IQ (moved subtopic)

Wait, this was a real list he already had? You're joking, right? Sklansky, did you have you been maintaining a list? How often do you revise it, is it on a schedule or just when you see someone do something particularly smart or particularly dumb?

I am engaging in what you might call inferential reasoning.

You stated it as a fact but, more to the point, it's something I can totally see him doing. He seems to have put more thought into it than I had imagined, hence my question in the first place.

I don't normally have to spell things out for you.

Second guessing myself, confidence has taken a bit of a hit since being placed all the way down there with the plebs, obviously.

This is among the funniest things I could even imagine happening.

+1 to being supremely amused

😀

Ok Sklansky, since you don't want to tell us whether you had a running list, last question.

There are some people missing from this list. Can you provide updated standings to include the following names, at least for those with whose work you feel you are sufficiently familiar:

well named
DVaut1
Trolly
Playbig2000
Deuces McCracken
David Sklansky
Ken Jennings

OK. I will answer your question seriously even though you don't deserve it.

1. I of course, didn't have a running list. But notable posts, even from long ago, I remember to this day.

2. Obviously there is no way anyone could claim to put these names in the precise order as I pretended to. And in case it wasn't obvious I gave everyone evidence by calling clovis and not bruce z a "tie" to accentuate the silliness. (I did it because while both are very smart, they are often at each other's throats because of their differing takes on poker hands. So, I had some fun by calling them equal which I knew would bother them as much as you pretend to be bothered by chez.

3. Some of the stuff I wrote had hidden messages. One thing in particular that was far more important than the others (and hopefully will not be divulged if recognized). In other cases, I also
sometimes accentuated an inside joke such as with commonwealth.

4. In spite of the above, the list is 80% on the square and never far off from my real opinion. But my opinion differs slightly from most in several ways. For instance, I think someone who can figure out a way to answer a calculus question without calculus is probably smarter than the typical A student in calculus 101. I also think that someone who can teach a failing algebra student to get an A probably is. (Because when he answers a question wrongly you should usually be able to instantly say "no, that would be the answer if you changed the question to ..... Or when explaining a concept you should be instantly able to say " but this wouldn't be true if I had changed the question by saying ....."Or after teaching a concept and having the student get it right, you should be able to follow up with another question that differs from the first in a way that he or she is about 90% to get right.

5. I am pretty harsh towards a terrible post. Suzzer once wrote that logic is only useful to help you argue a conclusion you have already come to rather than perhaps come to a different conclusion. Now he has to live with the consequences of that post forever.

6. In case there is any confusion, the list was for the politics forum only. Both the Probability forum and the GTO forum, and perhaps others, have posters who would supplant most, if not all of the highest names on my list.

7. As to your names I don't have enough memories of notable posts in either direction other than the fact that DVaut was especially complimentary towards me the few times we engaged. I believe you can still look that up.

I hope none were meant for me. I’ve never been a big fan of your “riddles”.

Did you actually find those Dvaut posts?

Ken Jennings is another one who would have little chance against you on the math SAT which was the subject matter of my challenge to extreme Christians who think that you don't go to hell if and only if (I got it right this time) you believe in Jesus. And that isn't even his belief.

As to me I don't know if you mean now or thirty years ago when I was 46. In any case here are two math problems that I did without math just a few years ago. If you can answer them without looking them up, I might move you all the way to ninth. (Can't go further because Angry Queer got an 800 on the math SAT and I'm not sure you could. Simplicitis is passable because of his unfortunate post mentioning mental illness and because he was only Magna cum Laude rather than Summa cum Laude. [I can tell you several major posts of almost everyone on my lists without looking them up back to the beginning of this website. For instance, I had to take points from our two math Phd's who would normally rank 1-2 because one of them, about 12 years ago, claimed he wasn't all that good at logic puzzles or in teaching math challenged students and the other one, about eight years ago, said he wasn't interested in the work All the Cheese was doing on Gabriel's horn.]) If enough interest is shown I'll demonstrate more of that memory.

Here are the two math/not math questions.

1. The cubes add up to squares. ! + 8 +27 +64 = 1,9,36,100 etc.

Rather than using mathematical induction to verify this, devise a geometric algorithm that unfolds the increasingly big "rubic's cubes" and adds to those mini cubes that have previously formed a square, to form a bigger square.

