Comment author: Irgy 30 July 2015 05:02:10AM *  1 point [-]

To my view, the 1/36 is "obviously" the right answer, what's interesting is exactly how it all went wrong in the other case. I'm honestly not all that enlightened by the argument given here nor in the links. The important question is, how would I recognise this mistake easily in the future? The best I have for the moment is "don't blindly apply a proportion argument" and "be careful when dealing with infinite scenarios even when they're disguised as otherwise". I think the combination of the two was required here, the proportion argument failed because the maths which normally supports it couldn't be used without at some point colliding with the partly-hidden infinity in the problem setup.

I'd be interested in more development of how this relates to anthropic arguments. It does feel like it highlights some of the weaknesses in anthropic arguments. It seems to strongly undermine the doomsday argument in particular. My take on it is that it highlights the folly of the idea that population is endlessly exponentially growing. At some point that has to stop regardless of whether it has yet already, and as soon as you take that into account I suspect the maths behind the argument collapses.

Edit: Just another thought. I tried harder to understand your argument and I'm not convinced it's enough. Have you heard of ignorance priors? They're the prior you use, in fact the prior you need to use, to represent a state of no knowledge about a measurement other than an invariance property which identifies the type of measurement it is. So an ignorance prior for a position is constant, and for a scale is 1/x, and for a probability has been at least argued to be 1/x(1-x). These all have the property that their integral is infinite, but they work because as soon as you add some knowledge and apply Bayes rule the result becomes integrable. These are part of the foundations of Bayesian probability theory. So while I agree with the conclusion, I don't think the argument that the prior is unnormalisable is sufficient proof.

Comment author: ksvanhorn 05 August 2015 04:48:02PM *  1 point [-]

Actually, no, improper priors such as you suggest are not part of the foundations of Bayesian probability theory. It's only legitimate to use an improper prior if the result you get is the limit of the results you get from a sequence of progressively more diffuse priors that tend to the improper prior in the limit. The Marginalization Paradox is an example where just plugging in an improper prior without considering the limiting process leads to an apparent contradiction. My analysis (http://ksvanhorn.com/bayes/Papers/mp.pdf) is that the problem there ultimately stems from non-uniform convergence.

I've had some email discussions with Scott Aaronson, and my conclusion is that the Dice Room scenario really isn't an appropriate metaphor for the question of human extinction. There are no anthropic considerations in the Dice Room, and the existence of a larger population from which the kidnap victims are taken introduces complications that have no counterpart when discussing the human extinction scenario.

You could formalize the human extinction scenario with unrealistic parameters for growth and generational risk as follows:

  • Let n be the number of generations for which humanity survives.

  • The population in each generation is 10 times as large as the previous generation.

  • There is a risk 1/36 of extinction in each generation. Hence, P(n=N+1 | n >= n) = 1/36.

  • You are a randomly chosen individual from the entirety of all humans who will ever exist. Specifically, P(you belong to generation g) = 10^g / N, where N is the sum of 10^t for 1 <= t <= n.

Analyzing this problem, I get

P(extinction occurs in generation t | extinction no earlier than generation t) = 1/36

P(extinction occurs in generation t | you are in generation t) = about 9/10

That's a vast difference depending on whether or not we take into account anthropic considerations.

The Dice Room analogy would be if the madman first rolled the dice until he got snake-eyes, then went out and kidnapped a bunch of people, randomly divided them into n batches, each 10 times larger than the previous, and murdered the last batch. This is a different process than what is described in the book, and results in different answers.

Comment author: Slider 28 July 2015 10:13:14PM 3 points [-]

The madman murders only almost always. It is possible but vanishingly unlikely that he just never rolls snake eyes (or he runs outside of the total population with the growth so he can't get a full patch). Option 1 doesn't care whether the doom ultimately happens while option 2 assumes that the doom will happen.

The proper enlish version of option two would be "Given that the dice came up snake eyes and that you were kidnapped at some point what is the probabilty that it did so while you were kidnapped?". Notice also that this is independent off what dice readings result in doom. That is if the world is only saved on snake eyes the chance is still "only" 9/10.

Comment author: ksvanhorn 29 July 2015 06:32:39PM 0 points [-]

Note that

P(you are in batch t | murders batch t & you are kidnapped)

cannot be 9/10 for all t; in fact, this probability must go to 0 in the limit as t -> infinity, regardless of what prior you use.

Comment author: Jiro 29 July 2015 02:29:15PM *  6 points [-]

Go to a casino. Bet $1 on something with a 50% chance of winning. If you win, you have won $1; try again. If you lose, double your bet size (which means that winning will leave you having won $1 total over the sequence of doubled bets) and repeat.

One argument says that in the long run, you will come out a winner, because every bet you make is part of a sequence and at the end of that sequence, you are $1 richer. Another argument says that in the long run, you will only break even, because each bet has a 50% chance of winning and a 50% chance of losing the same amount of money.

Of course, the answer is that you can't increase your bet infinitely, and when you stop increasing your bet, the statistical loss at the point where you stop increasing your bet exactly makes up for the statistical win all the other times you finished the sequence and won $1.

Furthermore, if you could increase your bet infinitely, this problem wouldn't happen, but if you could increase your bet infinitely, the expectation isn't well defined, because you are trying to compute it for a non-converging infinite series.

