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Archive for May, 2010

From Chapter 12 of Eliezer Yudkowsky’s fanfiction Harry Potter and the Method of Rationality:

You couldn’t change history. But you could get it right to start with. Do something differently the first time around.

This whole business with seeking Slytherin’s secrets… seemed an awful lot like the sort of thing where, years later, you would look back and say, ‘And that was where it all started going wrong.’

And he would wish desperately for the ability to fall back through time and make a different choice…

It was a rather counterintuitive thought… but…

But he could, there was no rule saying he couldn’t, he could just pretend he’d never heard that little whisper. Let the universe go on in exactly the same way it would have if that one critical moment had never occurred. Twenty years later, that was what he would desperately wish had happened twenty years ago, and twenty years before twenty years later happened to be right now. Altering the distant past was easy, you just had to think of it at the right time.

I think this is a very powerful idea — trying to anticipate ahead of time what decisions you might regret.

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Status quo bias can sometimes be motivated by guilt.

As new scientific discoveries are made, it’s inevitable that we will be informed that some of our past actions — actions that we thought were beneficial, were actually harmful.  However, for some people this leads to strong motivated skepticism in defense of their earlier actions.  This seems to arise out of guilt.

For example, I’ve noticed that some parents who chose to use formula instead of breastfeeding get very defensive when they are informed of the scientific evidence in favor of the latter.

This all seems quite misplaced.  There is nothing wrong with taking actions that were defensible with the knowledge available at the time.  There is, however,  something wrong with encouraging people to ignore the current evidence in order to help you not have to admit the sub-optimality of your previous decisions.

Rationalizing by ignoring counterfactuals

A common way people rationalize opinions that result form guilt bias is with anecdotal evidence.

Examples:

  • After hearing about the evidence of the benefits of breastfeeding, someone might argue “Well, my kids were fed formula, and they turned out fine.”
  • After being informed that a particular public school isn’t very good, someone might respond “my  cousin Charlie went there, and he’s a successful attorney now.”

These examples ignore the relevant counterfactuals.  It’s not a question of whether your kids turned out okay.  The wrong comparison is to compare your child, who was fed formula, to children who might be viewed as not having turned out okay.  The relevant comparison is with your counterfactual children.  What would your children be like had they been breastfed?  Would they have had fewer illnesses, stomach aches, etc?  Would they live longer?  Would they be happier or smarter?  It does not matter if Charlie is more successful than some of his peers.  What matters is how Charlie compares with his counterfactual twin that attended a better school.  While we can’t know the answers to those questions, randomized trials do tell us about average differences between counterfactual groups.

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Time seems to pass by more quickly as we get older, when we are having fun and when we are busy.

Megan McArdle discusses some research on this (link):

New research may cast doubt on this, however:

Age accounted for four per cent of the variance in how quickly participants said the last ten years had passed and just one per cent of the perception of time’s speed in general. By contrast, how busy and rushed people reported feeling accounted for ten per cent of the variance in subjective speed of time. Consistent with this, women reported feeling more rushed than men, on average, and they perceived time to go by more quickly.

So it’s not youth that slows our time perception, but the fact that grammar and middle school involve incredibly long stretches of boredom.

Time is relatively shorter as you age (if you are 10 years old a year is 1/10th of your life, whereas if you are 50 years old it’s 1/50th of your life).  I suspect that is the primary direct effect of age. The direct age effect seems to be small.

It appears that busy-ness trumps age.

Imagine there are 3 continuous variables: age, busy-ness level (how busy/rushed you feel), fun level (how much fun you are having)

It would be interesting to see the joint distribution of these three, and how they relate to perceived passage of time.

If you are busy with something fun, does time seem to pass by more quickly than if you are busy with something you don’t enjoy?  Is the effect additive?

How does the joint distribution of (busy,fun) change with age?  When you are young, you have more leisure time but less freedom.  Do you have more fun?  There must be some research on this, but I didn’t find anything.

