Bayes' Theorem Illustrated (My Way)

3rd Jun 2010

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New Comment

195 comments, sorted by Click to highlight new comments since: Today at 4:14 AM

Some comments are truncated due to high volume. (⌘F to expand all)

This is great. I hope other people aren't hesitating to make posts because they are too "elementary". Content on Less Wrong doesn't need to be advanced; it just needs to be Not Wrong.

In my defense, we've had other elementary posts before, and they've been found useful; plus, I'd really like this to be online somewhere, and it might as well be here.

It's quite interesting that people feel a need to defend themselves in advance when they think their post is elementary, but almost never feel the same obligation when the post is supposedly too hard, or off-topic, or inappropriate for other reason. More interesting given that we all have probably read about the illusion of transparency. Still, seems that inclusion of this sort of signalling is irresistible, although (as the author's own defense has stated) the experience tells us that such posts usually meet positive reception.

As for my part of signalling, this comment was not meant as a criticism. However I find it more useful if people defend themselves only after they are criticised or otherwise attacked.

614y

It's conceivable that people being nervous about posts on elementary subjects means that they're more careful with elementary posts, thus explaining some fraction of the higher quality.

514y

It is possible. However I am not sure that the elementary posts have higher average quality than other posts, if the comparison is even possible. Rather, what strikes me is that you never read "this is specialised and complicated, but nevertheless I decided to post it here, because..."
There still apparently is a perception that it's a shame to write down some relatively simple truth, and if one wants to, one must have a damned good reason. I can understand the same mechanism in peer reviewed journals, where the main aim is to impress the referee and demonstrate the author's status, which increases chances to get the article published. (If the article is trivial, it at least doesn't do harm to point out that the autor knows it.) Although this practice was criticised here for many times, it seems that it is really difficult to overcome it. But at least we don't shun posts because they are elementary.

A piece of advice I heard a long time ago, and which has sometimes greatly alleviated the boredom of being stuck in a conference session, is this: If you're not interested in what the lecturer is talking about, study the lecture as a demonstration of how to give a lecture.

By this method even an expert can learn from a skilful exposition of fundamentals.

514y

That's more or less the default visualization; unfortunately it hasn't proved particularly helpful to me, or at least not as much as the visualization presented here -- hence the need for this post.
The method presented in the post has a discrete, sequential, "flip-the-switch" feel to it which I find very suited to my style of thinking. If I had known how, I would have presented it as an animation.

I don't have a very advanced grounding in math, and I've been skipping over the technical aspects of the probability discussions on this blog. I've been reading lesswrong by mentally substituting "smart" for "Bayesian", "changing one's mind" for "updating", and having to vaguely trust and believe instead of rationally understanding.

Now I absolutely get it. I've got the key to the sequences. Thank you very very much!

Sigh. Of course I upvoted this, but...

The first part, the abstract part, was a joy to read. But the Monty Hall part started getting weaker, and the Two Aces part I didn't bother reading at all. What I'd have done differently if your awesome idea for a post came to me first: remove the jarring false tangent in Monty Hall, make all diagrams identical in style to the ones in the first part (colors, shapes, fonts, lack of borders), never mix percentages and fractions in the same diagram, use cancer screening as your first motivating example, Monty Hall as the second example, Two Aces as an exercise for the readers - it's essentially a variant of Monty Hall.

Also, indicate more clearly in the Monty Hall problem statement that whenever the host can open two doors, it chooses each of them with probability 50%, rather than (say) always opening the lower-numbered one. Without this assumption the answer could be different.

Sorry for the criticisms. It's just my envy and frustration talking. Your post had the potential to be so completely awesome, way better than Eliezer's explanation, but the tiny details broke it.

this does seem like the type of article that should be a community effort.. perhaps a wiki entry?

7[anonymous]14y

2nd'd. Here I would like to encourage you (=komponisto) to do a non-minor edit (although those are seldom here), to give the post the polish it deserves.

This is a fantastic explanation (which I like better than the 'simple' explanation retired urologist links to below), and I'll tell you why.

You've transformed the theorem into a spatial representation, which is always great - since I rarely use Bayes Theorem I have to essentially 'reconstruct' how to apply it every time I want to think about it, and I can do that much easier (and with many fewer steps) with a picture like this than with an example like breast cancer (which is what I would do previously).

Critically, you've represented the WHOLE problem visually - all I have to do is picture it in my head and I can 'read' directly off of it, I don't have to think about any other concepts or remember what certain symbols mean. Another plus, you've included the actual numbers used for maximum transparency into what transformations are actually taking place. It's a very well done series of diagrams.

If I had one (minor) quibble, it would be that you should represent the probabilities for various hypotheses occuring visually as well - perhaps using line weights, or split lines like in this diagram.

But very well done, thank you.

(edit: I'd also agree with cousin_it that the first half of the post is the stronger part. The diagrams are what make this so great, so stick with them!)

I don't get it really. I mean, I get the method, but not the formula. Is this useful for anything though?

Also, a simpler method of explaining the Monty Hall problem is to think of it if there were more doors. Lets say there were a million (thats alot ["a lot" grammar nazis] of goats.) You pick one and the host elliminates every other door except one. The probability you picked the right door is one in a million, but he had to make sure that the door he left unopened was the one that had the car in it, *unless* you picked the one with a car in it, which is a one in a million chance.

514y

It might help to read the sequences, or just read Jaynes. In particular, one of the central ideas of the LW approach to rationality is that when one encounters new evidence one should update one's belief structure based on this new evidence and your estimates using Bayes' theorem. Roughly speaking, this is in contrast to what is sometimes described as "traditional rationalism" which doesn't emphasize updating on each piece of evidence but rather on updating after one has a lot of clearly relevant evidence.
Edit: Recommendation of Map-Territory sequence seems incorrect. Which sequence is the one to recommend here?

214y

How to Actually Change your Mind and Mysterious Answers to Mysterious Questions

014y

Updating your belief based on different pieces of evidence is useful, but (and its a big but) just believing strange things based on imcomplete evidence is bad. Also, this neglects the fact of time. If you had an infinite amount of time to analyze every possible scenario, you could get away with this, but otherwise you have to just make quick assumptions. Then, instead of testing wether these assumptions are correct, you just go with them wherever it takes you. If only you could "learn how to learn" and use the Bayesian method on different methods of learning; eg, test out different heuristics and see which ones give the best results. In the end, you find humans already do this to some extent and "traditional rationalism" and science is based off of the end result of this method. Is this making any sense? Sure, its useful in some abstract sense and on various math problems, but you can't program a computer this way, nor can you live your life trying to compute statistics like this in your head.
Other than that, I can see different places where this would be useful.

714y

And so it is written, "Even if you cannot do the math, knowing that the math exists tells you that the dance step is precise and has no room in it for your whims."

214y

I may not be the best person to reply to this given that I a) am much closer to being a traditional rationalist than a Bayesian and b) believe that the distinction between Bayesian rationalism and traditional rationalism is often exaggerated. I'll try to do my best.
So how do you tell if a belief is strange? Presumably if the evidence points in one direction, one shouldn't regard that belief as strange. Can you give an example of a belief that should considered not a good belief to have due to strangeness that one could plausibly have a Bayesian accept like this?
Well yes, and no. The Bayesian starts with some set of prior probability estimates, general heuristics about how the world seems to operate (reductionism and locality would probably be high up on the list). Everyone has to deal with the limits on time and other resources. That's why for example, if someone claims that hopping on one foot cures colon cancer we don't generally bother testing it. That's true for both the Bayesian and the traditionalist.
I'm curious as to why you claim that you can't program a computer this way. For example, automatic Bayesian curve fitting has been around for almost 20 years and is a useful machine learning mechanism. Sure, it is much more narrow than applying Bayesianism to understanding reality as a whole, but until we crack the general AI problem, it isn't clear to me how you can be sure that that's a fault of the Bayesian end and not the AI end. If we can understand how to make general intelligences I see no immediate reason why one couldn't make them be good Bayesians.
I agree that in general, trying to generally compute statistics in one's head is difficult. But I don't see why that rules out doing it for the important things. No one is claiming to be a perfect Bayesian. I don't think for example that any Bayesian when walking into a building tries to estimate the probability that the building will immediately collapse. Maybe they do if the building is very rickety l

114y

Guessing here you mean locality instead of nonlocality?

014y

Yes, fixed thank you.

