I'm terrible at stats. Say there are 10 tokens in a container and you pull 9 of those tokens at random. You flip them over and find that these 9 tokens have a green dot on the bottom. I'm presuming that you can then claim some confidence that all the tokens in this set have green dots on the bottom. My best guess at the moment is to count all the permutations where you could get 9 green tokens in a row if there was 1 non-green token in the set and then compare that to all the permutations where you would get the 1 non-green token in the first 9 draws, which gave me 90%. If this isn't correct I'd like to get the correct methodology.

This isn't stats, it's philosophy of science or maths. Inductive reasoning suggests there's a 90% chance, but lots of educated people in the know will tell you that you don't have enough information. My personal take is 90%. There is no methodology. Anyone telling you there is either doesn't know what he's doing or is making assumptions that are outside of the information specified. imo

Philosophy, eh? I can do that. Can you think of any objections to the following:

I know that 90% of the time I'd get a non-green token if there was 1 in the set. By choosing the tokens with no bias (randomly) I ensure that it is necessary that I would receive a green token this percentage of the time. I also know that after 9 draws I have not received a non-green token. Since this would necessarily occur 10% of the time if there were a green token in the set it is then necessary that 90% of the time all tokens in the set are green.

I should leave this to the pros, because I may not be understanding this 100%, nor explaining it correctly, but I believe you've either missed out an assumption in your statement, or you're making an incorrect assumption based on the information you have so far. On the basis of seeing 9 green tokens, you're assuming that there's a connection between those ones and the remaining one, which isn't necessary.

There's no direct connection between the tokens seen and the token left, but I can definitely state that whenever there is a non-green token the "9 green token" scenario would occur 10% of the time. Since this scenario has occurred I can then conclude that 10% of the time there is a non-green token in the set and the other 90% the set is all-green. Or so the logic goes.

But before you pick up any tokens you have no reason to believe there's 0 or 10 green tokens or anything in between.

Please, someone come in and explain this in a way that's satisfactory, I'm mangling this one.

Right. This is a conclusion we can only draw after pulling 9 green tokens. The reason 1 non-green token comes into play is that this is the necessary condition to refute the claim "all tokens are green".

You might start by assuming the the probability of a token being green follows a uniform [0,1] distribution. Then adjust the density function of probability based on results.

Before you pull a token the density function of probability is shaped like a brick, the corners are (0,0), (0,1), (1,0), and (1,1). Pull one token green token and the density function will be shaped like a triangle whose corners are (0,0), (1,0), and (1,2). If the token wasn't green token the density function would be shaped like a triangle whose corners are (2,0), (0,0), and (1,0). If you pull 2 green tokens and 0 non-green tokens the density function will be shaped like part of a parabola. After n pulls (assuming all green tokens) the density function will be:

f(x) = C*x^n, if 0 < x < 1 and
f(x) = 0, otherwise.

We know that the integral of f(x) when evaluated from 0 to 1 is 1. We can use that to find C. C will equal n+1.

f(x) = (n+1)*x^(n+1)

The expected value of probability is the integral of x*f(x) evaluated from 0 to 1.

This is equal to the integral of x*(n+1)*x^n = (n+1)*x^(n+1) evaluated from 0 to 1 = (n+1)/(n+2).

If you have n successes and 0 failures the expected value of the probability of success is (n+1)/(n+2).

Ok, so that outputs the same answer as Baye's theorem (that may actually be Baye's theorem, I can't say). Could you give me a quick explanation of why that solution is preferable to using permutations? What is analysis using permutations not taking into account that this method does?

IMHO The answer is indeterminate. This problem is presented too much in vacuum. You can't make any reliable estimate of all the tokens being green-dot without having some known a priori probability of green-dot or non-green dot tokens. You have to know more about the tokens. As presented, we don't know that any non-green dot tokens even exist in the universe.

example 1. Say you know a pool of exactly billion tokens are known to exist and only one is non-green dot. Someone randomly scoops 10 into a container and the first 9 are green dot. How likely are they all green? 99% or more.

example 2. Say you know a pool of exactly billion tokens are known to exist and all but nine are non-green dot. Someone randomly scoops 10 into a container. By some chance the first nine are green dot. How likely are they all green dot? 0%.

