04 Jul 07
Originally posted by PalynkaWhy is that so?
The addition of a red ball is akin to saying "the oldest child is a boy", so this example isn't the best one for your point as it confirms the original solution.
You are claiming that given a bag that has two balls in and I tell you that one ball is red, the probability of the other being green is 2/3.
If I now tell you that I added the red one at some point then the probability changes to 1/2 ?
I am not saying you are wrong but just trying to understand it myself.
It seems that it has something to do with the fact that the one I am putting in could never have been green.
04 Jul 07
2 edits
Originally posted by twhiteheadYou are claiming that given a bag that has two balls in and I tell you that one ball is red, the probability of the other being green is 2/3.
Why is that so?
You are claiming that given a bag that has two balls in and I tell you that one ball is red, the probability of the other being green is 2/3.
If I now tell you that I added the red one at some point then the probability changes to 1/2 ?
I am not saying you are wrong but just trying to understand it myself.
It seems that it has something to do with the fact that the one I am putting in could never have been green.
I have to confess I'm struggling with the counter-intuitiveness in this one. You have to say "at least" one is red, not that one specific one is red.
Prob(RR)=1/4
Prob(GG)=1/4
Prob(~RR and ~GG)=1/2=Prob(RGu) u: unsorted
Prob(R*u)=3/4
=> Prob('RGu)/Prob(R*u)=2/3=Prob(the other one is green)
I think this is correct. 😕
When you say that you add one red, then you're specifying which one of the balls is red (e.g. the oldest child/the second ball), while if you say "at least one is red" you're not specifying which one.
Prob(RR)=1/4
Prob(GG)=1/4
Prob(GRu)=1/2=Prob(GR)+Prob(RG)=1/4+1/4
Prob(G*u)=3/4=Prob(GR)+Prob(RG)+Prob(GG)=3/4
Prob(G*)=1/2=Prob(GR)+Prob(GG)=1/4+1/4
Prob(*R)=1/2
Prob(G*|*R)=Prob(GR)/Prob(*R)=(1/4)/(1/2)=1/2
It is somewhat counter-intuitive (at least to me) but I think it's correct.
Edit - I think this example of yours is excellent as it illustrates clearly how refined the interpretation of new information must be.
04 Jul 07
I think that when I add the red ball to the bag, I am eliminating the possibility that I could have added a green ball and still resulted in a red and green ball in the bag.
The cool thing about it is that if I say anything about God, for example "God is bigger than a peanut" then I immediately reduce the probability that he exists.
Now if only I could actually get that green ball to change color simply by saying things about the red ball .....
04 Jul 07
5 edits
Originally posted by NemesioLet's use the notation from the Wikipedia page:
I think I got it 'at the buzzer' (right before you posted). The card example made it much clearer.
Let's see how Bayes works this out.
Nemesio
http://en.wikipedia.org/wiki/Prosecutor%27s_fallacy
Here's the first step.
Define proposition I: Leroy is innocent.
Next, establish P(I): The prior probability that Leroy is innocent; that is, the probability that a random lottery player is innocent before knowing anything else about him. Empircally, this is a very high probability -- very few people actually cheat at the lottery, so not knowing anything else about Leroy other than that you saw him buy a lottery ticket, your prior estimation of his innocence would be close to 1. Let's call it .95 in order to have a number to work with.
[As an interesting aside, note that as a matter of law, a criminal defendant in the United States must be presumed not guilty by the judge and jury until the evidence demonstrates otherwise. This high P(I) is the Bayesian component corresponding to that doctrine. If there were no such presumption, P(I) would be taken to be .5; if there were a presumption of guilt until the evidence demonstrates innocence, P(I) would be taken to be close to 0. And this latter being empirically false shows that legal systems that adopt that doctrine are doing things ass-backwards and implementing the prosecutor's fallacy as a matter of law.]
Make sense so far?