-7[anonymous]14y

414y

Only for the stated purpose of this website - to be "less wrong"! :) Quoting from Science Isn't Strict Enough:
As for the rest of your comment: I completely agree! That was actually the explanation that the OP, komponisto, gave to me to get Bayesianism (edit: I actually mean "the idea that probability theory can be used to override your intuitions and get to correct answers") to "click" for me (insofar as it has "clicked"). But the way that it's represented in the post is really helpful, I think, because it eliminates even the need to imagine that there are more doors; it addresses the specifics of that actual problem, and you can't argue with the numbers!

414y

Quite a bit! (A quick Google Scholar search turns up about 1500 papers on methods and applications, and there are surely more.)
The formula tells you how to change your strength of belief in a hypothesis in response to evidence (this is 'Bayesian updating', sometimes shortened to just 'updating'). Because the formula is a trivial consequence of the definition of a conditional probability, it holds in any situation where you can quantify the evidence and the strength of your beliefs as probabilities. This is why many of the people on this website treat it as the foundation of reasoning from evidence; the formula is very general.
Eliezer Yudkowsky's Intuitive Explanation of Bayes' Theorem page goes into this in more detail and at a slower pace. It has a few nice Java applets that you can use to play with some of the ideas with specific examples, too.

414y

That's awesome. I shall use it in the future. Wish I could multi upvote.

814y

The way I like to think of the Monty Hall problem is like this... if you had the choice of picking either one of the three doors or two of the three doors (if the car is behind either, you win it), you would obviously pick two of the doors to give yourself a 2/3 chance of winning. Similarly, if you had picked your original door and then Monty asked if you'd trade your one door for the other two doors (all sight unseen), it would again be obvious that you should make the trade. Now... when you make that trade, you know that at least one of the doors you're getting in trade has a goat behind it (there's only one car, you have two doors, so you have to have at least one goat). So, given that knowledge and the certainty that trading one door for two is the right move (statistically), would seeing the goat behind one of the doors you're trading for before you make the trade change the wisdom of the trade? You KNOW that you're getting at least one goat in either case. Most people who I've explained it to in this way seem to see that making the trade still makes sense (and is equivalent to making the trade in the original scenario).
I think the struggle is that people tend to dismiss the existance of the 3rd door once they see what's behind it. It sort of drops out of the picture as a resolved thing and then the mind erroneously reformulates the situation with just the two remaining doors. The scary thing is that people are generally quite easily manipulated with these sorts of puzzles and there are plenty of circumstances (DNA evidence given during jury trials comes to mind) when the probabilities being presented are wildly misleading as the result of erroneously eliminating segments of the problem space because they are "known".

314y

There's a significant population of people - disproportionately represented here - who consider Bayesian reasoning to be theoretically superior to the ad hoc methods habitually used. An introductory essay on the subject that many people here read and agreed with A Technical Explanation of Technical Explanation.

014y

One more application of Bayes I should have mentioned: Aumann's Agreement Theorem.

Wonderful. Are you aware of the Tuesday Boy problem? I think it could have been a more impressive second example.

"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

(The intended interpretation is that I have two children, and at least one of them is a boy-born-on-a-Tuesday.)

I found it here: Magic numbers: A meeting of mathemagical tricksters

I always much prefer these stated as questions - you stop someone and say "Do you have exactly two children? Is at least one of them a boy born on a Tuesday?" and they say "yes". Otherwise you get into wondering what the probability they'd say such a strange thing given various family setups might be, which isn't precisely defined enough...

514y

Very true. The article DanielVarga linked to says:
... which is just wrong: whether it is different depends on how the information was obtained. If it was:
... then there's zero new information, so the probability stays the same, 1/3rd.
(ETA: actually, to be closer to the original problem, it should be "Select one of your sons at random and tell me the day he was born", but the result is the same.)

114y

I think the only reasonable interpretation of the text is clear since otherwise other standard problems would be ambiguous as well:
"What is probability that a person's random coin toss is tails?"
It does not matter whether you get the information from an experimenter by asking "Tell me the result of your flip!" or "Did you get tails?". You just have to stick to the original text (tails) when you evaluate the answer in either case.
[[EDIT] I think I misinterpreted your comment. I agree that Daniel's introduction was ambiguous for the reasons you have given.
Still the wording "I have two children, and at least one of them is a boy-born-on-a-Tuesday." he has given clarifies it (and makes it well defined under the standard assumptions of indifference).

614y

Yesterday I told the problem to a smart non-math-geek friend, and he totally couldn't relate to this "only reasonable interpretation". He completely understood the argument leading to 13/27, but just couldn't understand why do we assume that the presenter is a randomly chosen member of the population he claims himself to be a member of. That sounded like a completely baseless assumption to him, that leads to factually incorrect results. He even understood that assuming it is our only choice if we want to get a well-defined math problem, and it is the only way to utilize all the information presented to us in the puzzle. But all this was not enough to convince him that he should assume something so stupid.

314y

For me, the eye opener was this outstanding paper by E.T. Jaynes:
http://bayes.wustl.edu/etj/articles/well.pdf
IMO this describes the essence of the difference between the Bayesian and frequentist philosophy way better than any amount of colorful polygons. ;)

114y

I get that assuming that genders and days of the week are equiprobable, of all the people with exactly two children, at least one of whom is a boy born on a Tuesday, 13/27 have two boys.

214y

True, but if you go around asking people-with-two-chidren-at-least-one-of-which-is-a-boy "Select one of your sons at random, and tell me the day of the week on which he was born", among those who answer "Tuesday", one-third will have two boys.
(for a sufficiently large set of people-with-two-chidren-at-least-one-of-which-is-a-boy who answer your question instead of giving you a weird look)
I'm just saying that the article used an imprecise formulation, that could be interpreted in different ways - especially the bit "if you supply the extra information that the boy was born on a Tuesday", which is why asking questions the way you did is better.

114y

The Tuesday Boy was loads of fun to think about the first time I came across it - thanks to the parent comment. I worked through the calculations with my 14yo son, on a long metro ride, as a way to test my understanding - he seemed to be following along fine.
The discussion in the comments on this blog on Azimuth has, however, taken the delight to an entirely new level. Just wanted to share.

114y

It seems to me that the standard solutions don't account for the fact that there are a non-trivial number of families who are more likely to have a 3rd child, if the first two children are of the same sex. Some people have a sex-dependent stopping rule.
P(first two children different sexes | you have exactly two children) > P(first two children different sexes | you have more than two children)
The other issue with this kind of problem is the ambiguity. What was the disclosure algorithm? How did you decide which child to give me information about? Without that knowledge, we are left to speculate.

414y

This issue is also sometimes raised in cultures where male children are much more highly prized by parents.
Most people falsely assume that such a bias, as it stands, changes gender ratios for the society, but its only real effect is that correspondingly larger and rarer families have lots of girls. Such societies typically do have weird gender ratios, but this is mostly due to higher death rates before birth because of selective abortion, or after birth because some parents in such societies feed girls less, teach them less, work them more, and take them to the doctor less.
Suppose the rules for deciding to have a child without selective abortion (and so with basically 50/50 odds of either gender) and no unfairness post-birth were: If you have a boy, stop; if you have no boy but have fewer than N children, have another. In a scenario where N > 2, two child families are either a girl and a boy, or two girls during a period when their parents still intend to have a third. Because that window is relatively small relative to the length of time that families exist to be sampled, most two child families (>90%?) would be gender balanced.
Generally, my impression is that parental preferences for one or the other sex (or for gender balance) are generally out of bounds in these kinds of questions because we're supposed to assume platonicly perfect family generating processes with exact 50/50 odds, and no parental biases, and so on. My impression is that cultural literacy is supposed to supply the platonic model. If non-platonic assumptions are operating then different answers are expected as different people bring in different evidence (like probabilities of lying and so forth). If real world factors sneak in later with platonic assumptions allowed to stand then its a case of a bad teacher who expects you to guess the password of precisely which evidence they want to be imported, and which excluded.
This issue of signaling which evidence to import is kind of subtle, and

114y

Actually, a Bayesian and a frequentist can have different answers to this problem. It resides on what distribution you are using to decide to tell me that a boy is born on Tuesday. The standard answer ignores this issue.

214y

I don't know much about the philosophy of statistical inference. But I am dead sure that if the Bayesian and the frequentist really do ask the same question, then they will get the same answer. There is a nice spoiler post where the possible interpretations of the puzzle are clearly spelled out. Do you suggest that some of these interpretations are preferred by either a frequentist or a Bayesian?

