My solution was to generate random sentences in my head. I then iterate through the letters of each sentence, and if the letter is M or later in the alphabet, I select right. If the letter is before M in the alphabet I select left.
How random can you be? (2019)
11–20 of 60 posts
Re: How random can you be? (2019)
#12Adversarially it can be defeated. However, when trying to be faithfully random it does pretty good
Re: How random can you be? (2019)
#13So I did 128 presses and it was 48% right at guessing me. I guess I'm quite random, since ~50% is what you'd expect for an actual coin toss.
150 presses and 51% here, just trying with my fingers on my smartphone :)
And now to 200 with 49%. Feeling not so subtley smug.
And now to 400 with 50%. Admit I'm getting bored now.
Now the question I have to answer, is that because of my professional experience and history, something inherent in my mind, or was it...just random?
edit: just consciously trying to be random, not using any deliberate or external tricks.
Second question: what percentage of the population behaves like me and is it qualitatively different from the population that doesn't?
I don't believe I actually am random, but whatever it is i'm doing the results are somewhat statistically unlikely. Is knowledge/experience with randomness itself sufficient to defeat this method?
Re: How random can you be? (2019)
#14My solution was to generate random sentences in my head. I then iterate through the letters of each sentence, and if the letter is M or later in the alphabet, I select right. If the letter is before M in the alphabet I select left.
That's likely to introduce bias because the probability of the two halves aren't equal... My initial instinct would be that the latter half would be less likely (since it has Q, X, Z), but maybe not - as a rule of thumb the most common letters are ETAOIN SHURDL - so 6 from the second half, and 6 from the first.
Re: How random can you be? (2019)
#15My solution was to generate random sentences in my head. I then iterate through the letters of each sentence, and if the letter is M or later in the alphabet, I select right. If the letter is before M in the alphabet I select left.
According to English letter frequencies [1] that gives a right probability of
3.0129+6.6544+7.1635+3.1671+0.1962+7.5809+5.7351+6.9509+3.6308+1.0074+1.2899+0.2902+1.7779+0.2722
= 48.7294 %
even though you summed 14 of the 26 letters (a 13-13 split would lower the probability to 45.7165%).
[1] https://www3.nd.edu/~busiforc/handouts/cryptography/letterfr...
Re: How random can you be? (2019)
#16My solution was to generate random sentences in my head. I then iterate through the letters of each sentence, and if the letter is M or later in the alphabet, I select right. If the letter is before M in the alphabet I select left.
That's likely to introduce bias because the probability of the two halves aren't equal... My initial instinct would be that the latter half would be less likely (since it has Q, X, Z), but maybe not - as a rule of thumb the most common letters are ETAOIN SHURDL - so 6 from the second half, and 6 from the first.
They minimized bias already by splitting 12-14, see my other comment.
Re: How random can you be? (2019)
#17Adversarially it can be defeated. However, when trying to be faithfully random it does pretty good
Yeah, it was only 44% right on my 100 presses but I was trying to beat it, not be random. You can sometimes get surprisingly long runs of wrong guesses out of it by repeating your guess once you get a wrong.
Re: How random can you be? (2019)
#18What's the easiest pseudo-random sequence one can calculate mentally in order to beat this?
This one doesn't really "beat" this, but here is a simple PRNG that you can trivially compute using mental arithmetics proposed by George Marsaglia [1]: 1. Select an initial random number (seed) between 1 and 58. (This is accomplished by mentally collecting entropy from the environment, e.g. counting a group of objects you don't knew the count of before) 2. Multiply the least significant digit by 6, add the most sign…
You need to escape the * in the formula or indent it as code:
42 -> 2*6+4=16 -> 6*6+1=37 -> ...Re: How random can you be? (2019)
#19It's incredibly easy to cheat this since you actually, in real-time, can see the feedback of how random it thinks you are.
Re: How random can you be? (2019)
#20It's incredibly easy to cheat this since you actually, in real-time, can see the feedback of how random it thinks you are.
And how exactly do you use this information?