04 Jul 07
Originally posted by DoctorScribblesI'm good so far.
Make sense so far?
My only concern would be how to definitively establish what my propositions would be in a given
situation. That is, when I incorrectly guessed what my two propositions would be, it was because
I was not clearly asking the right question. So, while I think I will be able to follow this example
as you do it, I'm not sure that in another situation (say the forthcoming cookie one), that I would
be able to ask the proper question such that my two propositions would be self-evident.
But, please continue.
Nemesio
05 Jul 07
1 edit
Originally posted by NemesioIn general, you model the propositions like this:
I'm good so far.
My only concern would be how to definitively establish what my propositions would be in a given
situation. That is, when I incorrectly guessed what my two propositions would be, it was because
I was not clearly asking the right question. So, while I think I will be able to follow this example
as you do it, I'm not sure that in anothe ...[text shortened]... stion such that my two propositions would be self-evident.
But, please continue.
Nemesio
Let B be the proposition that has been revealed to be true.
Let A be the proposition whose conditional probability you would like to determine, conditioned on the fact that B.
In the Hold' em example, what was revealed to be true is that a random card was an Ace, and we wanted to know what the likelihood was that he held Ace-Queen given that revealed information.
As we'll see in the rest of this problem, what has been revealed to be true is that Leroy holds a winning ticket, and what we would like to know is the probability that he came about it legitimately given the fact that he has it.
05 Jul 07
Originally posted by DoctorScribblesVery, very helpful.
Let B be the proposition that has been revealed to be true.
Let A be the proposition whose conditional probability you would like to determine, conditioned on the fact that B.
I will try to deduce the propositions for the next problem.
Nemesio
05 Jul 07
9 edits
We've defined proposition I, and have established a reasonable value for P(I).
Now, define proposition E: Leroy holds a winning lottery ticket.
Before we get into establishing values for P(E), P(E|I), and finally using Bayes to compute the probability of interest P(I|E), let's take a look at the Bayes equation conceptually to make note of the relation among all of these values.
Since Bayes' Theorem is an equation with four terms, it can be examined in a variety of ways. Here, it is illuminating to rearrange the right-hand side of the equation to an equivalent representation:
P(I|E) = [P(E|I)/P(E)] * P(I)
Looking at this equation, it should be more clear that when one revises a prior probability about proposition I based on newly revealed evidence, the prior probability P(I) simply gets scaled by a fraction, namely P(E|I)/P(E).
If that fraction is greater than one, the revised probability P(I|E) is greater than the prior; the new evidence supports and strengthens the prior finding.
If that fraction is less than one, the revised probability is less than the prior; the new evidence weighs against the prior finding.
So, the task of revising a probability estimate based on new information amounts to nothing more than deciding which of P(E|I) and P(E) is greater, and by what degree.
In our case, note that since P(I) is very large, P(E) would have to be much, much greater than P(E|I) in order to yield a revised probability that indicates Leroy is very unlikely to be innocent. That is, it would have to be true that the probability that Leroy would hold a winning lottery ticket is much, much greater than the probability that Leroy would hold a winning lottery ticket given that he was in fact a legitimate winner in order for a reasonable person to think it very likely that Leroy has done something fraudulent.
Without yet getting into values for P(E) and P(E|I), does that framework make sense? Does it help to clarify conceptually what Bayes' Theorem says with regard to revising probability estimates in light of new evidence?
05 Jul 07
Originally posted by DoctorScribblesI disagree completely. The presumption of innocence has nothing to do with the actual probability that an accused is innocent, and your claim that the probability that he is innocent is high is I believe totally unfounded. In fact I suspect that you are committing the prosecutors fallacy yourself.
[As an interesting aside, note that as a matter of law, a criminal defendant in the United States must be presumed not guilty by the judge and jury until the evidence demonstrates otherwise. This high P(I) is the Bayesian component corresponding to that doctrine. If there were no such presumption, P(I) would be taken to be .5; if there were a presump ...[text shortened]... are doing things ass-backwards and implementing the prosecutor's fallacy as a matter of law.]