114y

Well, essentially, focusing on that coin flip is a very Bayesian thing to do. A frequentist approach to this problem won't imagine the prior coin flip often. See Eliezer's post about this here. I agree however that a careful frequentist should get the same results as a Bayesian if they are careful in this situation. What results one gets depends in part on what exactly one means by a frequentist here.

010y

Just so it's clear, since it didn't seem super clear to me from the other comments, the solution to the Tuesday Boy problem given in that article is a really clever way to get the answer wrong.
The problem is the way they use the Tuesday information to confuse themselves. For some reason not stated in the problem anywhere, they assume that both boys cannot be born on Tuesday. I see no justification for this, as there is no natural justification for this, not even if they were born on the exact same day and not just the same day of the week! Twins exist! Using their same bizarre reasoning but adding the extra day they took out I get the correct answer of 50% (14/28), instead of the close but incorrect answer of 48% (13/27).
Using proper Bayesian updating from the prior probabilities of two children (25% boys, 50% one each, 25% girls) given the information that you have one boy, regardless of when he was born, gets you a 50% chance they're both boys. Since knowing only one of the sexes doesn't give any extra information regarding the probability of having one child of each sex, all of the probability for both being girls gets shifted to both being boys.

010y

No, that's not right. They don't assume that both boys can't be born on Tuesday. Instead, what they are doing is pointing out that although there is a scenario where both boys are born on Tuesday, they can't count it twice--of the situations with a boy born on Tuesday, there are 6 non-Tuesday/Tuesday, 6 Tuesday/non-Tuesday, and only 1, not 2, Tuesday/Tuesday.
Actually, "one of my children is a boy born on Tuesday" is ambiguous. If it means "I picked the day Tuesday at random, and it so happens that one of my children is a boy born on the day I picked", then the stated solution is correct. If it means "I picked one of my children at random, and it so happens that child is a boy, and it also so happens that child was born on Tuesday", the stated solution is not correct and the day of the week has no effect on the probability

010y

No, read it again. It's confusing as all getout, which is why they make the mistake, but EACH child can be born on ANY day of the week. The boy on Tuesday is a red herring, he doesn't factor into the probability for what day the second child can be born on at all. The two boys are not the same boys, they are individuals and their probabilities are individual. Re-label them Boy1 and Boy2 to make it clearer:
Here is the breakdown for the Boy1Tu/Boy2Any option:
Boy1Tu/Boy2Monday Boy1Tu/Boy2Tuesday Boy1Tu/Boy2Wednesday Boy1Tu/Boy2Thursday Boy1Tu/Boy2Friday Boy1Tu/Boy2Saturday Boy1Tu/Boy2Sunday
Then the BAny/Boy1Tu option:
Boy2Monday/Boy1Tu Boy2Tuesday/Boy1Tu Boy2Wednesday/Boy1Tu Boy2Thursday/Boy1Tu Boy2Friday/Boy1Tu Boy2Saturday/Boy1Tu Boy2Sunday/Boy1Tu
Seven options for both. For some reason they claim either BTu/Tuesday isn't an option, or Tuesday/BTu isn't an option, but I see no reason for this. Each boy is an individual, and each boy has a 1/7 probability of being born on a given day. In attempting to avoid counting evidence twice you've skipped counting a piece of evidence at all! In the original statement, they never said one and ONLY one boy was born on Tuesday, just that one was born on Tuesday. That's where they screwed up - they've denied the second boy the option of being born on Tuesday for no good reason.
A key insight that should have triggered their intuition that their method was wrong was that they state that if you can find a trait rarer than being born on Tuesday, like say being born on the 27th of October, then you'll approach 50% probability. That is true because the actual probability is 50%.

110y

You're double-counting the case where both boys are born on Tuesday, just like they said.
If you find a trait rarer than being born on Tuesday, the double-counting is a smaller percentage of the scenarios, so being closer to 50% is expected.

010y

I see my mistake, here's an updated breakdown:
Boy1Tu/Boy2Any
Boy1Tu/Boy2Monday Boy1Tu/Boy2Tuesday Boy1Tu/Boy2Wednesday Boy1Tu/Boy2Thursday Boy1Tu/Boy2Friday Boy1Tu/Boy2Saturday Boy1Tu/Boy2Sunday
Then the Boy1Any/Boy2Tu option:
Boy1Monday/Boy2Tu Boy1Tuesday/Boy2Tu Boy1Wednesday/Boy2Tu Boy1Thursday/Boy2Tu Boy1Friday/Boy2Tu Boy1Saturday/Boy2Tu Boy1Sunday/Boy2Tu
See 7 days for each set? They aren't interchangeable even though the label "boy" makes it seem like they are.
Do the Bayesian probabilities instead to verify, it comes out to 50% even.

010y

What's the difference between
and
?

010y

In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2.
That's why the "born on tuesday" is a red herring, and doesn't add any information. How could it?

110y

This sounds like you are trying to divide "two boys born on Tuesday" into "two boys born on Tuesday and the person is talking about the first boy" and "two boys born on Tuesday and the person is talking about the second boy".
That doesn't work because you are now no longer dealing with cases of equal probability. "Boy 1 Monday/Boy 2 Tuesday", "Boy 1 Tuesday/Boy 2 Tuesday", and "Boy 1 Tuesday/Boy 1 Monday" all have equal probability. If you're creating separate cases depending on which of the boys is being referred to, the first and third of those don't divide into separate cases but the second one does divide into separate cases, each with half the probability of the first and third.
As I pointed out above, whether it adds information (and whether the analysis is correct) depends on exactly what you mean by "one is a boy born on Tuesday". If you picked "boy" and "Tuesday" at random first, and then noticed that one child met that description, that rules out cases where no child happened to meet the description. If you picked a child first and then noticed he was a boy born on a Tuesday, but if it was a girl born on a Monday you would have said "one is a girl born on a Monday", you are correct that no information is provided.

010y

The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.
Rephrase it this way:
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
Now, to see why Tuesday is irrelevant, I'll re-state it thusly:
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a 1/7 chance that child 1 was born on each day of the week, and there is a 1/7 chance that child 2 was born on each day of the week. There is a 1/49 chance that both children will be born on any given day (1/7*1/7), for a 7/49 or 1/7 chance that both children will be born on the same day. That's your missing 1/7 chance that gets removed inappropriately from the Tuesday/Tuesday scenario.

110y

1/3 (you either got hh, heads/tails,or tails/heads). You didn't tell me THE FIRST came up heads. Thats where you are going wrong. At least one is heads is different information then a specific coin is heads.
This is a pretty well known stats problem, a variant of Gardern's boy/girl paradox. You'll probably find it an intro book, and Jiro is correct. You are still overcounting. Boy-boy is a different case then boy-girl (well, depending on what the data collection process is).
If you have two boys (probability 1/4), then the probability at least one is born on Tuesday (1-(6/7)^2). ( 6/7^2 being the probability neither is born on Tuesday). The probability of a boy-girl family is (2*1/4) then (1/7) (the 1/7 for the boy hitting on Tuesday).

010y

Lets add a time delay to hopefully finally illustrate the point that one coin toss does not inform the other coin toss.
I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?

110y

No one is suggesting one flip informs the other, rather that when you say "one coin came up heads" you are giving some information about both coins.
This is 1/2, because there are two scenarios, hh, ht. But its different information then the other question.
If you say "one coin is heads," you have hh,ht,th, because it could be that the first flip was tails/the second heads (a possibility you have excluded in the above).

010y

No, it's the exact same question, only the labels are different.
The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options - HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.
Here's the probability in bayesian:
P(BoyBoy) = 0.25 P(Boy) = 0.5 P(Boy|BoyBoy) = 1
P(BoyBoy|Boy) = P(Boy|BoyBoy)*P(BoyBoy)/P(Boy)
P(BoyBoy|Boy)= (1*0.25) / 0.5 = 0.25 / 0.5 = 0.5
P(BoyBoy|Boy) = 0.5
It's exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn't the monty hall problem. Knowing half the problem (that there's at least one boy) doesn't change the probability of the other boy, it just changes what our possibilities are.

