One advantage to having a couple dozen digits of pi memorized is having a psuedo random number generator at your disposal.
Useful for generating random rocks paper scissors moves, or in this case, random directions (odd = right, even = left)
Useful for generating random rocks paper scissors moves, or in this case, random directions (odd = right, even = left)
Yes, that's a well-established effect. Also, most people tend to choose "rock" for the very first throw of a session.
f: π(k) --> δ(k)
So even if Pi could be used as a pseudorandom generator (and it actually can't be), you'd lose that property by defining an injective map from your domain of inputs to the codomain of outputs.
0: 0 Rock
1: 1 Paper
2: 2 Scissors
3: 0 Rock
4: 1 Paper
5: 2 Scissors
6: 0 Rock
7: 1 Paper
8: 2 Scissors
9: 0 Rock
0/Rock: 4
1/Paper: 3
2/Scissors: 3
The common way of dealing with this is to "ignore" any number that would make the set biased. Here you would ignore 9 and you have an even distribution. So, you're playing 7 rounds of RPS and you go
3/R
1/P
4/P
1/P
5/S
9! SKIP! 2/S
6/RAs for why converting digits in this way matters - a lot of randomness is expressed by the entropy. It's harder for you to correctly guess the sequence {1,7,9,3,6,8,2,4} than it is to guess the sequence {1,1,1,1,0,0,0,0}.
If I ask you to guess a decimal digit I've chosen "randomly", you have a 1/10 chance of being correct. If I ask you to do the same for binary digits, you have a 1/2 chance of being correct.
Basically you want to think of these as subsequences, not individual numbers. If Pi is normal (which is a big if), then Pi is normal in every single base, including decimal or binary. But it's not generally true that a normal number generates another normal number by mapping each digit to the digit's parity.
If pi is [absolutely] normal though all sequences exist in it at equal frequency. Meaning that for any given sequence there is an infinite number of positions in pi to find it and that all the possible following digit sequences are equally likely.
So the computer could never know the next digit.
Aside: guessing a D16 roll seems way more likely than guessing a nibble of binary, and perhaps a little less likely than guessing 4 coin flips!??
And why is that a problem for generating random arrow presses?
{0,1,2,3,4,5,6,7,8,9}
{0,1}
You're selecting from a space of 10 digits for inputs, but the computer only has to guess from a space of 2 digits for outputs.
Nobody knows! Everyone in math is sure that it is, but nobody had found a proof yet. (It may be false...) An extension to this question is if pi is a "normal number". More details: https://en.wikipedia.org/wiki/Normal_number http://mathworld.wolfram.com/NormalNumber.html
[There are some technical details because pi is not a random number, but for the sake of simplicity, let's assume that pi is a random number.]
It's much easier if we'd live in a word that use base 8 instead of base 10.
Let's suppose that we have the sequence of digits of pi in base 8.The algorithm of the GP is to replace {0,2,4,6}->0 and {1,3,5,7}->1 to obtain a binary "random" sequence.
Your alternative is to write pi in binary, and use it as a "random" sequence. But if this is a good "random" sequence then you can pick every third digit and get another good "random" sequence. [Here good means something like iid with uniform distribution]
But if you start at the correct position, it's equivalent to pick every third number of the binary representation and to classify the digits in the base 8 representation as even or odd.
If you choose other starting points to pick every third digit, you get alternative maps:
* low and high: {0,1,2,3}->0 and {4,5,6,7}->1 (like in the roulette[1])
* crazy: {0,1,4,5}->0 and {2,3,6,7}->1
These other two selections produce also good "random" sequences.
The important part is that the projection that is selected maps the same number of elements to each element. In this case the three methods maps 4 elements to 1. This ensures that it maps iid with an uniform distribution to an iid with a uniform distribution.
Moreover, you can pick any arbitrary 4 numbers and map them to 0 and map the other 4 to 1 and it will work as well as the other three maps I used. (This is like the red/black option in the roulette[1].)
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Back to base 10. Any map that maps 5 number to 1 will maps iid with an uniform distribution to an iid with a uniform distribution. In particular the even/odd map that the GP is using is fine.
With this map you loose a lot of entropy, but since there is infinite entropy you can drop a lot of it and still keep infinite entropy. It's not as efficient as using the base 2, but it correct.
[1] An ilegal fair roulette, with 36 numbers, without the green 0.
You're correct that if Pi is normal, it's normal in all bases. But that's precisely the point I'm getting at - if Pi is normal, you need to use base 2 for this to work because your codomain is just {0,1}.
Mapping a number to another number such that each digit becomes its own parity is materially different from converting that number to base 2. They aren't the same thing whatsoever, and you can't generally take a normal number and create another normal number this way.
So even if we accept the reasonable conjecture that Pi is normal, you still need to map it to the same base as your codomain in order for its entropy to be preserved. The proposed function is injective and destroys entropy.
You can use any base that is a multiple of 2 (like 10) and then apply a parity function to the digits. If pi is normal, then the digit parity sequence is also normal.
Sure you lose some entropy, but in some sense the digits of pi have 0 entropy anyway since they can be calculated. And in the other sense of treating the digits of pi as an unknown random sequence, there is infinite entropy, so throwing away 70% of it doesn't matter.