Number 1 and Benford's Law - Numberphile

Why number 1 is the "leading digit" more often than you may expect?
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NUMBERPHILE

Videos by Brady Haran

[aaronphillips]

We the party of science and math now. Biden's numbers are breaking this law hard.

[numberphile]

sorry, next time I'll put in on a locked off tripod, then re-film everything and second time with cutaways of hands and stuff...

80s style!

[aCloudOfHaze]

SAVE YOUR COUNTRY. THIS IS NOT ACCEPTABLE.
RIGGED2020

[anujlahoty]

What a concept, delineated unequivocally in a brilliant manner

[eoghan.]

The way that it intuitively made sense to me as soon as he said the law was as follows:
As we go up through the numbers we could stop at any point (this is the highest value in our set). No matter what the highest value is, the probability of any value below it starting with 1 can never be beneath 11%. However in most cases it will be above 11%, because we'll have gone through all the 1s but not all the other numbers.

[rabidlorax]

Guys I need help, look at the data coming out of Detroit, how can I put it back in the margin of error? I don’t wanna go back to prison :(

[DELLIS]

Hello to all the Sleepy Joe supporters!!! Tough scene

[jetmax_]

It’s been 12 years and the leading digit of this videos count is finally a 9.

In less than a month it will be back to a 1

[AlexanderEVtrainer]

For some reason whenever someone tells me that my decisions can be predicted with a law or formula, I tend to think all rebellious like "Oh no they can't!" But this actually makes sense. Wow I love Numberphile!

[LAnonHubbard]

I like the way this guy explains things. He's more to the point than some other presenters. Thanks for telling me about Benford's Law which I find pretty cool.

[subh]

Simple explanation: Draw the logarithmic scale along an axis. Paint red the regions where the numbers start with 1. The red paint will occupy about 30% of the line.
So, the assumption behind the underlying distribution from which the numbers are being drawn is that it should be an uniform distribution over the logarithmic scale.
Thus, if f(x) is the actual probability density, we should have f(x) = 1/x. [∵ if y=ln(x), F(y)=1 is distribution over y. Relationship: \int F(y) dy = \int f(x) dx]

[jonadabtheunsightly]

This only works if the numbers are chosen in a way that makes lower-magnitude numbers more common than would be expected in an even distribution.

Powers of two (or of any number) are a very nice demonstration, because they increase in magnitude at a fairly smooth rate as they go.

If the space of available numbers is finite and all numbers in that space are equally likely, you instead get something more like the sort of leading-digit probabilities people would tend to expect. So for example, in a sampling of genuinely random numbers between 1 and a billion, about 90% of them will (in base 10) be between 100 million and 1 billion; roughly 9% will be between 10 million and 100 million; and so on. With this sort of distribution, all leading digits are just about equally likely. You can make this more obvious by including the leading zeros, at which point it is straightforward that each leading digit, including 0, occurs 10% of the time.

[jbrowsingj]

I love this channel.
Thank you Brady!

[stevenjones]

Most intuitive way to think about it for me is that the ten 9s of the 90s are glued to one hundred 1s of the 100s, the one hundred 9s of the 900s are glued to one thousand 1s of the 1000s, etc. So the 9s are always linking arms with 1s that are ten times as many.

[JamesJones-ztyx]

It may be hard these days to use this to give an intuitive feel for Benford's Law, but if you have a slide rule, take a look at it.

[Kaczankuku]

If we first start from one, the frequency of appearing one is really as it was showed in the video but we can also start from nine and go to lower and lower numbers, then the frequency will equal to the frequency of one at first condition.

[soumyadatta]

Intuitively, I thought it would be a saw tooth wave, but mirror image of what Steve have shown. I mean sharply increasing then falling with a decay constant, not the opposite. Because the moment you reach an integer of 10, probability rapidly increases and then continue to fall slowly upto next rapid increment.

[BretFromPhilly]

It's just a log ratio, so the probability of a number falling between 1 and 2 is the same as finding a number between 5 and 10, or 150 and 300, and that's log (2/1). It's what you would find with exponential growth or decay.

[lidoz]

If you start with #20 and move upward you will get the same percentage... #1 is 30% only because it's the first number you are starting with and #10-19 follows shortly after

[GiulianoConteDrums]

This concept only works for one because it's the first digit to get all the probability as the power of 10 increases. Once you get into the 200s, the "2" probability is just catching up to the "1" probability because it already had increased from all the 100s, and same goes for "3" "4" and so on.