“If I give you a jar full of beans, and don’t tell you anything about it, and you draw ten beans out of it, and they are all black, and you don’t know how many beans are in the jar in all, there’s not a practical* way to calculate a probability that all the beans are black. Even if you know how many beans there are in the jar, without knowing what colors could possibly be in the jar, you still have no practical* way to calculate that probability.”
My interlocutor replied in so many words with “you’re an idiot, google ‘urn problem’.”
I’m not seeing anything using that google search that would suggest that what I said above is wrong. Nor am I seeing anything that would suggest I’m right, though.
Am I wrong?
*I added “practically” in these two spots because one can imagine ways to calculate a probability using near-supernatural powers of observation or something. Just hedging.
I find this link on the “Law of Succession”. If I’m understanding correctly (and I’m probably not because it’s a bit over my level of knowledge) it’s saying in the scenario from the OP, we should say the probability that the next bean will be black is 11/12??? Is that right?
This still doesn’t tell me what the probability that all the beans are black is, if I don’t know how many beans there are. Does it? (Maybe something happens here involving limits…?)
ETA I just noticed the article says the rule doesn’t apply when the number of successes equals the number of tries. Still, suppose nine of the ten had been black–the rule (if I’m reading it right) would mean the chance of the next bean being black is 10/12. This, apparently, without me knowing anything at all about the composition of the beans in the jar? If a probability can be calculated here (and this simply no less) then I am very surprised. Is this right? How does this work?
Using Bayesian reasoning, there is always a way to compute this probability as you start with a prior distribution you could use with no further information. Drawing n successive black beans would update your priors to refine your estimate.
Well, that doesn’t seem to be what my interlocutor has in mind, but let me ask about this. I start with no knowledge, so the prior is 50/50 (right?).
I hate to ask for this, it may be a bit of a bore, but can you walk me through how the first black bean updates the probability using Bayes theorem? I don’t understand, for example, what P(B|A) means. I know what B and A are but I don’t know how to say what the probability of B (the background) is given A (drawing a black bean). I mean, if I had to guess, I’d say P(B|A) and P(B) are both 1/1.
If I’m reading what you’re saying in the other thread correctly, it seems like you’re saying that in a scenario like the one I mentioned (the first one, without the specified number of beans in the jar) we can calculate a 90% confidence level upper limit on the probability that the next bean will be black. Is that right?
Since we don’t know how many beans are in the jar, I would think we still can’t give a probability that all the beans in the jar are black. (For any realistic jar, though, we could estimate an upper limit to how many beans could be in there…)
Now what about the case where we do know how many beans are in the jar. Can we calculate a probability that the next bean will be black, or just an upper limit on the probability with some confidence level?
Of course, if we know how many beans are in the jar, and we assume a certain division of the color space into a finite number of color categories, we can in theory brute force a calculation based on the number of possible combinations of beans. Does drawing 10 black beans in a row on the first 10 draws just leave us with a probability of the next one being black identical to the brute-force calculation? Or does it give us more information than that?
Is that question for Pasta? If it’s for me, I’m sorry to say I have no idea what your question means–I’m unfortunately over my head here. I made some statements I thought were correct, based on intuitions I thought were well enough trained to think accurately about what I was talking about, but now I’m not so sure. Maybe it’s possible to figure some probabilities I thought were impossible to figure. I was just basically thinking in a case like the one I described, you don’t have a way to know whether your sample is representative, because you don’t know anything about the jar other than the sample you picked out of it. Is that not right?
On reading what my interlocutor said further on, he mentions the Laplace rule explicitly. Please tell me if this is right.
For the case with 1000 beans, where we don’t know what all colors may be in the jar, and where we draw ten beans, and those ten are all black, he confidently asserts that the rule of replacement shows the probability that the next bean will be black is .91. (He also refers to this as an estimate of the proportion of beans in the jar that are black.)
Is this correct? If so, why?
ETA: And if he is wrong on the “successes can’t be equal to number of tries” technicality, is he nevertheless “right in spirit” so to speak?
If he is right that the probability is .91, my next question is, what does this probability mean? Why should I be, for example, “91% confident” that the next one will be black? Or if I shouldn’t be, then what does the probability mean otherwise?
Unfortunately, I’m over my head as well (and I thought you just needed a prompting)?
Warning: I will muck the following arguments up.
Sampling with replacement: that means you pull a pebble out of an urn, observe its color, then put it back in the urn. Then shake the urn up and repeat. Do something like that and if you get 12 black pebbles in row, it seems to me that you can say something about the share of black pebbles in the urn. It’s far more likely to be 90%+ than 10% or less, right? I mean if the urn has 10% black pebbles then the odds of picking out 12 black pebbles in a row would be 10^-12. If the urn were 90% black pebbles the odds of 12 consecutive picks would be about 28% - not likely (you should have come across at least one of a different color) but not something that can be ruled out.