In other words it depends on external circumstances which have not been specified. Comments are welcome.

I don't understand what you mean by using permutations. I'll take a stab and you can tell me if it's right.

We'll say A is equal to the number of ways we could have pulled 9 green tokens and 0 non-green tokens, given that there are exactly 9 green tokens and 1 non-green token. And B is equal to the number of ways we could have pulled 9 green tokens and 0 non-green tokens, given that there are exactly 10 green tokens and 0 non-green tokens. Then we can say the probability that there are exactly 9 green tokens and 1 non-green token is A/(A+B).

This assumes that each token had a 50% chance of being green. If this is true, then Bayes' Theorem should be applied to account for the fact that it was more likely to have exactly 9 green tokens than exactly 10 green tokens to begin with. And once you run it through Bayes' Theorem the probability that the 10th token is green will be 0.5.

Ah, I get it. Baye's theorem wouldn't apply here, because we're not presuming that each token has a 50% chance of being green (I believe he uses two known sets, one with 9/10 of a type and one with 10/10 of a type and you then have to determine which set you are drawing from after 9 draws).

I've been working on this a little bit. You can't make ANY direct claims about the final token is until it is drawn but you can make claims about your method for selecting sets which allow you to make indirect claims about the final token. We're looking for a non-green token. We know that whenever we look at a set with a non-green token in it it will be the last token selected 10% of the time. If we presuppose that we have no bias towards examining sets with green or non-green tokens then you will be examining a set with a non-green token 10% of the time after you pull 9 greens and an all-green set 90% of the time.

Wait, I can make a way stronger claim than that:

I am unsure as to whether there are any non-yellow tokens in any set. Whenever I choose from a set of 10 tokens every token must be either yellow or non-yellow. Any time there is a non-yellow token in a set of 10 I will get it on the last pick 10% of the time. Since I have picked 9 tokens out of the set and they are all yellow the entire set must be all yellow 90% of the time or more.

While I agree with your comment regarding priors, I don't understand why we can't assume the prior probability is 50/50. We have no information that would lead us to believe otherwise, so as far as we know, the probability of any token being green (vs. yellow) is chance. Given two options and no information about the prior probabilties of the two options, assigning each to a 50% chance seems the most reasonable to me.

Sherman

There is a 1/10 chance that this would've happened if there were 9 green tokens in the box. There is a 1, certain, chance this would've happened if there were 10 green tokens in the box.

If we start, before picking tokens, with an assumption that it is y times more likely to have 10 than 9 green tokens, it is now 10y times more likely. Depending on the context of the problem, it may be reasonable to assume both are equally likely (y=1). In that case, 10 green tokens is now 10 times more likely than 9 green tokens, which equates to 10/11 chance. (90% is slightly wrong).

This is pretty much Bayes' Theorem.

I am curious what the answers to this case would be:

Someone hands you four playing cards without saying anything at all about them and for no particular reason known to you. You see the frst three are Ah,Ad,Ac. What are the chances the last one is an Ace?
100% (since people group cards of same rank)?
90+% (Bayes or permutations)?
50%(either A or non-A, right) ?
49/52(3 of 52 left in deck)?
1/52 (could benay of 52 cards in a deck)?
indeterminate?
why?

95% of the tokens are Green dotted, +/- 5%.

With the deck of cards we have some prior knowledge of what else is in the universe. Or at least some assumptions. With the tokens, we didn't even know that the first 9 are green dotted, until we flipped them over. They could have been red dots, or no-dots, or each have a small picture of Beatrice Dahl on.

But we DO know there is one more possible token in the pot.

If we're assuming that each token has a 50% chance at being green then y = 1/10. Under this assumption the final token only has a 50% chance of being green.

It doesn't make sense to assume that each token has a 50% chance of being green and having 10 green tokens is just as likely as having 9 green tokens.

Unfortunately, statistics can't tell you the probability of the color of the last token. What it can tell you is this:

• If all the tokens are green, you would draw 9 green tokens 100% of the time.
• If there was one red token when you started, then there was a 10% chance of drawing all the green tokens first.

If you have other information available to you, then stats can may be able to answer your question. For example, if you knew that whoever filled the container would put in one red token 50% of the time and no red tokens the other 50% of the time, then you could answer the question by applying Bayes Theorem.

Hope that helps!