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In the US at least, it’s common for attorneys to ask negative questions.  For example:

  • Isn’t it true that you killed your wife?
  • At the time of the accident you were drunk, were you not?
  • Is it not true that you forged his signature?

If you haven’t killed your wife, weren’t drinking and didn’t forge a signature, the logically correct answer to those questions is yes.   However, a yes answer would be interpreted as a confession to the crimes.

Lawyers generally have a strong incentive to obtain clear, unambiguous information.  So why use negative questions?

Perhaps negative questions are viewed as more polite?

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Suppose 50% of people in a population have an asymptomatic form of cancer. None of them know if they have it. One of them is randomly selected and a diagnostic test is carried out (the result is not disclosed to them). If they don’t have cancer, they are woken up once. If they do have it, they are woken up 9 times (with amnesia-inducing drug administered each time, blah blah blah). Each time they are woken up, they are asked their credence (subjective probability) for cancer.

Imagine we do this repeatedly, randomly selecting people from a population that has 50% cancer prevalence.

World A: Everyone uses thirder logic

Someone without cancer will say: “I’m 90% sure I have cancer”

Someone with cancer will say: “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.” “I’m 90% sure I have cancer.”

Notice, everyone says they are 90% sure they have cancer, even though only 50% of them actually do.

Sure, the people who have cancer say it more often, but does that matter? At an awakening (you can pick one), people with cancer and people without are saying the same thing.

World B: Everyone uses halfer logic

Someone without cancer will say: “I’m 50% sure I have cancer”

Someone with cancer will say: “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.” “I’m 50% sure I have cancer.”

Here, half of the people have cancer, and all of them say they are 50% sure they have cancer.

My question: which world contains the more rational people?

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In professional sports, players are sometimes criticized for showboating (doing a touchdown dance, trash talking, etc.).  It’s considered poor sportsmanship.  We don’t want our kids to behave like that.

However, in professional hockey grown men get into fist fights during the game.  This seems like a much bigger offense.  I would rather have my child do a little dance after scoring, than punch one of the players on the other team. Yet, I haven’t heard many people complain about poor sportsmanship in hockey.  Why?

Possible explanations:

  • fighting has always been a part of hockey, but showboating in professional sports is a modern phenomenon (status quo bias)
  • trash talking is seen as low status, as people associate it with ghetto basketball games
  • we view boasting as worse than violence

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When it comes to probability, you should trust probability laws over your intuition.  Many people got the Monty Hall problem wrong because their intuition was bad.  You can get the solution to that problem using probability laws that you learned in Stats 101 — it’s not a hard problem.  Similarly, there has been a lot of debate about the Sleeping Beauty problem.  Again, though, that’s because people are starting with their intuition instead of letting probability laws lead them to understanding.

The Sleeping Beauty Problem

On Sunday she is given a drug that sends her to sleep. A fair coin is then tossed just once in the course of the experiment to determine which experimental procedure is undertaken. If the coin comes up heads, Beauty is awakened and interviewed on Monday, and then the experiment ends. If the coin comes up tails, she is awakened and interviewed on Monday, given a second dose of the sleeping drug, and awakened and interviewed again on Tuesday. The experiment then ends on Tuesday, without flipping the coin again. The sleeping drug induces a mild amnesia, so that she cannot remember any previous awakenings during the course of the experiment (if any). During the experiment, she has no access to anything that would give a clue as to the day of the week. However, she knows all the details of the experiment.

Each interview consists of one question, “What is your credence now for the proposition that our coin landed heads?”

Two popular solutions have been proposed: 1/3 and 1/2

The 1/3 solution

From wikipedia:

Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3.

Yes, it’s true that only in a third of cases would heads precede her awakening.

Radford Neal (a statistician!) argues that 1/3 is the correct solution.

This [the 1/3] view can be reinforced by supposing that on each awakening Beauty is offered a bet in which she wins 2 dollars if the coin lands Tails and loses 3 dollars if it lands Heads. (We suppose that Beauty knows such a bet will always be offered.) Beauty would not accept this bet if she assigns probability 1/2 to Heads. If she assigns a probability of 1/3 to Heads, however, her expected gain is 2 × (2/3) − 3 × (1/3) = 1/3, so she will accept, and if the experiment is repeated many times, she will come out ahead.

