Using Statistics To Beat Wordle

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  • Опубликовано: 1 окт 2024

Комментарии • 14

  • @umachakraborty8568
    @umachakraborty8568 2 года назад +2

    Your voice is very low

  • @patrickjack2943
    @patrickjack2943 2 года назад +1

    PureWordle ! No math, no computer, no helpers, (except for notepad), no hacks, just notepad and the game in another window. Playing for 23 days, streak is 23. 3.6 guesses average. PureWordle. Their words, they're there.

  • @patricebetts6531
    @patricebetts6531 2 года назад +1

    It helps to know the list of previously used wordle words.

  • @shrugalic
    @shrugalic 2 года назад +2

    Great video, love all the detail!
    When I apply this strategy to the full combined word list of nearly 13k words, I also get LARES as the optimal first guess.
    However, when using the same nearly 13k words as guesses, but only look the resulting distribution of the subset of 2315 possible solutions to determine the optimal guess, the result is ROATE instead.
    This takes about 1-2 seconds to calculate using parallelized Rust (6-9 seconds for the combined list). Code is not ready to share yet, but I'll likely put it on GH eventually. Edit: lowered calculation times, because I underestimated the power of Rust's --release flag.

    • @RandomMathsInc
      @RandomMathsInc  2 года назад

      Nice! Interested to see it eventually! (I have heard of Rust from others but never actually written it myself)
      One other thing that I thought of that would really speed things up is to precompute all possible outcomes for all words (rather than just the first few steps) and just store that somewhere. If enough solving is done using the system it might be the most efficient way of going about it.

    • @shrugalic
      @shrugalic 2 года назад

      @@RandomMathsInc Rust is very fast, so I calculate the remaining set of solutions that are left when trying any guess on any solution. Only takes a few seconds to do this for all 12972 guesses with all 2315 solutions.
      For the next guess, I can then just lookup the set for guess2 + the secret, and intersect it with the set for guess1 + the secret, to get the remaining viable words.
      Rinse and repeat.

    • @shrugalic
      @shrugalic 2 года назад +1

      @@RandomMathsInc I probably posted this already, but can't find the comment, I assume it was deleted by YT. ;) My code is up on a website Microsoft bought, using an easy to guess username. ;) There's an email there to contact me if so desired.

  • @oskain
    @oskain 2 года назад

    There is a 100% winrate strategy that starts with Flame, Shunt, Brick, Podgy (in any order). I can link the video if you want but after those 4 words you can get the word in 2 guesses. Sometimes it requires some thought as you might need to guess a word to get rid of letters instead.

  • @brentvandenbossche
    @brentvandenbossche 2 года назад +1

    Really cool!

  • @Drewy62
    @Drewy62 2 года назад

    You can look at the two dictionaries the game uses by looking at the code in your browser. Possible words are about 2500 and possible guesses are like 5000. They use a different set of words for possible solutions and allowed guesses.

    • @RandomMathsInc
      @RandomMathsInc  2 года назад

      In hindsight, probably should have. I didn’t think of doing some code inspection of the game initially (might be in a follow up vid if I ever do one). Thanks for mentioning it tho!

  • @vikromnarula1746
    @vikromnarula1746 2 года назад

    Dopeee!

  • @lukevideckis2260
    @lukevideckis2260 2 года назад +1

    Shouldn't you calculate expected number of guesses instead of expected # of remaining words?

    • @shrugalic
      @shrugalic 2 года назад +2

      I think so, but it would be a *lot* more computationally expensive to do so. Naively it means playing all possible remaining games for each guess. Currently I'm trying to to something like this, but only for a limited number of promising guesses, which are suggested by this method.