So a yes might rule out 75% of remaining options (for example) which provides 2 bits of information.
Yes, that looks like a mistake -- a truth booth only has 2 outcomes, so it can produce at most 1 bit of information.
Regarding the other mentions on the page of information levels exceeding 1 bit: Those are OK, since they allow match-ups, which for 6 people have 7 possible outcomes, thus can yield up to log2(7) ≈ 2.81 bits.
If all you can get is a "true" or "false" you expect, at most, one bit of information.
Optimal play to reduce the search space in both follow the same general pattern - the next check should satisfy all previous feedback, and included entries should be the most probable ones, both of those previously tested, and those not. If entries are equally probable, include the one which eliminates the largest number of remaining possibilities if it is correct.
For wordle, «most probable» is mostly determined by letter frequency - while in Mastermind, it’s pure probability based on previous guesses. For instance, if you play a Mastermind variant with 8 pegs, and get a 2/8 in the first test - each of your 8 pegs had a 1/4 chance of being correct. So you select 2 at random to include in the next guess.
If you then get a 2/8 from the second - you would include 4 previous entries in the next guess, 2 entries from the first that was not used in the second, as well as 2 entries from the 2nd - because the chance you chose the correct entries twice, is less than the chance the two hits are from the 6 you changed.
> Optimal play to reduce the search space in both follow the same general pattern - the next check should satisfy all previous feedback, and included entries should be the most probable ones, both of those previously tested, and those not.
The "next check should satisfy all previous feedback" part is not exactly true. That's hard-mode wordle, but hard mode is provably slower to solve than non-hard-mode (https://www.poirrier.ca/notes/wordle-optimal/) where the next guess can be inconsistent with previous feedback.
I don't think that's a justified assumption. I wouldn't be surprised if wordle puzzles intentionally don't follow common letter frequency to be more interesting to guess. That's certainly true for people casually playing hangman.
Thank you! I might look into this once I break my current streak of the localised wordle clone I'm playing now.
I always try to use as many different bits for the first few rounds...
But then again, maybe I'm not so good at these kinds of games as I think.
I noticed that for any match up score of X, the following match up would keep exactly X pairs in common. So if they scored 4/10 one week, they would change 6 couples before the next one. Employing that approach alone performed worse than the contestants did in real life, so didn't think it was worth mentioning!
> Employing that approach alone performed worse than the contestants did in real life, so didn't think it was worth mentioning!
Yeah, this alone should not be sufficient. At the extreme of getting a score of 0, you also need the constraint that you're not repeating known-bad pairs. The same applies for pairs ruled out (or in!) from truth booths.
Further, if your score goes down, you need to use that as a signal that one (or more) of the pairs you swapped out was actually correct, and you need to cycle those back in.
I don't know what a human approximation of the entropy-minimization approach looks like in full. Good luck!
For a start, the setting is an emotive one. It's not just a numeric game with arbitrary tokens, it's about "the perfect romantic partner." It would take an unusually self-isolating human to not identify who they feel their perfect match should be and bias towards that, subconsciously or consciously. We (nearly) all seek connection.
Then, it's reality TV. Contestants will be chosen for emotional volatility, and relentlessly manipulated to generate drama. No-one is going to watch a full season of a handful of math nerds take a few minutes to progress a worksheet towards a solution each week coupled with whatever they do to pass the time otherwise.
In my hypothetical version of "Are you the one?", the math nerds would be giving commentary and explaining the math behind how they'll solve "Are you the one?" while also hilariously explaining how foolish the contestants' theories are.
Um, what about those of us who watch Blood on the Clocktower streams?
The linked post is a very thorough treatment of AYTO and a great read. I really like the "guess who" bit on how to maximize the value of guesses. It's a shame the participants aren't allowed to have pen and paper—it makes optimization a lot trickier! I'm impressed they do as well as they do.
[0]: https://danturkel.com/2023/01/25/math-code-are-you-the-one.h...
Loved your post, really enjoyed getting into the meat of it. I wanted to position mine to a layman, kept asking myself "can I explain this to my Dad?"
I think where the post falls short is the absence of a silver bullet that contestants can use to win the game sooner.
[0]: https://github.com/daturkel/pyto/blob/master/AYTO_S8.ipynb [1]: https://blogs.sas.com/content/operations/2018/08/14/are-you-...
Basically, using the entropy produces a game tree that minimises the number of steps needed in expectation -- but that tree could be quite unbalanced, having one or more low-probability leaves (perfect matchings) many turns away from the root. Such a leaf will randomly occur some small fraction of the time, meaning those games will be lost.
For concreteness, a game requiring 6 bits of information to identify the perfect matching will take 6 steps on average, and may sometimes require many more; a minimax tree of height 7 will always be solved in at most 7 steps. So if you're only allowed 7 steps, it's the safer choice.
> Basically, using the entropy produces a game tree that minimises the number of steps needed in expectation
It might be even worse than that for problems of this kind in general. You're essentially using a greedy strategy: you optimize early information gain.
It's clear that this doesn't optimize the worst-case, but it might not optimize the expected number of steps either.
I don't see why it couldn't be the case that an expected-steps-optimal strategy gains less information early on, and thus produces larger sets of possible solutions, but through some quirk those larger sets are easier to separate later.