Neal is correct (about the gambling problem).

These two arguments for the 1/3 solution appeal to intuition and make no obvious mathematical errors.   So why are they wrong?

Let’s first start with probability laws and show why the 1/2 solution is correct. Just like with the Monty Hall problem, once you understand the solution, the wrong answer will no longer appeal to your intuition.

The 1/2 solution

P(Beauty woken up at least once| heads)=P(Beauty woken up at least once | tails)=1.  Because of the amnesia, all Beauty knows when she is woken up is that she has woken up at least once.  That event had the same probability of occurring under either coin outcome.  Thus, P(heads | Beauty woken up at least once)=1/2.  You can use Bayes’ rule to see this if it’s unclear.

Here’s another way to look at it:

If it landed heads then Beauty is woken up on Monday with probability 1.

If it landed tails then Beauty is woken up on Monday and Tuesday.  From her perspective, these days are indistinguishable.  She doesn’t know if she was woken up the day before, and she doesn’t know if she’ll be woken up the next day.  Thus, we can view Monday and Tuesday as exchangeable here.

A probability tree can help with the intuition (this is a probability tree corresponding to an arbitrary wake up day):

If Beauty was told the coin came up heads, then she’d know it was Monday.  If she was told the coin came up tails, then she’d think there is a 50% chance it’s Monday and a 50% chance it’s Tuesday.  Of course, when Beauty is woken up she is not told the result of the flip, but she can calculate the probability of each.

When she is woken up, she’s somewhere on the second set of branches.  We have the following joint probabilities: P(heads, Monday)=1/2; P(heads, not Monday)=0; P(tails, Monday)=1/4; P(tails, Tuesday)=1/4; P(tails, not Monday or Tuesday)=0.  Thus, P(heads)=1/2.

Where the 1/3 arguments fail

The 1/3 argument says with heads there is 1 interview, with tails there are 2 interviews, and therefore the probability of heads is 1/3.  However, the argument would only hold if all 3 interview days were equally likely.  That’s not the case here. (on a wake up day, heads&Monday is more likely than tails&Monday, for example).

Neal’s argument fails because he changed the problem. “on each awakening Beauty is offered a bet in which she wins 2 dollars if the coin lands Tails and loses 3 dollars if it lands Heads.”  In this scenario, she would make the bet twice if tails came up and once if heads came up.  That has nothing to do with probability about the event at a particular awakening.  The fact that she should take the bet doesn’t imply that heads is less likely.  Beauty just knows that she’ll win the bet twice if tails landed.  We double count for tails.

Imagine I said “if you guess heads and you’re wrong nothing will happen, but if you guess tails and you’re wrong I’ll punch you in the stomach.”  In that case, you will probably guess heads.  That doesn’t mean your credence for heads is 1 — it just means I added a greater penalty to the other option.

Consider changing the problem to something more extreme.  Here, we start with heads having probability 0.99 and tails having probability 0.01.  If heads comes up we wake Beauty up once.  If tails, we wake her up 100 times.  Thirder logic would go like this:  if we repeated the experiment 1000 times, we’d expect her woken up 990 after heads on Monday, 10 times after tails on Monday (day 1), 10 times after tails on Tues (day 2),…., 10 times after tails on day 100.  In other words, ~50% of the cases would heads precede her awakening. So the right answer for her to give is 1/2.

Of course, this would be absurd reasoning.  Beauty knows heads has a 99% chance initially.  But when she wakes up (which she was guaranteed to do regardless of whether heads or tails came up), she suddenly thinks they’re equally likely?  What if we made it even more extreme and woke her up even more times on tails?

Implausible consequence of 1/2 solution?

Nick Bostrom presents the Extreme Sleeping Beauty problem:

This is like the original problem, except that here, if the coin falls tails, Beauty will be awakened on a million subsequent days. As before, she will be given an amnesia drug each time she is put to sleep that makes her forget any previous awakenings. When she awakes on Monday, what should be her credence in HEADS?