110y

No, it isn't. You should consider that you are disagreeing with a pretty standard stats question, so odds are high you are wrong. With that in mind, you should reread what people are telling you here.
Now, consider "I flip two coins" the possible outcomes are hh,ht,th,tt
I hope we can agree on that much.
Now, I give you more information and I say "one of the coins is heads," so we Bayesian update by crossing out any scenario where one coin isn't heads. There is only 1 (tt)
hh,ht,th
So it should be pretty clear the probability I flipped two heads is 1/3.
Now, your scenario, flipped two coins (hh,ht,th,tt), and I give you the information "the first coin is heads," so we cross out everything where the first coin is tails, leaving (hh,ht). Now the probability you flipped two heads is 1/2.
I don't know how to make this any more simple.

-110y

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
I know it's not the be all end all, but it's generally reliable on these types of questions, and it gives P = 1/2, so I'm not the one disagreeing with the standard result here.
Do the math yourself, it's pretty clear.
Edit: Reading closer, I should say that both answers are right, and the probability can be either 1/2 or 1/3 depending on your assumptions. However, the problem as stated falls best to me in the 1/2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.

110y

Did you actually read it? It does not agree with you. Look under the heading "second question."
I did the math in the post above, enumerating the possibilities for you to try to help you find your mistake.
Edit, in response to the edit:
Which is exactly analogous to what Jiro was saying about the Tuesday question. So we all agree now? Tuesday can raise your probability slightly above 50%, as was said all along.
And you are immediately making the exact same mistake again. You are told ONE child is a boy, you are NOT told the FIRST child is a boy. You do understand that these are different?

010y

Re-read it.

-210y

The relevant quote from the Wiki:
We have no general population information here. We have one man with at least one boy.

110y

I'm not at all sure you understand that quote. Lets stick with the coin flips:
Do you understand why these two questions are different: I tell you- "I flipped two coins, at least one of them came out heads, what is the probability that I flipped two heads?" A:1/3 AND "I flipped two coins, you choose one at random and look at it, its heads.What is the probability I flipped two heads" A: 1/2

010y

For the record, I'm sure this is frustrating as all getout for you, but this whole argument has really clarified things for me, even though I still think I'm right about which question we are answering.
Many of my arguments in previous posts are wrong (or at least incomplete and a bit naive), and it didn't click until the last post or two.
Like I said, I still think I'm right, but not because my prior analysis was any good. The 1/3 case was a major hole in my reasoning. I'm happily waiting to see if you're going to destroy my latest analysis, but I think it is pretty solid.

010y

Yes, and we are dealing with the second question here.
Is that not what I said before?
We don't have 1000 families with two children, from which we've selected all families that have at least one boy (which gives 1/3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information. The probability that the second is a boy is 1/2, so the probability that both are boys is 1/2.
The possible options for the "Boy born on Tuesday" are not Boy/Girl, Girl/Boy, Boy/Boy. That would be the case in the selection of 1000 families above.
The possible options are Boy (Tu) / Girl, Girl / Boy (Tu), Boy (Tu) / Boy, Boy / Boy (Tu).
There are two Boy/Boy combinations, not one. You don't have enough information to throw one of them out.
This is NOT a case of sampling.

210y

As long as you realize there is a difference between those two questions, fine. We can disagree about what assumptions the wording should lead us to, thats irrelevant to the actual statistics and can be an agree-to-disagree situation. Its just important to realize that what the question means/how you get the information is important.
If we have one family with two children, of which one is a boy, they are (by definition) a member of the set "all families that have at least one boy." So it matters how we got the information.
If we got that information by grabbing a kid at random and looking at it (so we have information about one specific child), that is sampling, and it leads to the 1/2 probability.
If we got that information by having someone check both kids, and tell us "at least one is a boy" we have different information (its information about the set of kids the parents have, not information about one specific kid).
If it IS sampling (if I grab a kid at random and say "whats your Birthday?" and it happens to be Tuesday), then the probability is 1/2. (we have information about the specific kid's birthday).
If instead, I ask the parents to tell me the birthday of one of their children, and the parent says 'I have at least one boy born on Tuesday', then we get, instead, information about their set of kids, and the probability is the larger number.
Sampling is what leads to the answer you are supporting.

010y

The answer I'm supporting is based on flat priors, not sampling. I'm saying there are two possible Boy/Boy combinations, not one, and therefore it takes up half the probability space, not 1/3.
Sampling to the "Boy on Tuesday" problem gives roughly 48% (as per the original article), not 50%.
We are simply told that the man has a boy who was born on tuesday. We aren't told how he chose that boy, whether he's older or younger, etc. Therefore we have four possibilites, like I outlined above.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
If so, is not the probability for some combination of Boy/Boy 1/2? If not, why not? I don't see it.
BTW, contrary to my previous posts, having the information about the boy born on Tuesday is critical because it allows us (and in fact requires us) to distinguish between the two boys.
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can't reason that it pushes it to exactly 50%.

210y

No. For any day of the week EXCEPT Tuesday, boy and girl are equivalent. For the case of both children born on Tuesday you have for girls: Boy(tu)/Girl(tu),Girl(tu)/Boy(tu), and for boys: boy(tu)/boy(tu).
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says "I was born on a Tuesday," that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called "sampling")
Do you agree that if, instead, the parents give you the information "one of my two kids is a boy born on a Tuesday", that this is a different sort of information, information about the set of their children, and not about a specific child?

010y

I agree with this.
I agree with this if they said something along the lines of "One and only one of them was born on Tuesday". If not, I don't see how the Boy(tu)/Boy(tu) configuration has the same probability as the others, because it's twice as likely as the other two configurations that that is the configuration they are talking about when they say "One was born on Tuesday".
Here's my breakdown with 1000 families, to try to make it clear what I mean:
1000 Families with two children, 750 have boys.
Of the 750, 500 have one boy and one girl. Of these 500, 1/7, or roughly 71 have a boy born on Tuesday.
Of the 750, 250 have two boys. Of these 250, 2/7, or roughly 71 have a boy born on Tuesday.
71 = 71, so it's equally likely that there are two boys as there are a boy and a girl.
Having two boys doubles the probability that one boy was born on Tuesday compared to having just one boy.
And I don't think I'm confused about the sampling, because I didn't use the sampling reasoning to get my result*, but I'm not super confident about that so if I am just keep giving me numbers and hopefully it will click.
*I mean in the previous post, not specifically this post.

210y

This is wrong. With two boys each with a probability of 1/7 to be born on Tuesday, the probability of at least one on a Tuesday isn't 2/7, its 1-(6/7)^2

-110y

How can that be? There is a 1/7 chance that one of the two is born on Tuesday, and there is a 1/7 chance that the other is born on Tuesday. 1/7 + 1/7 is 2/7.
There is also a 1/49 chance that both are born on tuesday, but how does that subtract from the other two numbers? It doesn't change the probability that either of them are born on Tuesday, and both of those probabilities add.

410y

The problem is that you're counting that 1/49th chance twice. Once for the first brother and once for the second.

010y

I see that now, it took a LOT for me to get it for some reason.

210y

You overcount, the both on Tuesday is overcounted there. Think of it this way- if I have 8 kids do I have a better than 100% probability of having a kid born on Tuesday?
There is a 1/7x6/7 chance the first is born on Tuesday and the second is born on another day. There is a 1/7x6/7 chance the second is born on Tuesday and the first is born on another day. And there is a 1/49 chance that both are born on Tuesday.
All together thats 13/49. Alternatively, there is a (6/7)^2 chance that both are born not-on-Tuesday, so 1-(6/7)^2 tells you the complementary probability.

110y

Wow.
I've seen that same explanation at least five times and it didn't click until just now. You can't distinguish between the two on tuesday, so you can only count it once for the pair.
Which means the article I said was wrong was absolutely right, and if you were told that, say one boy was born on January 17th, the chances of both being born on the same day are 1-(364/365)^2 (ignoring leap years), which gives a final probability of roughly 49.46% that both are boys.
Thanks for your patience!
ETA: I also think I see where I'm going wrong with the terminology - sampling vs not sampling, but I'm not 100% there yet.

010y

"The first coin comes up heads" (in this version) is not the same thing as "one of the coins comes up heads" (as in the original version). This version is 50%, the other is not.

010y

How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?
If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.
The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn't matter that you don't know which is which, one flip doesn't have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a 50/50 split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don't get both of them!).
Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It's a fair coin, it stays 50%.

310y

Flip two coins 1000 times, then count how many of those trials have at least one head (~750). Count how many of those trials have two heads (~250).
Flip two coins 1000 times, then count how many of those trials have the first flip be a head (~500). Count how many of those trials have two heads (~250).
By the way, these sorts of puzzles should really be expressed as a question-and-answer dialogue. Simply volunteering information leaves it ambiguous as to what you've actually learned ("would this person have equally likely said 'one of my children is a girl' if they had both a boy and girl?").

