Predicting one of 2 characters you will randomly press

sctb 9 years ago |

Recent discussion: https://news.ycombinator.com/item?id=11824164

throwaway2016a 9 years ago |

For me it is pretty consistently 50% +/- 5 even after a minute. It doesn't seem to work well if you repeat the same key a lot (i.e don't flip to often).

overlordalex 9 years ago | |

This is actually due to a psychological thing where humans don't find long strings of repeated letters/digits "random" and hence will flip more frequently than you'd expect from a machine.

I also recall there being a problem in the use of 5-grams which lead to a sequence in which it guessed 0%, but I'm struggling to find it now.

js8 9 years ago | | |

If you would find the sequence I describe in my other comment, I think you would almost get that result, because you would (except at the beginning of the sequence) always pick the k-gram that has been seen one time less.

widforss 9 years ago | |

This is definetly not working. I'm certain that one can write a program like this, but I'm consistenly getting 50 % even though I use different strategies every time.

oneeyedpigeon 9 years ago | |

I got 52% after a good lot of keypresses (not sure exactly how many). I chose to not look at the screen at all during that time - I think that probably helps.

Forge36 9 years ago | | |

Mine started at 25% and worked it's way up to 54%. Which also involved me holding down keys for a few seconds. I'm not sure if it works or if Hacker New's user base is skewed towards people who are slightly better at being random. (or if that's cause by self reporting, would someone whose best was 80% want to come forward?)

Edit: Interesting tidbit: I pre-seeded at 100% by holding down F and it trended down to ~68% before I decided I shouldn't spend too much time on this. Maybe we didn't use it long enough?

udkl 9 years ago | | |

Anything < 50% is a fail for this program.

There is an inherent 50% probability of each character occurring next. i.e if you were to guess randomly, you would be right 50% of the time given enough trials.

LoSboccacc 9 years ago | | |

got it down to 50% but then started learning or I started focusing, it went as up as 56% before I got bored

andrezsanchez 9 years ago |

I did something like this a long time ago in a little competition for beating others' rock paper scissors programs, except I used the other players' moves relative to my own (e.g. my program could catch on to the other player choosing the move that would beat my last move) as well as their current pattern of winning or losing.

vorotato 9 years ago |

It works if you're trying to be random, however if you're trying to react and respond to its patterns you can win.

rohanprabhu 9 years ago |

So, I tried a simple python one-liner:

[os.urandom(1)[0] < 128 for x in xrange(0, 100)]

and typed 'd' for True and 'f' for False in the array, bringing the accuracy of the predictor down to 53%. Theoretically, I'm guessing doing it for large enough numbers should make it exactly 50%.

mikeash 9 years ago | |

If it doesn't come down to 50% for sufficiently large numbers, you've found a break in your OS's urandom and someone is about to be famous!

dzdt 9 years ago |

Doesn't work from my smartphone. Sounds like a fun challenge though, would like to try.

pi-rat 9 years ago | |

Seems like it's written using ES6 syntax, which many browsers can't handle yet. Safari stopped at a lambda ( () => ) here.

amake 9 years ago | |

I'm on OS X with Firefox 47.0.1 and it doesn't work for me either :(

justinsaccount 9 years ago | | |

I'm on OS X with Firefox 47.0.1 and it works fine. It doesn't display anything until the 6th keypress.

mikeash 9 years ago | | |

Doesn't work in Safari for me, but it does work in Chrome (after pressing keys six times).

justinsaccount 9 years ago | | |

Yeah, it uses this style function definition (ES6?) in a few places:

  document.addEventListener("DOMContentLoaded", () => {

Safari doesn't support that.

js8 9 years ago |

Is there, for a given k, an easy to remember sequence (easy to remember algorithm with little state) that has all the k-grams with the same frequency?

For k=1: 01 repeating

For k=2: 0011 repeating

For k=3: 00010111 repeating

justinpombrio 9 years ago | |

They're called De Bruijn sequences[1], and there's an algorithm for constructing them. I'm not sure how easy it is to run in your head, though.

[1] https://en.wikipedia.org/wiki/De_Bruijn_sequence#Algorithm

WillKirkby 9 years ago | |

probably relevant: https://en.wikipedia.org/wiki/De_Bruijn_sequence

edit: dangit, beaten to the punch.

tokenizerrr 9 years ago |

Very cool. Am getting ~50% accuracy so I'm glad to see my free will is still intact!

andrewclunn 9 years ago |

40%? Lame.