Calling Out Cribs and Starrman

Originally posted by DoctorScribbles
Do you want to do the cookie example?

Or are you struggling with the idea that the theorem expresses?

Er...both?

I'm wondering if the sorts of examples I can understand aren't ones that are too simple that I don't
see the point. That is, I felt like I could figure out the aces/queens example without using a
complicated formula (I might have done it wrong, but I felt it was possible to figure out).

I suppose after I get the application, I'd like to see how it would be used in a complicated scenario
that I wouldn't be able to work out.

Nemesio

Originally posted by Nemesio
Er...both?

I'm wondering if the sorts of examples I can understand aren't ones that are too simple that I don't
see the point. That is, I felt like I could figure out the aces/queens example without using a
complicated formula (I might have done it wrong, but I felt it was possible to figure out).

I suppose after I get the application, I'd like ...[text shortened]... it would be used in a complicated scenario
that I wouldn't be able to work out.

Nemesio

Let's work through the Lottery Winner version of the Prosecutor's Fallacy then. That should be at a good level that you can both understand the solution and see the benefit of using Bayesian analysis instead of direct analysis.

http://en.wikipedia.org/wiki/Prosecutor%27s_fallacy


Here's the scenario. Leroy has just won the lottery and is accused of cheating. The fallacious prosecutor presents the following case:

It is only 1% likely that a person who buys a lottery ticket actually wins.
This means that 99% of players are actually losers.
Thus, it is 99% likely that Leroy actually lost, so it is 99% likely that he holds a winning ticket through fraudulent means.

If you can spot the fallacy in the prosecutor's argument, correctly and generally articulating that fallacy should in essence say the same thing that Bayes' theorem says.

Want to take a shot at it? Or do you want to see Bayes in action with the numbers?

Originally posted by DoctorScribbles
Here's the scenario. Leroy has just won the lottery and is accused of cheating. The fallacious prosecutor presents the following case:

It is only 1% likely that a person who buys a lottery ticket actually wins.
This means that 99% of players are actually losers.
Thus, it is 99% likely that Leroy actually lost, so it is 99% likely that he holds a winning ticket through fraudulent means.

Proposition A: There is only a 1% chance that Leroy hit the lottery.
Proposition B: 99% of players are losers.

a)P(A) = 1%
b)P(B) = 99%

...

Am I on the right track?

Nemesio

Originally posted by Nemesio
That is, I felt like I could figure out the aces/queens example without using a
complicated formula (I might have done it wrong, but I felt it was possible to figure out).

You're absolutely right. Any conditional probability that can be arrived at using Bayes can be arrived at independently as a prior probability using non-Bayesian methods. The Hold'em problem was a good illustration of that.

There are two main practical benefits of Bayesian analysis. One is that it sometimes offers a shortcut, since for any given model, the priors are typically easier to arrive at than the conditionals.

Another is that the relationship the theorem guarantees can serve as a sanity check when you have arrived at your probability via other means, since the relation must hold regardless of the method used to compute the four factors.

Originally posted by Nemesio
Proposition A: There is only a 1% chance that Leroy hit the lottery.
Proposition B: 99% of players are losers.

a)P(A) = 1%
b)P(B) = 99%

...

Am I on the right track?

Nemesio

Let's try to tackle it conceptually first.

Does anything about the prosecutor's argument strike you as fishy?

Originally posted by DoctorScribbles
Here's the scenario. Leroy has just won the lottery and is accused of cheating. The fallacious prosecutor presents the following case:

It is only 1% likely that a person who buys a lottery ticket actually wins.
This means that 99% of players are actually losers.
Thus, it is 99% likely that Leroy actually lost, so it is 99% likely that he holds a winning ticket through fraudulent means.

Originally posted by DoctorScribbles
Does anything about the prosecutor's argument strike you as fishy?

The move from the 99% that he lost to the point where it is 99% likely that he holds a fraudulent ticket.

?

Nemesio

Originally posted by Nemesio
Originally posted by DoctorScribbles
[b]Does anything about the prosecutor's argument strike you as fishy?

The move from the 99% that he lost to the point where it is 99% likely that he holds a fraudulent ticket.

?

Nemesio[/b]

Let us grant that the prosecutor is correct on that point, that if Leroy actually lost, then it is certainly true that he has forged a winning ticket. That is, strike that last clause from the argument if you wish. I just included it to give a criminal feel to the argument.

Does everything else seem sound to you? What I have in mind is something more conceptual rather than an attack on a specific step in the prosecutor's argument.

Here's a hint: probabilities are always measures of information.

Originally posted by DoctorScribbles
Does everything else seem sound to you? What I have in mind is something more conceptual rather than an attack on a specific step in the prosecutor's argument.

Here's a hint: probabilities are always measures of information.

Well, I'm not sure I can express it properly, but it seems intuitive that if there's a 1% chance that
someone could win, then he could have been that guy.

Nemesio

Originally posted by DoctorScribbles
Here's a hint: probabilities are always measures of information.