110y

Yeah, probably the biggest thing I don't like about this particular question is that the answer depends entirely upon unstated assumptions, but at the same time it clearly illustrates how important it is to be specific.

010y

No there's not. The cases where the second child is a boy and the second child is a girl are not equal probability.
If you picked "heads" before flipping the coins, then the probability is 1/3. There are three possibilities: HT, TH, and HH, and all of these possibilities are equally likely.
If you picked "heads" and "Tuesday" before knowing when you would be flipping the coins, and then flipped each coin on a randomly-selected day, and you just stopped if there weren't any heads on Tuesday, then the answer is the same as the answer for boys on Tuesday. If you flipped the coin and then realized it was Tuesday, the Tuesday doesn't affect the result.
If you picked the sex first before looking at the children, the sex of one child does influence the sex of the other child because it affects whether you would continue or say "there aren't any of the sex I picked" and the sexes in the cases where you would continue are not equally distributed.

010y

Which boy did I count twice?
Edit:
BAny/Boy1Tu in the above quote should be Boy2Any/Boy1Tu.
You could re-label boy1 and boy2 to be cat and dog and it won't change the probabilities - that would be CatTu/DogAny.

014y

Note before you start calculating this: There's a distinction between the "first" and the "second" child made in the article. To avoid the risk of having to calculate all over again, take this into account if you want to compare your results to theirs.
I calculated the probability without knowing this, so that I just counted BG and GB as one scenario, where there's one girl and one boy. That means that without the Tuesday fact the probability of another boy is 1/2, not 1/3.
(I ended up at a posterior probability of 2/3, by the way.)

014y

I'm so happy: I've just got this one right, before looking at the answer. It's damn beautiful.
Thanks for sharing.

114y

Same here. It was a perfect test, as I've never seen the Tuesday Boy problem before. Took a little wrangling to get it all to come out in sane fractions, and I was staring at the final result going, "that can't be right", but sure enough, it was exactly right.
(Funny thing: my original intuition about the problem wasn't that far off. I was simply ignoring the part about Tuesday, and focusing on the prior probability that the other child was a boy. It gives a close, but not-quite-right answer.)

When I tried to explain Bayes' to some fellow software engineers at work, I came up with http://moshez.wordpress.com/2011/02/06/bayes-theorem-for-programmers/

514y

I was thinking about that too. Its actually something that comes up with the literature on "lying with statistics" where (either accidentally or out of a more or less subconscious attempt to convince) figures representing numbers are rescaled by a linear factor by the author causing super linear adjustments of area which is what the reader's visual processing really responds to.
Generally, textbooks recommend boring linear scales (basically bar charts) with an unbroken reference line off to the side to compare numbers. However, if you really want to use images with area (and for this article they're a brilliant addition) then the correct thing to do is to decide the number you're representing as the area and work back to the number you have access to for a given figure (like a circle radius or a pentagon edge length or whatever) and adjust that number to your calculated value.

Thank you Komponisto,

I have read many explanations of Bayesian theory, and like you, if I concentrated hard enough I could follow the reasoning , but I could never reason it out for myself. Now I can. Your explanation was perfect for me. It not only enabled me to "grok" the Monty Hall problem, but Bayesian calculations in general, while being able to retain the theory.

Thank you again, Ben

Thank you very much for this. Until you put it this way, I could not grasp the Monty Hall problem; I persisted in believing that there would be a 50/50 chance once the door was opened. Thank you for changing my mind.

The Monty Hall problem seems like it can be simplified: Once you've picked the door, you can switch to instead selecting the two alternate doors. You know that one of the alternate doors contains a goat (since there's only one car), which is equivalent to having Monty open one of those two doors.

The trick is simply in assuming that Monty is actually introducing any new information.

Not sure if it's helpful to anyone else, but it just sort of clicked reading it this time :)

This is really brilliant. Thanks for making it all seem so easy: for some reason I never saw the connection between an update and rescaling like that, but now it seems obvious.

I'd like to see this kind of diagram for the Sleeping Beauty problem.

I tend to think out Monty Hall like this: The probability you have chosen the door hiding the car is 1/3. Once one of the other two doors is shown to hide a goat, the probablity of the third door hiding the car must be 2/3. Therefore you double your chances to win the car by switching.

Wow, that was great! I already had a fairly good understanding of the Theorem, but this helped cement it further and helped me compute a bit faster.

It also gave me a good dose of learning-tingles, for which I thank you.

In figure 10 above, should the second blob have a value of 72.9% ? I noticed that the total of all the percents are only adding up to 97% with the current values. I calculated as follows: New 100% value: 10% + 35% + 3% = 48% H1 : 10% / 48% = 20.833% H2: 35% / 48% = 72.9166% H3: 3% / 48% = 6.25% Total: 99.99%

Also, I found this easy to understand visually. Thanks.

114y

My mathematics agrees - 72.9%. Good catch!

A presentation critique: psychologically, we tend to compare the relative *areas* of shapes. Your ovals in Figure 1 are scaled so that their *linear* dimensions (width, for example) are in the ratio 2:5:3; however, what we see are ovals whose areas are in ratio 4:25:9, which isn't what you're trying to convey. I think this happens for later shapes as well, although I didn't check them all.

314y

Really? I'd have said the exact opposite. For example in this post, the phrase "half the original's size" means that the linear dimensions are halved. This issue also come up in the production of bubble charts, where the size of a circle represents some value. When I look at a bubble chart it is often unclear whether the data is intended to be represented by the area or the radius.
It is certainly easier for me to compare linear dimensions than areas.

214y

Hence the popularity of bar charts, where the area and the linear dimension are coupled. But the visual impact is a function of area, more than length, even if it is hard to quantify in the eye - but quantification should be done by the quantitative numbers, not by graphical estimation.
(How many sentences can I start with conjunctions? Let me count the ways...)

Great visualizations.

In fact, this (only without triangles, squares,...) is how I've been intuitively calculating Bayesian probabilities in "everyday" life problems since I was young. But you managed to make it even clearer for me. Good to see it applied to Monty Hall.

Thank you Komponisto! Apparently, my brain works similar to yours on this matter. Here is a video by Richard Carrier explaining Bayes' theorem that I also found helpful.

Perhaps a better title would be "Bayes' Theorem Illustrated (My Ways)"

In the first example you use shapes with colors of various sizes to illustrate the ideas visually. In the second example, you using plain rectangles of approximately the same size. If I was a visual learner, I don't know if your post would help me much.

I think you're on the right track in example one. You might want to use shapes that are easier to estimate the relative areas. It's hard to tell if one triangle is twice as big as another (as measured by area), but it's easie...

I find that Monty Hall is easier to understand with N doors, N > 2.

N doors, one hides a car. You pick a door at random yielding 1/N probability of getting car. Host now opens N-2 doors which are not your door, all containing goats. The probability the other door left has a car is not (N-1)/N.

Set N to 1000 and people generally agree that switching is good. Set N to 3 and they disagree.

The challenge with Bayes' illustrations is to simultaneously.show 1) relations 2) ratios. The suggested approach works well. I am suggesting to combine Venn diagrams and Pie charts:

http://oracleaide.wordpress.com/2012/12/26/a-venn-pie/

Happy New Year!

This problem is not so difficult to solve if we use a binomial tree to try to tackle it. Not only we will come to the same (correct) mathematical answer (which is brilliantly exposed in the first post in this thread) but logically is more palatable.

I will exposed the logically, semantically derived answer straight away and then I will jump in to the binomial tree for the proof-out.

The probability of the situation exposed here, which for the sake of being brief I’m going to put it as “contestant choose a door, a goat is revealed in a different door, then co...

Thanks for posting this. Your explanations are fascinating and helpful. That said, I do have one quibble. I was misled by the Two Aces problem because I didn't know that the two unknown cards (2C and 2D) were precluded from also being aces or aces of spaces. It might be better to edit the post to make that clear.

While on topic, GREAT demo of conditional probability.

http://www.cut-the-knot.org/Curriculum/Probability/ConditionalProbability.shtml

Why would you need more than plain English to intuitively grasp Monty-Hall-type problems?

Take the original Monty Hall 'Dilemma'. Just imagine there are two candidates, A and B. A and B both choose the same door. After the moderator picked one door A always stays with his first choice, B always changes his choice to the remaining third door. Now imagine you run this experiment 999 times. What will happen? Because A always stays with his initial choice, he will win 333 cars. But where are the remaining 666 cars? Of course B won them!

