Large-world networks and species, social, and internet collapse

A Small-World Network is found in many naturally and and artificial situations. At the other extreme, if a network is totally connected, a complete digraph? You get zero information. This can be translated into species collapse, zero advertising revenue, and others such as ineffectual internet search.

Can this happen? In animal populations a fully connected graph is when everyone is closely related. The gene pool is stagnant. That population dies. In the internet, it could happen if everyone is on FaceBook and knows everyone else, or they are on Twitter, and follows everyone else. Another example, in the Mathematical field, a complete graph would mean that everyone would have an Erdős number of 1.

Is there a graph measure that correlates to the above? The more connected the less information, an Entropy?

Information loss, heat death. Time for another Big Bang.


  • October 3, 2015: Continuing the above line of thought, perfect order leads to zero information.
  • October 5, 2015: A Zombie Apocalypse is just a food network that is maximally connected, your next meal is sitting right next to you.


Pondering via the web

This morning got up early and sat at my desk while having a coffee. I got multiple windows up on my multi 24″ monitor setup and my adequate quadcore Windows personal computer. Perused the latest news and tech stories.

Then I opened my Feedly account and looked at my Science feeds. “The Reference Frame” has a new post. I like that blog. Most of the time I don’t know what he is writing about, and the anti-climate change diatribes sound like Fox News, but in all it has some tidbits for us science aficionados. Like today, he is discussing something about intelligence. Which leads me to recall that there was another attempt to present an “equation” of intelligence by another sciencecy type, Bart Kosko. Bart was discussing creativity. Creativity is the measure of the number of results or responses to the same number of data or stimulus, or something like that.

So, I searched for the Kosko reference. Couldn’t find it. I did find this quote, which is very good: ““Whenever the invisible hand isn’t operating the iron fist is”” — Bart Kosko. Then I sidetracked to the Subsethood theorem. Then to “The Sample Mean“.

Which of course, I now had to browse the web to make sure I really knew what the sample mean is. This led to views of various sites and of course Wikipedia (which led to all sorts of side tracks into other things).

Anyway, back to the sample mean. While reading up on it, I thought well (not listing the paths that led here) if the standard deviation is so weak when the data has huge outliers, why not do the same computations using the Median measure? I ran Excel and did a few. Not bad. If data is pretty ‘regular’ it is close to the STD. Hmmm. Maybe this should be part of statistics? Well, more browsing and of course it turns out that it is. It is called the Median Absolute Deviation (MAD). MAD can even be added to Excel.

Well, that was an hour. Not wasted. Certainly better than watching TV. Now back to the task at hand, working on something that may make a good Kickstarter project.

More links

The multiplicative inverse of zero

This is silly post, but was inspired while reading something which mentioned that zero has no multiplicative inverse. Since grade school I’ve always been puzzled by how math has weird corner cases like this, another: 1 divided by 3, never ends! Anyway, what if zero did have an inverse? Is there some Abstract Algebra thingy that has such a property?

So, lets call the inverse “j”, not ∞, it already has some meanings. Then we have: j0=1. Or, 0=1/j. This means that 0/0 = 1. I.e., (1/j)/(1/j) = 1*j/j*1 = 1.

Is there a physical meaning? What could 1/0 mean? Since multiplication can be interpreted as a scaling, then what happens when you scale something to zero? Well, using the analogy of a map or engineering drawing, a scaling doesn’t destroy a physical measurement, it allows a smaller or large value represent the other value. In a floor plan we may see something like 1/4 inch = 1 foot. In a full size scale, the multiplier is 1. As the multiplier becomes smaller we can represent larger measures. At zero scale, we represent infinity. Thus, scaling infinity to zero, still scales the whole thing, which is 1.

Yeah, doesn’t really work, unless your a mystic. Oh well, like I said silly. I won’t quit my day job.


Are lottery odds incorrect, and your chances drastically worse?

A simple application of elementary probability shows that lottery odds as reported by the operators are incorrect. Since there are independent events the multiplication rule applies.

A simple application of elementary probability shows that lottery odds as reported by the operators are incorrect.

Note: I have delayed posting this for a while since the results I come up with seem incorrect. Hopefully someone can respond and tell me where the problem lies.