2. If a roulette ball is slathered with paint and is put into the double zero slat and then moves one slat at a time clockwise or counter clockwise depending on a coin flip, it will eventually paint every slat. That might take several hundred flips. But when it is over there will a last slat painted. And the probability that a particular slat is the last one painted is, counterintuitively, 1/37 in all cases. Can you prove that? This problem was on this website so I trust you not to look it up. The high falootin Phds there, either couldn't answer it or used complicated math. I proved it with two sentences of pure logic (which they admitted was valid and kind of embarrassed them.) I'll give you 48 hours. Ninth place is on the line.

Still not going to say where I should be ranked except to say that I would be in trouble if the ranking included a particular person who is close to a regular poster here. That person did, I believe post a few times years ago but now doesn't. If the present poster I am referring to reads this and knows who I am talking about, they have my permission to divulge.

I have now written two good posts at your request. At the very least I expect in return that picture of the youngish man with the beard that other people receive in similar circumstances. No one has ever done that for me and it is one of my two main goals at this point. (The other is that when I die I become the only person ever to make BOTH of Unstuck Politics "Bump This Thread When,,, I thought I had it locked up but now I am a little scared that Dick Cheyney will beat me to it.

I don't think you did get it right. What you wrote means that believing in Jesus is necessary and sufficient for "not going to hell". While it is necessary, it is not sufficient. If you wanted to keep "only if", you need to add the other requirement - "you do not go to hell if and only if you believe in Jesus and have lived a moral life" (or whatever that requirement is, I don't keep track). Otherwise it's just "if".

However, I think your intention is to underscore that you if you don't believe in Jesus, you will automatically go to hell. The correct way to state this would just be the straightforward "if you don't believe in Jesus, you will go to hell". Note that this does not imply "if you believe in Jesus, you will not go to hell", whereas what you wrote does.

They don't add up to just any squares. The sum of the first N cubes adds up to the sum of the first N numbers squared. As most will remember, the sum of the first N natural numbers is [N(N+1)]/2, also known as the Nth triangular number. The sum of the first N cubes is this number squared.

I've seen quite a neat geometric demonstration of why this work in a youtube video, and there are also algebraic proofs for it which aren't inductive. I will happily concede that I would not come up with the geometric proof myself as my pattern recognition is terrible, and I always score very poorly on those online IQ tests which have nontrivial "which pattern comes next" questions. I can post an algebraic proof which is not inductive if you like.

I'll have a think about this now.

I have absolutely no idea what you're talking about here.

WRT to the math SAT: I was not educated in the US, so I have never taken a SAT. I did take something which IIRC was a practice math SAT when I was about 17 for shits and giggles and I scored pretty well. Can't remember if I scored 100% or not though, was a long time ago. If it wasn't 100%, it was close to it.

charge DS with rank ordering people's intelligence and he'll literally enter god mode

I think this nit has to be squared up a little. It should be "If you wanted to keep "if", you need to add the other requirement..." and "Otherwise it's just "only if".

By stating "if" alone, you are implying it is sufficient in and of itself whereas if you state "only if", it is implying that it is just one requirement, but a necessary one.

Yes, I think you're correct. Thanks!

--Where would Toothsayer go on the list? I was always a fan, but I also understand not for everyone.

He disagreed with me when I said that it should be illegal to set fire to a thousand-dollar bill in front of poor people to light your cigarette. I think he also disagreed when I said that Jefferson was scum for owning slaves. He sounded smart but he was wrong about too many things to make the top twenty if at all.

This took a bit longer than I care to admit. I'll go through my thought process.

I found it easier to imagine a pie chart with sectors numbered 1-6 and an arrow that starts in sector 1 and paints the sector it enters. Heads = move clockwise, tails = move anticlockwise.

First, we recognise that if we're currently in sector N, we need to have thrown N-1 more heads than tails or 6-N more tails than heads to get there. For example, if we're currently in sector 4, the only way we could have got here is by throwing 3 more heads than tails or 2 more tails than heads.

Next, we recognise that we can't have an excess of 3 heads without first having an excess of 2 heads and 1 head. This corresponds to the same as saying that if we're currently in sector 4, we must have already visited sectors 2 and 3 or sectors 5 and 6 or both.