Or conduct the experiment...

I can testify that this isn't anywhere near as obvious to most people than it is to you. I, for one, had to have other people explain it to me the first time I ran into the problem, and even then it took a small while.

114y

I think the very problem in understanding such issues is shown in your reply. People assume too much, they read too much into things. I never said it has been obvious to me. I asked why you would need more than plain English to understand it and gave some examples on how to describe the problem in an abstract way that might be ample to grasp the problem sufficiently. If you take things more literally and don't come up with options that were never mentioned it would be much easier to understand. Like calling the police in the case of the trolley problem or whatever was never intended to be a rule of a particular game.

314y

Well, yeah. But if I recall, I did have a plain English explanation of it. There was an article on Wikipedia about it, though since this was at least five years ago, the explanation wasn't as good as it is in today's article. It still did a passing job, though, which wasn't enough for me to get it very quickly.

214y

Yesterday, when falling asleep, I remembered that I indeed used the word 'obvious' in what I wrote. Forgot about it, I wrote the plain-English explanation from above earlier in a comment to the article 'Pigeons outperform humans at the Monty Hall Dilemma' and just copied it from there.
Anyway, I doubt it is obvious to anyone the first time. At least anyone who isn't a trained Bayesian. But for me it was enough to read some plain-English (German actually) explanations about it to come to the conclusion that the right solution is obviously right and now also intuitively so.
Maybe the problem is also that most people are simply skeptical to accept a given result. That is, is it really obvious to me now or have I just accepted that it is the right solution, repeated many times to become intuitively fixed? Is 1 + 1 = 2 really obvious? The last page of Russel and Whitehead's proof that 1+1=2 could be found on page 378 of the Principia Mathematica. So is it really obvious or have we simply all, collectively, come to accept this 'axiom' to be right and true?
I haven't had much time lately to get much further with my studies, I'm still struggling with basic Algebra. I have almost no formal education and try to educate myself now. That said, I started to watch a video series lately (The Most IMPORTANT Video You'll Ever See) and was struck when he said that to roughly figure out the doubling time you simply divide 70 by the percentage growth rate. I went to check it myself if it works and later looked it up. Well, it's NOT obvious why this is the case, at least not for me. Not even now that I have read up on the mathematical strict formula. But I'm sure, as I will think about it more, read more proofs and work with it, I'll come to regard it as obviously right. But will it be any more obvious than before? I will simply have collected some evidence for its truth value and its consistency. Things just start to make sense, or we think so because they work and/or are consistent

214y

If you're interested, here is a good explanation of the derivation of the formula. I don't think it's obvious, any more than the quadratic formula is obvious: it's just one of those mathematical tricks that you learn and becomes second nature.

014y

I'm not sure I'm completely happy with that explanation. They use the result that ln(1+x) is very close to x when x is small. This is due to the Taylor series expansion of ln(1+x) (edit:or simply on looking at the ratio of the two and using L'Hospital's rule), but if one hasn't had calculus, that claim is going to look like magic.

214y

Here are more examples:
* Why a negative times a negative should be a positive.
* Intuition on why a^-b = 1/(a^b) (and why a^0 =1)
Those explanations are really great. I've missed such in school when wondering WHY things behave like they do, when I was only shown HOW to use things to get what I want to do. But what do these explanations really explain. I think they are merely satisfying our idea that there is more to it than meets the eye. We think something is missing. What such explanations really do is to show us that the heuristics really work and that they are consistent on more than one level, they are reasonable.

114y

Well, it's an approximation, that's all. Pi is approximately equal to 355/113 - yeah, there's good mathematical reasons for choosing that particular fraction as an approximation, but the accuracy justifies itself. [edited sentence:] You only need one real revelation to not worry about how true Td = 70/r is: that the doubling time is a smooth line - there's no jaggedy peaks randomly in the middle. After that, you can just look how good the fit is and say, "yeah, that works for 0.1 < r < 20 for the accuracy I need".

011y

Although it's late, I'd like to say that XiXiDu's approach deserves more credit and I think it would have helped me back when I didn't understand this problem. Eliezer's Bayes' Theorem post cites the percentage of doctors who get the breast cancer problem right when it's presented in different but mathematically equivalent forms. The doctors (and I) had an easier time when the problem was presented with quantities (100 out of 10,000 women) than with explicit probabilities (1% of women).
Likewise, thinking about a large number of trials can make the notion of probability easier to visualize in the Monty Hall problem. That's because running those trials and counting your winnings looks like something. The percent chance of winning once does not look like anything. Introducing the competitor was also a great touch since now the cars I don't win are easy to visualize too; that smug bastard has them!
Or you know what? Maybe none of that visualization stuff mattered. Maybe the key sentence is "[Candidate] A always stays with his first choice". If you commit to a certain door then you might as well wear a blindfold from that point forward. Then Monty can open all 3 doors if he likes and it won't bring your chances any closer to 1/2.

-1114y

I have a small issue with the way you presented the Monty python problem. In my opinion, the setup could be a little clearer. The Bayesian model you presented holds true iff you make an assumption about the door you picked; either goat (better) or car (less wrong). If you pick a door at random with no presuppositions (I believe this is the state most people are in), then you have no basis to decide to switch or not switch, and have a truly 50% chance either way. If instead you introduce the assumption of goat, when the host opens up the other goat, you know you had a 2/3 chance to pick a goat. With both goats known or presumed, the last door must be the car with an error rate of 1/3.

114y

As far as I can see, one-third of the time the first door you picked had the car. What happens afterward cannot change that one-third. The only way it could change your one-third credence is if sometimes Monty Hall did one thing and sometimes another, depending on whether you picked the car.

114y

While the overall probabilities for the game will never change, the contestant’s perception of the current state of the game will cause them to affect their win rate. To elaborate on what I was saying, imagine the following internal monologue of a contestant:
I’ve eliminated one goat. Two doors left. One is a goat, one is a car. No way to tell which is which, so I’ll just randomly pick one.
IMO, this is probably what most contestants believe when faced with the final choice. Obviously, there is a way to have a greater success rate, but this person is evaluating in a vacuum. If contestants were aware of the actual probabilities involved, I think we would see less “agonizing” moments as the contestants decide if they should switch or not. By randomly picking door A or B, irrespective of the entirety game, you’ve lost your marginal advantage and lowered your win rate. That being said, if they still “randomly” pick switch every their win rate will be the expected, actual probability.
Edit: The same behaviour can be seen in Deal or No Deal. If for some insane reason, they go all the way to the final two cases, the correct choice is to switch. I don’t know exactly how many cases you have to choose from, but the odds are greatly against you that you picked the 1k case. If the case is still on the board, the chick is holding it. Yet, people make the choice to switch based entirely on the fact that there are two cases and 1 has 1k and the other has 0.01. They think they have a 50/50 shot, so they make their odds essentially 50/50 by randomly choosing. In other words, they might as well as have flipped a coin to make the decision between the two cases.

914y

Actually, I just realized... there is no reason to swap on Deal or No Deal. The reason why you swap in Monty Hall is that Monty knows which door has the goats and there is no chance he will open a door to reveal a car. But in Deal or No Deal the cases that get opened are chosen by the contestant with no knowledge of what is inside them. It's like if the contestant got to pick which of the two remaining doors to open instead of Monty, there is a 1/3 chance the contestant would open the door with the car leaving her with only goats to choose from. The fact the the contestant got lucky and didn't open the door with the car wouldn't tell her anything about which of the two remaining doors the car is really behind.
ETA: Basically Deal or No Deal is just a really boring game.

014y

Well, it's exciting for those who like high-stakes randomness. And there are expected utility considerations at every opportunity for a deal (I don't remember if there's a consistent best choice based on the typical deal).

214y

I was talking about this in my other comment.

014y

Maybe it could be interesting if you treat it as a psychology game - trying to predict, based on the person's appearance, body language, and statements, whether they will conform to expected-utility or not?

514y

The way the deals work going down to the final two cases can end up the best strategy. Basically they weight the deals to encourage higher ratings. As long as there is a big money case in play they won't offer the contestant the full average of the cases- presumably viewers like to watch people play for the big money so the show wants these contestants to keep going. If all the big money cases get used up the banker immediately offers the contestant way more than they are worth to get them off the stage and make way for a new contestant.
(This was my conclusion after watching until I thought I had figured it out. I could be reading this into a more complex or more random pattern.)

