conclusions drawn are only as good as the distribution chosen.
Aside: A quasi-joke that I’ve always thought summarizes the issue of prior knowledge well… (Note that a “Bayesian” is one who has no problem drawing a conclusion that requires a prior (which the Bayesian will choose, possibly rather subjectively), and a “Frequentist” is one who avoids making strict inferences (avoiding the need for priors entirely) ) :
A Frequentist is one who uses impeccable logic to draw conclusions with which everyone agrees and about which no one cares. A Bayesian is one who answers questions in which everyone is interested by using assumptions that no one believes.
Everybody here knows that you can’t find the actual probability of drawing a red marble from the problem as stated in the OP, except by using the original numbers of red and blue marbles as variables.
However, given the added information that every number is equally likely and that there is no limit on the original numbers, we can conclude that the extra 3 red and 2 blue marbles will make no difference.
Which means that the probability will depend entirely on the distribution already in the bag, and since we were given the added information that every number is equally probable in that distribution, 1/2 of the times we perform this experiment we will draw a red marble.
I believe you are right provided one more change is made to the problem as specified in the OP: instead of “a bag”, we need a large number of bags, the marbles in which are distributed as you say. Each trial consists of chosing a bag, then a marble from that bag.
But I’d call this tolerably obvious. If the distribution of marbles into bags is done so there is no pattern to it, the result of chosing a bag and then a marble should not be different from simply dumping all the marbles into a heap and chosing one blindly.
I guess this is “added information” in a way, in that I didn’t explicitly state these conditions in my OP. But they are exactly the conditions I intended to be taken as implicit–since I didn’t specify that any number is more likely than others I assumed you guys would take this to mean that every number is equally likely, and since I didn’t specify a limit on the original numbers, I assumed you guys would take this to mean that, well, no limit is specified.
Lazy Dragon, thanks for your post. I think you cleared up alot of the issue.
After all that has gone on above, I think the answer is: “1/2, sort of.”
I guess this is “added information” in a way, in that I didn’t explicitly state these conditions in my OP. But they are exactly the conditions I intended to be taken as implicit–since I didn’t specify that any number is more likely than others I assumed you guys would take this to mean that every number is equally likely, and since I didn’t specify a limit on the original numbers, I assumed you guys would take this to mean that, well, no limit is specified.
Lazy Dragon, thanks for your post. I think you cleared up alot of the issue.
After all that has gone on above, I think the answer is: “1/2, sort of.”
Thank you:) My conlusion would be ‘slightly above 1/2 versus slightly below 1/2.’ Since I’ve already written the following, I’ll post it anyway:
This is certainly an interesting question, and my first thought would be to ask which variable I’m supposed to be working with. If it’s x, I’d choose a flat PD (probability distribution) for x, or ditto for x[sup]2[/sup]. The way I understand it choosing a flat PD simply amounts to presupposing no knowledge about x, that is, I’m simply saying that I have no reason to believe it will be close to 1 rather than close to 0. Similarly, if x[sup]2[/sup] is the variable in question, then it would make sense to choose a flat PD for that, while if I’m working with both the answer is less obvious. Are you saying that choosing a flat PD does something more than ‘assigning no preference’? How can I inject less information? In deciding the probability of drawing a red vs a blue marble from a bag, I could get by with slightly less, saying merely that absent more information the PD of the proportion red/blue is symmetric.
No we don’t. And that is exactly why I’m saying what I’m saying. Given the number of marbles in the world it is highly unlikely that red and blue ones are precisely equally numerous. But which is it? Either there are more red or there are more blue. That’s two possibilities. What I’m saying is that if you don’t know anything else, then both are equally likely from your point of view.
And yet, that’s what I’m arguing. Probabilities are always calculated from given information. If you have picked a marble from a 50/50 distribution of red and blue marbles, but haven’t yet looked at it, what is the probability that when you look, you’ll find a red/blue one? 50/50? Unknowable? Your friend looked before you did. Does that change anything for you?
I apologize if it seems that I’m just saying the same things over. I do find this somewhat hard to explain. But my main point is that probability must be with respect to a point of view, meaning that the PD is different from the point of view of your friend who looked at the marble (for whom it is in fact decided) than it is for you. What is random from one point of view is determined from another. (Is my shoe size a random variable?)
Looking at this again it doesn’t entirely make sense. I think I should be saying that the PD of the percentage of red (or blue) marbles is symmetric around 50%.
How well this works as a prediction of red vs blue once you start picking obviously depends on how close it is to the actual numbers of marbles. Maybe there are more red than blue. But then, there could be more blue, and you just didn’t know it.
I can’t accept this. If we have no information, then the only sensible choice is to admit that we have no information. If we’re given that the original distribution of marbles is a result of a stochastic (random) process, then that would be true, but we are not told in the problem statement that it’s the result of randomness. For all we know, the bag could have been filled by Monty Hall, trying to trick us, who knows that some would assume a uniform distribution and therefore pick something drastically different.
I honestly don’t see any information in the OP to support the notion of symmetry around 50%.
The best you can say is that there is no info that would support the conclusion that there are more red than blue, or more blue than red marbles. That sounds like symmetry. But there is equally no information to support the notion of 50% blue, 50% red. That’s the nature of no information - you just don’t know.
And I believe that ignorance of the relative probability of two outcomes, while it does make you unable to choose the one that’s more likely, does not make them equally likely.
The answers seem obvious to me: the chance of a blue marble is 50%, and this chance is not altered by anyone looking at the marble after I’ve selected it and before I look. After I look at the red marble I actually chose, there’s no reason to change the view that the chance of a blue was 50% at the time I chose. I can certainly say that there’s zero chance that this marble I now hold is blue, but that has nothing to say about the probability of selecting a marble from the bag.
Suppose the distribution was 90/10 blue/red, but all I was told was that I should select a marble that would be either blue or red. I might easily imagine that the distribution is equal, but that does not make it so and does not affect my 10% chance of selecting a red marble.
It’s true that he knows there’s a 100% chance the marble is red, whereas all I know is that the marble came from a population that is evenly blue/red. But all that says is something obvious: that knowledge of an outcome affects how you’d bet. If you and I watch a video of a football game I’ve seen and you haven’t, you probably don’t want to bet with me on how it turns out.
When we return to the OP we find there is no information about the distribution of marbles in the bag. My belief is that this means we can conclude very little about the chances of drawing a blue marble. The chances depend solely on the unknown distribution, and not at all on my state of mind. There isn’t any reason to conclude that the chances are near 50%.
Yes, but which would he pick? That’s random to us. And maybe it would be someone else filling the bag. Someone with entirely different motives. Anyone could have been filling the, bag–that’s just another random variable, another layer of randomness. I think you are focusing too much on real, metaphysical randomness. For the purposes of probability calculations there is no difference between lack-of-information randomness and “real” randomness, and most calculations done by statisticians are performed on the first kind rather than the second.
And that is in fact all that I’m claiming. The symmetry around 50% that I’m talking about is simply saying that it is equally likely to have a 60/40 as a 40/60-distribution (or 70/30 to 30/70.) I think the double use of the word ‘distribution’ is confusing, sorry about that. One use of that word refers to the relative number of each color marble in the bag, while the probability distribution (PD) describes the probability of these (non-PD) distributions. So I’m not claiming there are 50% of each, just saying 80/20 is no more likely than 20/80. A flat (uniform) PD claims something slightly stronger: that every (non-PD) distribution is equally likely.
Sorry about the confused language. I should try to think of different words for the two kinds of distributions I’m talking about.