If I follow the logic expressed at What are the odds of hitting the same number twice on a roulette table? and apply it to the lottery situation then the lottery is computed correctly by the lottery operators. The critical aspect is that the first selection’s probability doesn’t matter to the person taking the risk. The second part of the cited page gives the odds for a specific number being picked twice. I think that case applies to an external observer to a particular lottery game instance. Huh?

About a year ago while speaking with my brother Robert on the phone he casually mentioned a joke he made. He said lotteries are so funny since the winning number has to be picked twice in order to win. First by the lottery company and then by the player. We both laughed.

But, then later I did a double take, huh? That is true, you have to pick the number set, and then a few days later, the lottery company will also pick a set. If the sets match then you win. Ok, that makes sense. But, if seen this way, the probability value they give for winning couldn’t be correct. Could it? They only give the odds of picking any set, not the winning set. It has to be much harder to win, thus, the probability much lower.

How to compute the Probability?
In probability theory there are rules for combination of events. If the events in an “experiment” are independent, you just multiply the probability values of each: P(A and B) = P(A intersection B) = P(A)P(B).

Further details are on this High School wiki page:

Multiplying probabilities

Probabilities are multiplied together whenever an event occurs in multiple “stages” or “steps.” For example, consider rolling a single die twice; the probability of rolling a 6 both times is calculated by multiplying the probabilities for the individual steps involved. Intuitively, the first step is simply the first roll, and the second step is the second roll. Therefore, the final probability for rolling a 6 twice is as follows:

P(rolling a 6 twice) = P(rolling a 6 the first time) X P(rolling a 6 the second time) = 1/6 X 1/6 = 1/36 approx 2.8%

Similarly, note that the multiplication of probabilities is often associated with the use of the word “and” — whenever we say that some event E is equivalent to all of the events X, Y, and Z occurring, we use multiplication to combine their probabilities (if they are independent).

More info on this wikipedia entry: Probability, Mathematical treatment

Does this apply to the lotteries, like Powerball? There are two events, though separated by days. The consumer, player, picks a set, then later the operator picks there own set. And, they are independent, neither event is dependent on the other. So, the problem, to me, is interpreting the “experiment”. I contend that the whole game, which takes place over a few days is one thing, an experiment, and so the multiplication rule applies.

Rolling a 195,249,054 sided Spherical polyhedron
Another way of relating them is to use two die rolls. But now instead of a die with six faces we use N faces, where N is the total number of possible number sets we could pick in a lottery game. This “die” is really a form of Spherical polyhedron. Lets say N is 195,249,054 possible unique numbers, which correspond to each possible set. So when we roll two dice the total probability would be (1/195,249,054 X 1/195,249,054). Remember, these are “normal” die, just having a ginormous number of faces.

The above is not even mentioned in the Lottery math references, for example, this Wikipedia entry, Lottery mathematics. So some conceptual misunderstanding on my part is very likely.

Lets take an actual example, the Powerball lottery states on their “Powerball – Prizes and Odds” page that to win the Grand Prize the odds are: 1 in 195,249,054. This is derived by application of math stuff to determine the combinations of the five white balls (1-59) and a red ball (1-39).

If we apply the multiplication rule the actual probability of winning is:

1 in 38,122,193,087,894,916

That’s 1 in 38 quadrillion. Big difference!
In scientific notation: 3.8122193087894916 x 1016

What is the Expected Value now?

This analysis couldn’t be correct. First, the number is too large, there are too many winners. Second, I have never heard of anything like this. Surely if this were the case it would be news. So where is the mistake?

I think it has something to do with the “same set of numbers”. Then its not just a simple multiplication of probability? If I find out, I’ll update this post.

So what?
You should not be paying the “idiot tax”. True, but when the prize reaches 100 million I bet there are some math professors out there buying a ticket too. Further, it is an interesting math subject.

This article analyzes the occurrence of a lottery draw that duplicated the same numbers and argues that my kind of instinctive analysis above is incorrect. Adventures in Probability. So, perhaps, the way to look at this issue is to compute the probability of the same winning combination being picked twice in a row? What the article says that it is 3.8 X10^16 but this has to be multiplied by the amount of combinations, so:
(1/((3.8 x 10^16) * 195,249,054))*195249054 = 1/195249054. That same as what the lottery provider quotes!

I don’t get it yet. Then why is the two dice example where each die has ‘N’ faces not calculated in the same way?

Further Reading

Off Topic

Groovy program to print the product:

x = new Long('195249054');
printf('%,d',(x * x));