314y

Actually, people are much more likely to not switch if they think they are both equal. I've just finished reading Jason Rosenhouse's excellent book on the Monty Hall Problem and there's a strong tendency for people not to switch. Apparently people are worried that they'll feel bad if they switch and that action causes them to lose. (The book does a very good job discussing a lot about both the problem and psych studies about how people react to it or different versions. I strongly recommend it.)

114y

On the actual show, sometimes Monty Hall did one thing and sometimes another, so far as I am told. We're not talking about actual behavior of contestants in an actual contest, we're talking about optimal behavior of contestants in a fictional contest.
Edit: I'm sorry, I really don't know what you're arguing. Am I making sense?

114y

Well the Monty Hall problem as stated never occurred on Let's Make a Deal. He's even on record saying he won't let a contestant switch doors after picking, in response to this problem.

014y

Robin's description is correct. I'm not sure what you're saying.
ETA: this thread has gotten ridiculous. I'm deleting the rest of my comments on it. The best source for info on Monty Hall is youtube. He does everything. One thing that makes it rather different is that it is usually not clear how many good and bad prizes there are.

414y

I'm really shocked by the reactions of the mathematicians. I remember solving that problem in like the third week of my Intro to Computer Science Class. And before that I had heard of it and thought through why it was worth switching. I didn't realize it caused so much confusion as recently as 20 years ago.

214y

The problem causes a lot of confusion. There are studies which show that this is in fact cross-cultural. It seems to deeply conflict with a lot of heuristics humans use for working out probability. See Donald Granberg, "Cross-Cultural Comparison of Responses to the Monty Hall Dilemma" Social Behavior and Personality, (1999), 27:4 p 431-448. There are other relevant references in Jason Rosenhouse's book "The Monty Hall Problem." The problem clashes with many common heuristics. It isn't that surprising that some mathematicians have had trouble with it. (Although I do think it is surprising that some of the mathematicians who have had trouble were people like Erdos who was unambiguously first-class)

314y

Wow! I looked this up and it turns out it's described in a book I read a long time ago, The Man Who Loved Only Numbers (do a "Search Inside This Book" for "Monty Hall"). Edit: In this book, the phrase "Book proof" refers to a maximally elegant proof, seen as being in "God's Book of Proofs".
I encountered the problem for the first time in a collection of vos Savant's Parade pieces. It was unintuitive of course, but most striking for me was the utter unconvincibility of some of the people who wrote to her.

214y

Yes, my fallback if my intuition on a probability problem seems to fail me is always to code a quick simulation - so far, it's always taken on about a minute to code and run. That anyone bothered to write her a letter, even way back in the 70's, is mind-boggling.

014y

Yeah it's remarkable isn't it?
I suppose the thing about the Monty-Hall problem which makes it 'difficult' is that there is another agent with more information than you, who gives you a systematically 'biased' account of their information. (There's an element of 'deceitfulness' in other words.)
An analogy: Suppose you had a coin which you knew was either 2/3 biased towards heads or 2/3 biased towards tails, and the bias is actually towards heads. Say there have been 100 coin tosses, and you don't know any of the outcomes but someone else ("Monty") knows them all. Then they can feed you 'biased information' by choosing a sample of the coin tosses in which most outcomes were tails. The analogous confusion would be to ignore this possibility and assume that Monty is 'honestly' telling you everything he knows.

114y

Expert confidence. I read vos Savants book with all the letters she got and like how the problem seems to really be a test for the mental clarity and politeness of the actors involved.
Anyone knows of problems that get similarly violent reactions from experts?

214y

From Wikipedia:
The citation is from a letter from Monty himself, available online here.
I'm not sure how the article you linked to is relevant. It does describe an instance of Monty Hall actually performing the experiment, but it was in his home, not on the show.

0[anonymous]14y

exactly as Robin said.

214y

thomblake's remark was relevant too, though - from what I said, you might imagine that Monty Hall let people switch on the show. All the clarifications are relevant and good.

0[anonymous]14y

Yes, you might imagine that, and you'd probably be right. Thom's quote is evidence against that claim, but very weak.

014y

Aaargh! And I had upvoted that, believing a random Internet comment over a reliable offline source! That's a little embarrassing.
The article is awesome, by the way. Thanks!

214y

See response here

014y

You are making perfect sense; it’s me that is not. I had thought to clarify the issue for people that might still not “get it” after reading the article. Instead, I’ve only muddied the waters.

(This post is elementary: it introduces a simple method of visualizing Bayesian calculations. In my defense, we've had other elementary posts before, and they've been found useful; plus, I'd really like this to be online somewhere, and it might as well be here.)I'll admit, those Monty-Hall-type problems invariably trip me up. Or at least, they do if I'm not thinking

verycarefully -- doing quite a bit more work than other people seem to have to do.What's more, people's explanations of how to get the right answer have almost never been satisfactory to me. If I concentrate hard enough, I can usually follow the reasoning, sort of; but I never quite "see it", and nor do I feel equipped to solve similar problems in the future: it's as if the solutions seem to work only in retrospect.

Minds work differently, illusion of transparency, and all that.

Fortunately, I eventually managed to identify the source of the problem, and I came up a way of thinking about --

visualizing-- such problems that suits my own intuition. Maybe there are others out there like me; this post is for them.I've mentioned before that I like to think in very abstract terms. What this means in practice is that, if there's some simple, general, elegant point to be made,

tell it to me right away. Don't start with some messy concrete example and attempt to "work upward", in the hope that difficult-to-grasp abstract concepts will be made more palatable by relating them to "real life". If you do that, I'm liable to get stuck in the trees and not see the forest. Chances are, I won't have much trouble understanding the abstract concepts; "real life", on the other hand......well, let's just say I prefer to start at the top and work downward, as a general rule. Tell me how the trees relate to the forest, rather than the other way around.

Many people have found Eliezer's Intuitive Explanation of Bayesian Reasoning to be an excellent introduction to Bayes' theorem, and so I don't usually hesitate to recommend it to others. But for me personally, if I didn't know Bayes' theorem and you were trying to explain it to me, pretty much the worst thing you could do would be to start with some detailed scenario involving breast-cancer screenings. (And not just because it tarnishes beautiful mathematics with images of sickness and death, either!)

So what's the right way to explain Bayes' theorem to me?

Like this:

We've got a bunch of hypotheses (states the world could be in) and we're trying to figure out which of them is true (that is, which state the world is actually in). As a concession to concreteness (and for ease of drawing the pictures), let's say we've got three (mutually exclusive and exhaustive) hypotheses -- possible world-states -- which we'll call H

_{1}, H_{2}, and H_{3}. We'll represent these as blobs in space:Figure 0Now, we have some prior notion of how probable each of these hypotheses is -- that is, each has some

prior probability. If we don't know anything at all that would make one of them more probable than another, they would each have probability 1/3. To illustrate a more typical situation, however, let's assume we have more information than that. Specifically, let's suppose our prior probability distribution is as follows: P(H_{1}) = 30%, P(H_{2})=50%, P(H_{3}) = 20%. We'll represent this by resizing our blobs accordingly:Figure 1

That's our

priorknowledge. Next, we're going to collect someevidenceandupdateour prior probability distribution to produce aposteriorprobability distribution. Specifically, we're going to run a test. The test we're going to run has three possible outcomes: Result A, Result B, and Result C. Now, since this test happens to have three possible results, it would be really nice if the test just flat-out told us which world we were living in -- that is, if (say) Result A meant that H_{1}was true, Result B meant that H_{2}was true, and Result 3 meant that H_{3}was true. Unfortunately, the real world is messy and complex, and things aren't that simple. Instead, we'll suppose that each result can occur under each hypothesis, but that the different hypotheses have different effects on how likely each result is to occur. We'll assume for instance that if Hypothesis H_{1}is true, we have a 1/2 chance of obtaining Result A, a 1/3 chance of obtaining Result B, and a 1/6 chance of obtaining Result C; which we'll write like this:P(A|H

_{1}) = 50%, P(B|H_{1}) = 33.33...%, P(C|H_{1}) = 16.166...%and illustrate like this:

Figure 2(Result A being represented by a triangle, Result B by a square, and Result C by a pentagon.)