If a crime is committed and every person within a radius of say 100 killometres is rounded up and accused then the probability of their innocence is high, but if only one is accused then there is a much lower probability that he is innocent.
Similarly, if, in a lottery, on average only 1 in 1000 players cheat, and if 1000 tickets were issued and one person comes forward, then the probability of his innocence is not 999/1000 but much lower. It is infact dependent on the odds of actually winning, so if the odds of winning were very low then the odds of him being a cheater are very high.
05 Jul 07
19 edits
Originally posted by twhiteheadIn fact I suspect that you are committing the prosecutors fallacy yourself.
I disagree completely. The presumption of innocence has nothing to do with the actual probability that an accused is innocent, and your claim that the probability that he is innocent is high is I believe totally unfounded. In fact I suspect that you are committing the prosecutors fallacy yourself.
If a crime is committed and every person within a radius ...[text shortened]... is high, but if only one is accused then there is a much lower probability that he is innocent.
No, you are the one committing the prosecutor's fallacy. You are saying that being accused constitutes evidence against innocence, and that to accurately determine P(I|E) one ought not use a high value for P(I).
How can I be committing the prosecutor's fallacy when I haven't even yet given my estimate for P(I|E); when I haven't even yet said how likely I think it is that Leroy is a fraud given that he is holding a winning ticket?
your claim that the probability that he is innocent is high is I believe totally unfounded
I claim that P(I) is very high because that corresponds with empirical truth --- very, very few people that I pass on the street are murderers or lottery frauds. You deny this; you must believe that most people you pass on the street are in fact criminals. How many lottery frauds have you met?
If a crime is committed and every person within a radius of say 100 killometres is rounded up and accused then the probability of their innocence is high, but if only one is accused then there is a much lower probability that he is innocent.
This is false. Why do you think merely being accused constitutes evidence that ought to weigh against a person's innocence? Further, for the two probabilities that you describe, which do you think corresponds to P(I) in the model?
Similarly, if, in a lottery, on average only 1 in 1000 players cheat, and if 1000 tickets were issued and one person comes forward, then the probability of his innocence is not 999/1000 but much lower.
This is false. Show the calculations that lead to your finding.
It is infact dependent on the odds of actually winning, so if the odds of winning were very low then the odds of him being a cheater are very high.
This is the prosecutor's fallacy in all its glory! It's so blatant.
I'd be interested in hearing you state the prosecutor's fallacy in your own words, if you do in fact think that it is a fallacy.
05 Jul 07
Originally posted by DoctorScribblesMy mistake. I misunderstood what you were saying.
I claim that P(I) is very high because that corresponds with empirical truth --- very, very few people that I pass on the street are murderers or lottery frauds.
This is false. Why do you think merely being accused constitutes evidence that ought to weigh against a person's innocence? Further, for the two probabilities that you describe, which do you think corresponds to P(I) in the model?
Yes being accused does constitute evidence and does increase the probability you are guilty.
If you disagree then explain why the percentage of accused who are found guilty is higher than the percentage of all people that are actually guilty. For example if 5% of all people accused of murder are guilty of the crime then that far exceeds the percentage of people in the general population that are guilty of murder and far, far exceeds the percentage of people in the general population that are guilty of a particular murder.
A lottery is similar in that a cheater will always come forward and thus if someone comes forward it constitutes evidence towards him being a cheater.
By evidence I mean "information which causes an increase in the probability of guilt."
05 Jul 07
Originally posted by twhiteheadDo you agree that the probability of a randomly picked player being innocent is 999/1000?
Similarly, if, in a lottery, on average only 1 in 1000 players cheat, and if 1000 tickets were issued and one person comes forward, then the probability of his innocence is not 999/1000 but much lower.