I'm not following your logic. Consider the setup we actually have:
1. You get to ask a series of yes/no questions. (Ignoring the matchups.)
2. Each question can produce, in expectation, up to one bit of information.
3. In order to achieve this maximum expectation, it is mathematically necessary to use questions that invariably do produce exactly one bit of information, never more or less. If your questions do not have this property, they will in expectation produce less than the maximum amount of information, and your expected number of steps to the solution will increase, which contradicts your stated goal of minimizing that quantity.
You get the minimum number of steps needed in expectation by always using questions with maximum entropy. Yes. But those questions never have any variation in the amount of entropy they produce; a maximum entropy strategy can never take more - or fewer - steps than average.¹
¹ Unless the number of bits required to solve the problem is not an integer. Identifying one of three options requires 1.585 bits; in practice, this means that you'll get it after one question 1/3 of the time and after two questions the other 2/3 of the time. But identifying one of 128 options requires 7 bits, you'll always get it after 7 questions, and you'll never be able to get it after 6 questions. (Assuming you're using a strategy where the expected number of questions needed is 7.)
This game is constructed such that the questions you can ask are not arbitrary, so you cannot choose them to always produce one bit of entropy (you need to frame your questions as ten matchups in parallel, using all the contestants exactly once) and the number of bits you need may indeed not be an integer.
Because you can't choose your questions to partition the state space arbitrarily, that affects not just the question you ask today, but also previous days: you want to leave yourself with a partitionable space tomorrow no matter what answers you get today.
In the Guess Who analogy, it's against the rules or at least the spirit to ask "does your character have a name which is alphabetically before Grace?". That would allow a strategy which always divides the state space exactly in two.
That is one case where root-to-leaf path lengths can vary, though it's not obvious to me that it exhausts all such cases -- in particular, even if we have "ideal leaves" (numbering a power of 2, and each equally likely), it's not clear that there is always a question we can ask that divides a given node's leaves exactly in half.
Discussing "events" (ie, Truth Booth or Match Up) together muddles the analysis a bit.
I agree with Medea above that a Truth Booth should give at most 1 bit of information.
So if only MUs, we're talking around 10 events - meaning you could get enough information on MUs alone to win the game! Conversely, it would take about 20 events to do this just for TBs.
It's not super obvious from the graphs, but you can just about notice that the purple dots drop a bit lower than the pink ones!
Hope this helps
It's interesting how close 22.5 is to the 21.8 bits of entropy for 10!, and that has me wondering how often you would win if you followed this strategy with 18 truth booths followed by one match up (to maintain the same total number of queries).
Simulation suggests about 24% chance of winning with that strategy, with 100k samples. (I simplified each run to "shuffle [0..n), find index of 0".)
I am thinking about making a website for it when the next season starts.
Also: in Germany at least they have 10 x 10 candidates from the start, but sometimes they add a 11th or even 12th of one gender so that there are double matches (e.g. 1 woman has two man as match and needs to find one of it to succeed). This raises the possible combination quite a bit.
Yes there's a gender fluid season and a season where someone had > 1 match, as well as people leaving part way through the season (apparently perfect matches are interchangeable...). All very interesting spins on the core problem to solve; would be really interested if anyone tries to tackle those seasons.
My lived childhood is old enough to be someone's "research."
My wife and I bought the game - it's a great turn based came you can play whilst having trashy reality shows on in the background!
Other than that, in my research I came across a boardgame called Mastermind, which has been around since the 70s. This is a very similar premise - think of it as "Guess Who?" on hard mode.https://www.mcsweeneys.net/articles/the-mastermind-box-cover...
> A truth booth is where a male & female ...
The use of "male" and "female" as nouns sounds very unnatural. "A man and a woman" would be a little less jarring, imo.
It’s mostly about how to elicit the information from the contestants. Once you have the probabilities from them, it seems relatively straightforward.
An obvious one is the traitors, but I dunno if there's much you can do with this one as the contestants rarely gain much concrete information.
"Deal or no deal" / "let's make a deal" would have interesting game theory approaches - probably has a lot of parallels with Monty Hall?
Countdown (UK) - solving the maths puzzles on here using integer programming would be cool
Well I guess free money except for that one. In that one, one of the contestants, Danny, did the math for optimizing their remaining Truth Booths and Match Ups to get it down to a 50/50 shot.
The interactive visualizations on this post are fantastic. More technical content should be presented this way. Makes complex probability much more intuitive.
Would've been a great Pudding post imo, but oh well, happy to find this nice devblog instead.
Really impressive imo. I don't remember the last time I was this engaged reading an article on HN.
> 0 0.3679
> 1 0.3679
> 2 0.1839
> 3 0.0613
> 4 0.0153
> 5 0.0031
For 0, it's a well known [1] result 1/e, I remember a puzzle where people left their hat and then pick one randomly.
Looking at the table it looks like the general formula is 1/(e*n!) that is a Possion distribution. Compare with https://en.wikipedia.org/wiki/Poisson_distribution#Examples_...
Anyway, I'm not sure if my observation helps too much to solve the original problem.
[1] at least I know it :)
One thing the show runners do subtle alterations that makes the logic much harder.
The Traitors has to do lots of these tricks when not playing the Celebrity edition because there's a self-selection for the sort of person who has already played Werewolf/Avalon-type games.