Hmm. We don't know the likelihood that he would make a false claim?

Originally posted by Nemesio
Well, I'm not sure I can express it properly, but it seems intuitive that if there's a 1% chance that
someone could win, then he could have been that guy.

Nemesio

Yes! We're heading in the right direction now.

Here's what strikes me as fishy about the argument: it presents no evidence, no information, to distinguish poor Leroy from anybody else who plays the lottery. Every legitimate winner would be guilty under this prosecutor's argument.

Now, here's where in the deduction the fallacy lies. It's an instance of comparing apples to oranges between the second and third lines.

In the second line, it is true that 99% of players are losers, given simply the information that they are players.

But in the third line, the prosecutor is neglecting to take into account all information available about this player, the one known to possess what appears to be a winning ticket.

It's akin to taking a card off a shuffled deck, seeing that it is black (imagine that!), and concluding that it is a Club with probability 1/4 since 1/4 of all cards are Clubs. As you're neglecting the information that this card is black, the prosecutor is neglecting the information that this guy holds an apparent winning ticket. It should be clear that once you see the card is black, you should compute the probability that it is a Club given that it is black, thereby revising your prior probability; similarly, the prosecutor should instead be computing the probability that Leroy is a loser given that he holds an apparent winning ticket.

In both cases, you're failing to revise your prior probability estimates in light of new information; you're remaining ignorant.

If that makes sense conceptually, then we'll show how Bayes resolves the issue.

Originally posted by DoctorScribbles
If that makes sense conceptually, then we'll show how Bayes resolves the issue.

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

Originally posted by Nemesio
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

Another cool probability scenario is this one:
http://en.wikipedia.org/wiki/Boy_or_Girl

If a family has two children and you know one is a boy then the probability that the other is a girl is 2/3

But if I tell you something about the boy, for example I tell you he is the youngest, then the probability changes to 1/2, or if I tell you that he is the oldest, the probability changes to 1/2, or if I tell you he is the tallest, the fattest, the cleverest etc etc.
So why does the probability of the other child being a girl get affected by what we know about the boy?
The real problem is that we often have the illusion that probability dictates reality and we somehow feel that if the probability for a child being a girl is changed then reality must have changed which is actually not the case, what changes is the information we have about the child.

Although the wikipedia article doesn't mention it, the real problem is that the 2/3 probability is based on information that isn't actually presented in the original problem in that it assigns an order to the children which I personally don't think is valid.

Originally posted by twhitehead
Although the wikipedia article doesn't mention it, the real problem is that the 2/3 probability is based on information that isn't actually presented in the original problem in that it assigns an order to the children which I personally don't think is valid.

Can you expand on this?

Particularly why don't you think its valid and what is the information that isn't presented in the original problem.

Originally posted by Palynka
Can you expand on this?

Particularly why don't you think its valid and what is the information that isn't presented in the original problem.

The 2/3 probability is based on the assumption that the items in question are sorted ie that Boy/Girl is different from Girl/Boy and that each of the following are equally likely:
Boy/Girl, Girl/Boy, and Boy/Boy.
The initial information however does not include that and thus we must assume that Boy/Girl, Girl/Boy are equivalent and that the Boy/Girl and Boy/Boy are equally likely thus making the required probability 1/2.
Another way to look at it is to say that the sorting is assumed but then we should also differentiate between the cases Boy1/Boy2 and Boy2/Boy1 again giving us a probability of 1/2.

To illustrate:
if I have a bag which contains two balls either of which may be red or green.
If I tell you that one is Red, then the probability of the other being green is 1/2 and not 2/3.
You may disagree. If so consider this scenario:
I have a bag containing 1 ball which may be red or green. The probability for it being green is clearly 1/2.
I add a red ball to the bag. I now have a bag that contains two balls and one of them is red. Could the addition of a red ball really change the probability of the original ball being green?

Think about it.

Originally posted by twhitehead
The 2/3 probability is based on the assumption that the items in question are sorted ie that Boy/Girl is different from Girl/Boy and that each of the following are equally likely:
Boy/Girl, Girl/Boy, and Boy/Boy.
The initial information however does not include that and thus we must assume that Boy/Girl, Girl/Boy are equivalent and that the Boy/Girl and ...[text shortened]... of a red ball really change the probability of the original ball being green?

Think about it.

The 2/3 probability is based on the assumption that the items in question are sorted ie that Boy/Girl is different from Girl/Boy and that each of the following are equally likely:
Boy/Girl, Girl/Boy, and Boy/Boy.

Not simply on the fact that they are sorted but that Girl/Boy (unsorted) is more likely than Boy/Boy or Girl/Girl. Sorting just helps to illustrate this fact as it becomes obvious why it is more likely. If you have doubts just think about the possibilities of BB or GG. They're both 1/4 each, so GB (unsorted) must be 1/2.

Could the addition of a red ball really change the probability of the original ball being green?
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.

Oops, have to go.