All this problem is is the same idea applied to probability of death instead of expectation of win. If the madman ever runs out of people, the overall probability depends exactly on what the madman does when he runs out of people (since it's not as well defined as it is for bets). If the madman never runs out of people, the probability involves a non-converging infinite series and so is not well defined.

If this is a metaphor for extinction, then when the madman runs out of people, he keeps rolling the dice on the remaining people until it eventually comes up snake eyes, in which case the chance of extinction is 100%. On the other hand, they can last arbitrarily long given an arbitrarily small probability of extinction.

Comment author: ksvanhorn 29 July 2015 06:11:04PM 1 point [-]

An earlier version of my analysis (the previous blog post) looked at the case of finite n and found, as you suggest, that the possibility of running out of people to kidnap is an important consideration. You can choose the number of batches n to be so large that it is virtually certain a priori that the madman will eventually murder:

P(eventually murders) = 1 - epsilon for some small epsilon

However, it turns out that conditioning on the fact that you are kidnapped changes the probability dramatically:

P(eventually murders | you are kidnapped) = about 10/9 * 1/36

The reason for this is that there are about 9 times as many people in the final batch as in all other batches combined, so the fact that you are kidnapped is strong evidence that the madman is on his last batch of potential victims.

Comment author: ksvanhorn 11 March 2014 12:10:46AM 4 points [-]

I'm not sure that "jack of all trades" is a helpful identity, given the known benefits of economic specialization. Remember the origin of that term: "Jack of all trades, and master of none." It's often more useful to be really, really good at one thing and trade for what you need in other areas.

It can often be useful to have a "T-shaped" expertise, though: some level of familiarity with a wide variety of topics, and deep expertise in one area. The cross bar of the T helps you when your existing expertise and skills are not enough -- you know enough to find someone who can help you, or to know what new skills / knowledge you need to pick up. (Or, perhaps more importantly, you know what you don't know.)

Comment author: ksvanhorn 20 September 2013 04:26:41AM 1 point [-]

Do you know why this book is on the MIRI course list? What is the connection to Friendly AI?

Comment author: ksvanhorn 03 September 2013 07:12:11PM 0 points [-]

I've certainly found this to be a useful strategy when dealing with complicated problems in software development. Sometimes a problem is just too big, and I can't quite see how all the pieces need to fit together. If I allow myself to leave some important design problems unresolved while I work on the parts that I do understand well enough to write, I often find that the other pieces then fall into place straightforwardly.

Comment author: JQuinton 13 August 2013 05:37:02PM 12 points [-]

"Absence of evidence isn't evidence of absence" is such a ubiquitous cached thought in rationalist communities (that I've been involved with) that its antithesis was probably the most important thing I learned from Bayesianism.

Comment author: ksvanhorn 22 August 2013 06:56:02PM 4 points [-]

I find it interesting that Sir Arthur Conan Doyle, the author of the Sherlock Holmes stories, seems to have understood this concept. In his story "Silver Blaze" he has the following conversation between Holmes and a Scotland Yard detective:

Gregory (Scotland Yard detective): "Is there any other point to which you would wish to draw my attention?"

Holmes: "To the curious incident of the dog in the night-time."

Gregory: "The dog did nothing in the night-time."

Holmes: "That was the curious incident."

Comment author: ksvanhorn 26 September 2012 06:49:14PM 0 points [-]

There are many publicly available data sets and plenty of opportunities to mine data online, yet we see little if any original analysis based on them here. We either don't have norms encouraging this or we don't have enough people comfortable with statistics doing so.

In my case, I'm comfortable with statistics but don't know where to find the data for questions that interest me. The fact that much research is nearly inaccessible if you're not affiliated with a university or other large institution is also a problem.

Comment author: Benquo 20 June 2012 03:45:28AM 6 points [-]

Your understanding of mathematical expectation seems accurate, though the wording could be simplified a bit. I don't think that you need the "many worlds" style exposition to explain it.

One common way of thinking of expected values is as a long-run average. So If I keep playing a game with an expected loss of $10, that means that in the long run it becomes more and more probable that I'll lose an average of about $10 per game.

Comment author: ksvanhorn 20 June 2012 07:08:32PM 3 points [-]

You could write a whole book about what's wrong with this "long-run average" idea, but E. T. Jaynes already did: Probability Theory: The Logic of Science. The most obvious problem is that it means you can't talk about the expected value of a one-off event. I.e., if Dick is pondering the expected value of (time until he completes his doctorate) given his specific abilities and circumstances... well, he's not allowed to if he's a frequentist who treats probabilities and expected values as long-run averages; there is no ensemble here to take the average of.

Expected values are weighted averages, so I would recommend explaining expected values in two parts:

  • Explain the idea of probabilities as degree of confidence in an outcome (the Bayesian view);

  • Explain the idea of a weighted average, and note that the expected value is a weighted average with outcome probabilities as the weights.

You could explain the idea of a weighted average using the standard analogy of balancing a rod with weights of varying masses attached at various points, and note that larger masses "pull the balance point" towards themselves more strongly than do smaller masses.

Comment author: ksvanhorn 10 June 2012 04:55:28AM 1 point [-]

Soldiers in the American military are, of course, an untouchable target

I know I'm playing with fire, but... touch. (Warning: deliberately inflammatory polemic.)

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