He argues:

The adherent of the 1/2 view will maintain that Beauty, upon awakening, should retain her credence of 1/2 in HEADS, but also that, upon being informed that it is Monday, she should become extremely confident in HEADS:
P+(HEADS) = 1,000,001/1,000,002

This consequence is itself quite implausible. It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads.

It’s correct that, upon awakening on Monday (and not knowing it’s Monday), she should retain her credence of 1/2 in heads.

However, if she is informed it’s Monday, it’s unclear what she conclude.  Why was she informed it was Monday?  Consider two alternatives.

Disclosure process 1:  regardless of the result of the coin toss she will be informed it’s Monday on Monday with probability 1

Under disclosure process 1, her credence of heads on Monday is still 1/2.

Disclosure process 2: if heads she’ll be woken up and informed that it’s Monday.  If tails, she’ll be woken up on Monday and one million subsequent days, and only be told the specific day on one randomly selected day.

Under disclosure process 2, if she’s informed it’s Monday, her credence of heads is 1,000,001/1,000,002.  However, this is not implausible at all.  It’s correct.  This statement is misleading: “It is, after all, rather gutsy to have credence 0.999999% in the proposition that an unobserved fair coin will fall heads.”  Beauty isn’t predicting what will happen on the flip of a coin, she’s predicting what did happen after receiving strong evidence that it’s heads.

ETA (5/9/2010 5:38AM)

If we want to replicate the situation 1000 times, we shouldn’t end up with 1500 observations.  The correct way to replicate the awakening decision is to use the probability tree I included above. You’d end up with expected cell counts of 500, 250, 250, instead of 500, 500, 500.

Suppose at each awakening, we offer Beauty the following wager:  she’d lose $1.50 if heads but win $1 if tails.  She is asked for a decision on that wager at every awakening, but we only accept her last decision. Thus, if tails we’ll accept her Tuesday decision (but won’t tell her it’s Tuesday). If her credence of heads is 1/3 at each awakening, then she should take the bet. If her credence of heads is 1/2 at each awakening, she shouldn’t take the bet.  If we repeat the experiment many times, she’d be expected to lose money if she accepts the bet every time.

The problem with the logic that leads to the 1/3 solution is it counts twice under tails, but the question was about her credence at an awakening (interview).

ETA (5/10/2010 10:18PM ET)

Suppose this experiment were repeated 1,000 times. We would expect to get 500 heads and 500 tails. So Beauty would be awoken 500 times after heads on Monday, 500 times after tails on Monday, and 500 times after tails on Tuesday. In other words, only in a third of the cases would heads precede her awakening. So the right answer for her to give is 1/3.

Another way to look at it:  the denominator is not a sum of mutually exclusive events.  Typically we use counts to estimate probabilities as follows:  the numerator is the number of times the event of interest occurred, and the denominator is the number of times that event could have occurred.

For example, suppose Y can take values 1, 2 or 3 and follows a multinomial distribution with probabilities p1, p2 and p3=1-p1-p2, respectively.   If we generate n values of Y, we could estimate p1 by taking the ratio of #{Y=1}/(#{Y=1}+#{Y=2}+#{Y=3}). As n goes to infinity, the ratio will converge to p1.   Notice the events in the denominator are mutually exclusive and exhaustive.  The denominator is determined by n.

The thirder solution to the Sleeping Beauty problem has as its denominator sums of events that are not mutually exclusive.  The denominator is not determined by n.  For example, if we repeat it 1000 times, and we get 400 heads, our denominator would be 400+600+600=1600 (even though it was not possible to get 1600 heads!).  If we instead got 550 heads, our denominator would be 550+450+450=1450.  Our denominator is outcome dependent, where here the outcome is the occurrence of heads.  What does this ratio converge to as n goes to infinity?  I surely don’t know.  But I do know it’s not the posterior probability of heads.

explanation for the title of this post (link)

Long debate about this here

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