Next, we recognise that our stopping condition is when we have reached both sector N and sector 7-N from opposite directions as this means we have visited each sector at least once. The exception to this is when we stop in sectors 2 or 6, as this means we have been progressing in one direction only.

This corresponds to saying that our stopping conditions are when we've seen an excess of:

5 heads or
4 heads and 1 tail or
3 heads and 2 tails or
2 heads and 3 tails or
1 head and 4 tails or
5 tails or
4 tails and 1 head

etc.

Each of these conditions corresponds to a specific sector that we stop in. For example, if we stop when we've seen an excess of 3 heads and 2 tails (in that order), we will end up stopping in sector 5.

Next, we need to make an argument that all these possibilities are equiprobable. Let's look at a couple of examples.

In order to stop in sector 5, we need to see:

An excess of 3 heads then an excess of 2 tails or
An excess of 1 tail then an excess of 4 heads

In order to stop in sector 4 we need to see:

An excess of 2 heads then an excess of 3 tails or
An excess of 2 tails then an excess of 3 heads

In order to stop in sector 6 we need to see:

An excess of 5 heads or
An excess 4 heads then an excess of 1 tail

But tails and heads are interchangeable. Seeing an excess of 2 heads then an additional excess of 3 heads (i.e. a total excess of 5 heads) is no different to seeing an excess of 2 heads then subsequently an excess of 3 tails or an excess of 1 head and subsequently an excess of 4 tails. Coins don't have a memory about prior excesses. (This part is definitely a bit hand-wavy, I'd like to formalise this a bit better).

But, assuming that we've shown that stopping when we've seen an excess of N heads and 5-N tails has the same probability independent of N, I believe this is what was to be shown.

Note that I sanity checked my reasoning with a "biased coin" test - if my logic "worked" on a coin that landed heads 99% of the time (i.e. every number would still be equiprobable), then my logic would be wrong, and this did in fact prevent me from going down a dead end along the way (using symmetry of the wheel to overlay a previously established "walk" onto a different starting point).

TLDR: To stop on N, we need to flip an excess of N heads and 37-N tails (or vice versa) and then stop, and when this happens is independent of N. Since there are 37 possible N's, this implies that they are equiprobable.

Wait, I think there is some faulty reasoning in all that. I forgot that the excess needs to be over the whole string, not just from a given starting point. So to stop at 6, we would need an excess of 1 tail then subsequent excess of 6 heads etc. So, we will stop whenever we've seen a net excess of 5 heads or 5 tails. I think the general principle still holds - we need to go forward 5, or back 1 and forward 6, or back 2 and forward 7 etc for a net movement/excess of 5 in either direction.

Didn't check your logic because my way is much shorter.

For a slat to be the last one with paint, two things must happen to it. The ball has to hit one of its sides but then, rather than go in on the next flip, it must turn around and then go all the way around the wheel to its other side. Once it does that it will usually quickly go into the slat from that second side. But no matter which side it enters from, it will be last slat filled. Each slat to be last needs a parlay. First be hit on one side before the experiment is over. Then it must make its way all the way to its other side without first falling into the original side. The first event is 100% for all slats (all slats will have the experience of "feeling" the ball on one side before the other side.) Once that happens, the second event (getting all the way around without ever falling into the original slat) is a 1/37 shot (just as it would be if someone with one dollar was playing a freezeout with someone with 37 dollars for a dollar a pop.) The parlay is 100% times 1/37 for all slats (including the two the ball touches at the beginning which are adjoining double zero).

I don't think I fully follow. Does this take into account that some slats have a head start? Let's pretend that the slats are numbered clockwise 1-37 for ease. At the start of the experiment, slats 2 and 37 already have had one of their sides touched, so aren't they "ahead" of, say, slat 18? I don't see how your logic accounts for this.

I'm not 100% on my own logic either. This is making me want to write a sim to check if your answer is actually correct now! I was shoehorning my logic to work into what you told me the answer should be, so it was rather hand-wavy at a key juncture.

You are forgetting that the numbers not next to 00 can get their first feel from either side. Put another way when a number far away from double zero gets it first feel, at least half of its competitors have been knocked out of the competition. But you don't need to know that. All you need to know is that all slats will get their 36-1 chance. It's not like there is a possibility that the ones further away will sometimes be shut out before they get that chance.