[BTW: I initially thought you were a math guy. If my explanation is too basic/too ignorant, accept my apologies.]
Anyway, a Poisson distribution is applied to waiting times. It has more to do with statistics that probability though. I think.
I recall a forecast of the future probability of nuclear war, grounded solely on the observation that we’ve lived for 60+ years (almost 70 now) without a nuclear exchange. You can form other sorts of baseline probabilities in that manner.
I’ve heard of LaPlace, but am wholly unfamiliar with the LaPlace rule.
Not at all. I’m a mathy kind of personality, but I never ended up learning that much actual math. I gave up when it became the slightest bit difficult (as was my habit back then) when I took Calculus in high school–a move I regret to this day.
Okay, yes, that makes sense. I can see the argument that the next bean will probably be black. I’m still not sure it helps us calculate a probability that all the beans are black. What do you think? It seems plausible to me that if you iterate calculations on “next-beans” through 990 iterations, you’ll end up with exactly the same probability as if you’d simply calculated one divided by the total number of possible bean configurations. (That 91 percent for the first bean would go lower for the second bean (while we’re still sitting at 10 beans and calculating out the future) and lower still for the third, and so on.)
If you want to use Laplace’s rule, then you stated the original question wrong and incompletley. Laplace’s rule applies for independent random variables, so you need to specify that the draws are done with replacement. Secondly the simple version of Laplace’s rule that the probability that the next ball is black gives a probability of (s+1)/(n+2) if you’ve seen s successes in n trials. For you that would be 11/12. However, this rule only applies if both success and failure are possible, and if they are then the probability that all balls are black is 0 as otherwise failure is impossible.
Using Bayesian analysis to answer the question are all balls black is gong to give strange results. If you use any natural prior on the percentage, p, of balls that are black, it will be a continuous density function with a zero probability (but not zero density) at every value of p. So the probability that p = 1 exactly is zero.
I think the only way your question would likely have an answer under standard probability analysis would be to assume a prior distribution that number of ball in the urn is N (Perhaps a Poisson distribution) and for each N assume a prior for the number of black balls is 0, 1, 2, …, N. That would end up giving you a non-zero probability that all balls are black.
However, that would be quite a difficult problem, and I doubt very much it is the question you were asked to consider.
To be clear, it’s not me that wanted to use Laplace’s rule, it’s someone else that was telling me it applies here. It sounds like you’re saying it doesn’t really, since the scenario is ill-defined for that rule.
Some more questions.
To get real stark and clear, let’s focus on the following scenario (different from the one in the OP though very similar):
There is a jar with 1000 beans. You’re allowed to randomize it. You don’t know anything about what kinds of beans could be in the jar. You draw ten beans from the jar at random, replacing each one after it is drawn. All ten beans drawn are black.
A. What is the probability that the next bean will be black?
B. What is the probability that all the beans in the jar are black?
I would have thought the answer to B would be 1/2^990. Is that right?
For A, I would have thought the answer is very close to 50/50, if there is any meaningful answer at all. Is that right?
10 from 1000 is the same as 1 from 100. We know therefore that 1% of the beans are black, so there is at least a 1% chance that you pull a black bean next. I don’t think that you can infer any more about the beans than that.
Refining probabilities requires initial assumptions. In the “Laplace rule of succession” you assume that the portion of black balls is some unknown p, and that p is initially assumed to be uniformly distributed over [0,1]. (This might be called the default assumption of maximum ignorance or maximum entropy.)
If sampling with replacement, 10 blacks and 0 non-blacks yields a Laplacian estimate of (10+1)/(10+2) = .91667
Sampling without replacement, it might seem that you need to estimate another unknown: the initial total number of balls. However, the Laplace estimate will give the correct estimate in this case as well, regardless of the initial count, provided the urn has at least one more ball to draw and given the assumption that the portions were equally likely; i.e. that for a 99-ball urn, the black counts 0, 1, 2, …, 99 were each chosen initially with probability 1%.
Proof of this claim is left as an exercise for the Board’s mathematicians.
I know diddly squat about this problem, it’s been decades since I did statistics at any intensity, but here’s my thoughts:
This is maybe like sampling a bunch of people for an opinion poll. you can say “the result is X with a confidence of 4%”
Basically, you’ve drawn randomly. What percentage of the urn has to be black such that this sample is 3-sigma or less chance of happening with random draw?
For example, if the urn was 50% black, 50% other (with replacement, so 50-50 each time) this is like “what are the odds of flipping a coin 10 times and getting Heads each time?”
1/2^10 is 1/1024 Well below 3 sigma.