Originally posted by DoctorScribbles
This is false. Show the calculations that lead to your finding.
Now suppose that the probability of a randomly picked player winning the lottery by normal means is 1 in a million. So the probability of having a winner is 1 in 1000. (Am I right?)
Surely the probability of someone coming forward and being innocent is then a combination of the two probabilities and will be lower than 999/1000. Sorry but I dont know how to calculate the actual probability.
05 Jul 07
4 edits
Originally posted by DoctorScribblesThis is false. Why do you think merely being accused constitutes evidence that ought to weigh against a person's innocence?
[b]In fact I suspect that you are committing the prosecutors fallacy yourself.
No, you are the one committing the prosecutor's fallacy. You are saying that being accused constitutes evidence against innocence, and that to accurately determine P(I|E) one ought not use a high value for P(I).
How can I be committing the prosecutor's fallacy w e prosecutor's fallacy in your own words, if you do in fact think that it is a fallacy.[/b]
I think he's alluding to something like Type I and Type II errors. You know very well that there is a trade-off between these two, so the choice of H0 may not be indifferent if the trade-off between Type I and Type II errors is fixed. I think you are correct because one can adapt that trade-off whenever you change H0 to guilty. (edit - I'm not sure if testing a hypothesis can be classified as Bayesian, though... can it?)
Scribs:This is simply false. Show your calculations that lead to your finding.
B: Player cheated
A: Player comes forward
P(B)=1/1000
P(A)=1/1000+999/1000*1/1000 (he either won legitimately with prob 1/1000 or he didn't and cheated)
P(A|B)=1 (or else he hasn't cheated)
P(B|A)=1/1000/(P(A)) = 0.5002... > 1/1000
Probability of his innocence is 1-P(B|A) < 999/1000
But I think it becomes apparent why this isn't similar to his previous example.
This is the prosecutor's fallacy in all its glory! It's so blatant.
Yes and no. The odds are higher than the average odds of cheating but are also lower than the odds of losing. (if odds of winning are lower than prob of cheating). How much depends on the relative weight between them.
P(won) = 1/1000 < P(B|A) < 999/1000 = P(lost )
The fallacy, in my opinion, is believing that the probability he's guilty is close to P(lost).
05 Jul 07
2 edits
Originally posted by twhiteheadBecause being accused is often empirically correlated with the existence of actual incriminating evidence, and in Bayesian analysis, prior probabilities must be established from a position of ignorance with regard to the evidence compelling you to revise them. In Bayesian analysis, all of that incriminating evidence is accounted for in the P(E) and P(E|I) terms, to be ultimately reflected in the revised P(I|E) term.
Yes being accused does constitute evidence and does increase the probability you are guilty.
If you disagree then explain why the percentage of accused who are found guilty is higher than the percentage of all people that are actually guilty. [/b]
If you establish a prior P(I) based upon the same evidence you use to construct P(E) and P(E|I), you are misusing Bayes' and the equation will no longer be true.
If you establish a prior P(I) based upon some evidence, you must exclude that evidence from E when establishing P(E), for E is always evidence external to any evidence used to construct P(I). In the case at hand, incorporating the evidence of the winning ticket into P(I) leaves you with no other evidence at all, which means you can't use Bayes' at all, which makes the fallacy even worse since the prosecutor refers to conditional probabilities that don't even exist in his model.
05 Jul 07
3 edits
Originally posted by PalynkaYou are miscalculating P(A). Recall that P(A) is a prior probability. You must do it from a position of ignorance about the truth of A itself.
B: Player cheated
A: Player comes forward
P(B)=1/1000
P(A)=1/1000+999/1000*1/1000 (he either won legitimately with prob 1/1000 or he didn't and cheated)
Rethink your formulation of A and your calculation of P(A) for a second.
A is "The player comes forward."
Then when you calculate P(A), your stated dichotomy assumes that he did in fact come forward, when that's the very thing whose probability you're supposed to be assessing.