# Where does your birthday land on pi?

#1

#2

A mathematical question for someone versed in this kind of stuff:

So piâ€™s decimal representation is infinite and never-repeating - Does it follow that it contains all numerical combinations?

(i.e. is there somewhere deep in the googols of numerals a stretch of pi with a thousand 7s in a row? Million 7s? Can it be proven or disproven? If not, what precludes certain patterns from forming?)

#3

It depends on whether or not itâ€™s a normal number. AFAIK, theyâ€™re not yet sure that pi is normal, but they think it is.

#4

Mine lands on Pi â€śa la modeâ€ť.

#5

Yay!

Happy Pi Day!

#6

A more difficult problem is to find the smallest decimal number that never appears in the digits of pi.

#7

Todayâ€™s Pi/pie themed Something Positive made me LOL:

#8

Reminds me of this clip from â€śPerson of Interestâ€ťâ€“

#9

Pah, automatically reduced the year to two digits. I want to find my birthday in ISO 8601 format, the only true date format.

#10

AmenÂ

#11

My ISO 8601 birthdate is much earlier in pi than my MMDDYY birthdate.

#12

So what is the likely answer?

#13

Thatâ€™s essentially a subset of the problem I stated. If pi is truly infinite with no hard limit on possible combinations, it should contain all numbers. Unless weâ€™re getting into some Cantor stuff about various dimensions of infinityâ€¦

#14

I noticed that, too. Isnâ€™t truncating to YY what got us into the Y2K mess? Good lord! Does Pi have a Y2K problem we havenâ€™t remediated?!?!?!?

OK. Iâ€™ve calmed down again now.

#15

I havenâ€™t done the math, but I think the problem is that, on average, you have to go 100 times further into pi to find a given 8-digit number than a given 6-digit number.

So the probable reason that theyâ€™re not doing 4-digit years is that that it would require that much more computing power.

#16

The likely answer is that pi is normal, on the grounds that most real numbers appear to be normal. But thereâ€™s no proof as yet.

Should? I donâ€™t know why that would follow. A sense of duty?

#17

Because otherwise it would settle into a loop. Since it doesnâ€™t loop (as has been proven), it â€śneeds to exploreâ€ť* the phase space of combinations further and further. Which on the plane of infinity means it either: i) explores all combinations; or ii) skips some and keeps going further. (Which is the core of my question. Does it skip? If so, why and what combinations?)

Or so my logic tells me. Admittedly though, Iâ€™m not a mathematician. Which is why I asked for an answer from a mathematician.

#18

I meant: Based on the assumption that pi is most likely normal, what is the answer to my original question?

#19

Many numbers donâ€™t settle into a loop, and yet donâ€™t contain many possible strings. For example the Thue-Morse sequence is non-periodic but contains no triples (three-in-a-row) of any item.

#20

I can point you to the wikipedia snippet which relates the most closely to your expression above, which says:

Even though there will be sequences such as 10, 100, or more consecutive tails (binary) or fives (base 6) or even 10, 100, or more repetitions of a sequence such as tail-head (two consecutive coin flips) or 6-1 (two consecutive rolls of a die), there will also be equally many of any other sequence of equal length. No digit or sequence is "favored".

But Iâ€™m not convinced myself that any arbitrary sequence of digits will eventually turn up in any given normal number (such as pi, if it is normal).