If Hypothesis H

_{2}is true, we'll assume there's a 10% chance of Result A, a 70% chance of Result B, and a 20% chance of Result C:Figure 3

(P(A|H_{2}) = 10% , P(B|H_{2}) = 70%, P(C|H_{2}) = 20%)Finally, we'll say that if Hypothesis H

_{3}is true, there's a 5% chance of Result A, a 15% chance of Result B, and an 80% chance of Result C:Figure 4(P(A|H

_{3}) = 5%, P(B|H_{3}) = 15% P(C|H_{3}) = 80%)Figure 5 below thus shows our knowledge prior to running the test:

Figure 5Note that we have now carved up our hypothesis-space more finely; our possible world-states are now things like "Hypothesis H

_{1}is true and Result A occurred", "Hypothesis H_{1}is true and Result B occurred", etc., as opposed to merely "Hypothesis H_{1}is true", etc. The numbers above the slanted line segments -- thelikelihoodsof the test results, assuming the particular hypothesis -- representwhat proportionof the total probability mass assigned to the hypothesis H_{n}is assigned to the conjunction of Hypothesis H_{n}and Result X; thus, since P(H_{1}) = 30%, and P(A|H_{1}) = 50%, P(H_{1}& A) is therefore 50% of 30%, or, in other words, 15%.(That's really all Bayes' theorem is, right there, but -- shh! -- don't tell anyone yet!)

Now, then, suppose we run the test, and we get...Result A.

What do we do? We

cut off all the other branches:Figure 6So our updated probability distribution now looks like this:

Figure 7...except for one thing: probabilities are supposed to add up to 100%, not 21%. Well, since we've

conditionedon Result A, that means that the 21% probability mass assigned to Result A is now the entirety of our probability mass -- 21% is the new 100%, you might say. So we simply adjust the numbers in such a way that they add up to 100%and the proportions are the same:Figure 8There! We've just performed a Bayesian update. And that's what it

looks like.If, instead of Result A, we had gotten Result B,

Figure 9then our updated probability distribution would have looked like this:

Figure 10Similarly, for Result C:

Figure 11Bayes' theoremis the formula that calculates these updated probabilities. Using H to stand for a hypothesis (such as H_{1}, H_{2}or H_{3}), and E a piece of evidence (such as Result A, Result B, or Result C), it says:P(H|E) = P(H)*P(E|H)/P(E)

In words: to calculate the updated probability P(H|E), take the portion of the prior probability of H that is allocated to E (i.e. the quantity P(H)*P(E|H)), and calculate what fraction this is of the total prior probability of E (i.e. divide it by P(E)).

What I like about this way of visualizing Bayes' theorem is that it makes the importance of prior probabilities -- in particular, the difference between P(H|E) and P(E|H) --

visually obvious. Thus, in the above example, we easily see that even though P(C|H_{3}) is high (80%), P(H_{3}|C) is much less high (around 51%) -- and once you have assimilated this visualization method, it should be easy to see that even more extreme examples (e.g. with P(E|H) huge and P(H|E) tiny) could be constructed.Now let's use this to examine two tricky probability puzzles, the infamous Monty Hall Problem and Eliezer's Drawing Two Aces, and see how it illustrates the correct answers, as well as how one might go wrong.

The Monty Hall ProblemThe situation is this: you're a contestant on a game show seeking to win a car. Before you are three doors, one of which contains a car, and the other two of which contain goats. You will make an initial "guess" at which door contains the car -- that is, you will select one of the doors, without opening it. At that point, the host will open a goat-containing door from among the two that you did not select. You will then have to decide whether to stick with your original guess and open the door that you originally selected, or switch your guess to the remaining unopened door. The question is whether it is to your advantage to switch -- that is, whether the car is more likely to be behind the remaining unopened door than behind the door you originally guessed.

(If you haven't thought about this problem before, you may want to try to figure it out before continuing...)

The answer is that it

isto your advantage to switch -- that, in fact, switchingdoublesthe probability of winning the car.People often find this counterintuitive when they first encounter it -- where "people" includes the author of this post. There are two possible doors that could contain the car; why should one of them be more likely to contain it than the other?

As it turns out, while constructing the diagrams for this post, I "rediscovered" the error that led me to incorrectly conclude that there is a 1/2 chance the car is behind the originally-guessed door and a 1/2 chance it is behind the remaining door the host didn't open. I'll present that error first, and then show how to correct it. Here, then, is the

wrongsolution:We start out with a perfectly correct diagram showing the prior probabilities:

Figure 12The possible hypotheses are Car in Door 1, Car in Door 2, and Car in Door 3; before the game starts, there is no reason to believe any of the three doors is more likely than the others to contain the car, and so each of these hypotheses has prior probability 1/3.

The game begins with our selection of a door. That itself isn't evidence about where the car is, of course -- we're assuming we have no particular information about that, other than that it's behind one of the doors (that's the whole point of the game!). Once we've done that, however, we will then have the opportunity to "run a test" to gain some "experimental data": the host will perform his task of opening a door that is guaranteed to contain a goat. We'll represent the result Host Opens Door 1 by a triangle, the result Host Opens Door 2 by a square, and the result Host Opens Door 3 by a pentagon -- thus carving up our hypothesis space more finely into possibilities such as "Car in Door 1 and Host Opens Door 2" , "Car in Door 1 and Host Opens Door 3", etc:

Figure 13Before we've made our initial selection of a door, the host is equally likely to open either of the goat-containing doors. Thus, at the beginning of the game, the probability of each hypothesis of the form "Car in Door X and Host Opens Door Y" has a probability of 1/6, as shown. So far, so good; everything is still perfectly correct.

Now we select a door; say we choose Door 2. The host then opens either Door 1 or Door 3, to reveal a goat. Let's suppose he opens Door 1; our diagram now looks like this:

Figure 14But this shows equal probabilities of the car being behind Door 2 and Door 3!

Figure 15Did you catch the mistake?

Here's the

correctversion:As soon as we selected Door 2, our diagram should have looked like this:Figure 16With Door 2 selected, the host no longer has the

optionof opening Door 2; if the car is in Door 1, hemustopen Door 3, and if the car is in Door 3, hemustopen Door 1. We thus see that if the car is behind Door 3, the host is twice as likely to open Door 1 (namely, 100%) as he is if the car is behind Door 2 (50%); his opening of Door 1 thus constitutes some evidence in favor of the hypothesis that the car is behind Door 3. So, when the host opens Door 1, our picture looks as follows:Figure 17which yields the correct updated probability distribution:

Figure 18## Drawing Two Aces

Here is the statement of the problem, from Eliezer's post:

(Once again, you may want to think about it, if you haven't already, before continuing...)

Here's how our picture method answers the question:

Since the person holding the cards has at least one ace, the "hypotheses" (possible card combinations) are the five shown below:

Figure 19Each has a prior probability of 1/5, since there's no reason to suppose any of them is more likely than any other.

The "test" that will be run is selecting an ace at random from the person's hand, and seeing if it is the ace of spades. The possible results are:

Figure 20Now we run the test, and get the answer "YES"; this puts us in the following situation:

Figure 21The total prior probability of this situation (the YES answer) is (1/6)+(1/3)+(1/3) = 5/6; thus, since 1/6 is 1/5 of 5/6 (that is, (1/6)/(5/6) = 1/5), our updated probability is 1/5 -- which happens to be the same as the prior probability. (I won't bother displaying the final post-update picture here.)

What this means is that the test we ran did not provide any additional information about whether the person has both aces beyond simply knowing that they have at least one ace; we might in fact say that the result of the test is screened off by the answer to the first question ("Do you have an ace?").

On the other hand, if we had simply asked "Do you have the ace of spades?", the diagram would have looked like this:

Figure 22which, upon receiving the answer YES, would have become:

Figure 23The total probability mass allocated to YES is 3/5, and, within that, the specific situation of interest has probability 1/5; hence the updated probability would be 1/3.

So a YES answer in this experiment, unlike the other, would provide evidence that the hand contains both aces; for if the hand contains both aces, the probability of a YES answer is 100% -- twice as large as it is in the contrary case (50%), giving a likelihood ratio of 2:1. By contrast, in the other experiment, the probability of a YES answer is only 50% even in the case where the hand contains both aces.

This is what people who try to explain the difference by uttering the opaque phrase "a random selection was involved!" are actually talking about: the difference between

Figure 24and

.

Figure 25The method explained here is far from the only way of visualizing Bayesian updates, but I feel that it is among the most intuitive.

(

I'd like to thank my sister,Vive-ut-Vivas, for help with some of the diagrams in this post.)