“Are you the one?” is free money
blog.owenlacey.dev
blog.owenlacey.dev
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-...
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!
This is incorrect, the correct strategy is mostly to check the most probable match (the exception being if the people in that match has less possible pairings remaining than the next most probable match).
The value of confirming a match, and thus eliminate all other pairings involving those two from the search space, is much higher than a 50/50 chance of getting a no match and only excluding that single pairing.
Do you have any hard evidence, or just basing this on vibes? Because your proposed strategy is emphatically not how you maximize information gain.
Scaling up the problem to larger sizes, is it worth explicitly spending an action to confirm a match that has 99% probability? Is it worth it to (most likely) eliminate 1% of the space of outcomes (by probability)? Or would you rather halve your space?
This isn't purely hypothetical, either. The match-ups skew your probabilities such that your individual outcomes cease to be equally probable, so just looking at raw cardinalities is insufficient.
If you have a single match out of 10 pairings, and you've ruled out 8 of them directly, then if you target one of the two remaining pairs, you nominally have a 50/50 chance of getting a match (or no match!).
Meanwhile, you could have another match-up where you got 6 out of 10 pairings, and you've ruled out 2 of them (thus you have 8 remaining pairs to check, 6 of which are definitely matches). Do you spend your truth booth on the 50/50 shot (which actually will always reveal a match), or the 75/25 shot?
(I can construct examples where you have a 50/50 shot but without the guarantee on whether you reveal a match. Your information gain will still be the same.)
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.
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.
The faster you can discard all words containing «e» because of a negative match, the better.
If you want to be really optimal, you’ll use their list of possible words to calculate the actual positional frequencies and pick the highest closest match based on this - that’s what «mostly» was meant to imply, but the general principle of how to reduce the search space quickly is the same
> 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.
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.
For example, if your first guess on wordle is BOUND and you learn that the word is _OUND, you know the answer is one of FOUND, HOUND, MOUND, POUND, ROUND, SOUND, WOUND. Satisfying all previous feedback leaves you checking those one at a time and losing with probability 2/7. Or you could give up the 1-in-7 chance of winning in 2 and trade it for certainly winning in either 3 or 4: HARMS checks four of those options, and WHOOP identifies the remaining three.
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...
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.
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's true but not relevant; needing a non-integer number of bits makes it possible for some games to end one turn faster than other games, but it doesn't make it possible for a maximum-entropy strategy to have more variation than that one maybe-necessary-maybe-not turn. The scenario described by my parent comment still isn't possible.
> 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)
You're confusing two types of questions. The question where you're targeting one bit of information is "are Matt and Felicia a match?", just one pair.
A full proposed matchup of n pairs has n+1 possible answers and so your goal is to produce log₂ (n+1) bits. (Better conceptualized as one base-n+1 "bit".) I agree that it's not obvious how to do this or to what extent it's possible.
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.
> 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 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.
If all you can get is a "true" or "false" you expect, at most, one bit of information.
If you ask the question "which of A, B, or C is true?" then you're not asking a yes/no question, and it's not surprising that you expect to gain more than 1 bit of information.
edit: just saw that the article fixed this recently, and the values are now <1
A true answer for a potential match is actually a state update for all of the (n-1) edges connecting either contestant, that's 2(n-2) edges that can be updated to be false. Some of these may already be known from previous rounds' matchups but that's still more than a single binary.
If both of these are equally likely, you gain one bit of information, the maximum possible amount. If you already have other information about the situation, you might gain _less_ than one bit on average (because it confirms something you already knew, but doesn't provide any new information), but you can't gain more.
But “Yes” obviously gives me much more than one bit of the information I need to know the answer.
Just to make this unambiguous: If you ask me to guess a number between one and one billion, and by fantastic luck I guess right, your “yes/no” answer obviously gives me more than one bit of information as to the right answer.
That's not what I see.
https://news.ycombinator.com/item?id=46282007 They have an example that calculates the expected information gained by truth booths and all of the top ones are giving more than one bit. How can this be? It is a yes/no question a max of 1 bit should be possible
https://news.ycombinator.com/item?id=46282343 the expected information (which is the sum of the outcome probability times the outcome information for each of the two possible outcomes) is always less than or equal to one.
The specific comment you replied to had one sentence that didn't say "expected" or "average", but the surrounding sentences and comments give context. The part you objected to was also trying to talk about averages, which makes it not false.
Can’t gain more!
The core confusion is this idea that the answer to a yes/no question can’t provide more than one bit of information, no matter what the question or answer. This is false. The question itself can encode multiple bits of potential information and the answer simply verifies them.
"you might gain _less_ than one bit on average [...], but you can't gain more."
On. Average.
That's a true statement. Can't gain more than one bit on average.
One bit, however, is not “the maximum possible amount” you can gain from an oracular answer to a yes/no question. The OP covers exactly this point re: the “Guess Who?” game.
That's why people are talking about the maximum expected value.
Say you're trying to guess the number on a 6-sided die that I've rolled. If I wanted to outright tell you the answer, that would be 2.58 bits of information I need to convey. But you're trying to guess it without me telling, so suppose you can ask a yes or no question about the outcome. The maximum of the _expected_ information add is 1 bit. If you ask "was it 4 or greater?", then that is an optimal question, because the expected information gain is min-maxed. That is, the minimum information you can gain is also the maximum: 1 bit. However, suppose you ask "was it a 5?". This is a bad question, because if the answer is no, there are still 5 numbers it could be. Plus, the likelihood of it being 'no' is high: 5/6. However, despite these downsides, it is true that 1/6 times, the answer WILL be yes, and you will gain all 2.58 bits of information in one go. The downside case more than counteracts this and preserves the rules of information theory: the _expected_ information gain is still < 1 bit.
EDIT: D'oh, nevermind. Re-reading the post, it's definitely talking about >1 bit expectations of potential matchings. So I don't know!
The match-ups can however give more information, as it isn't giving a yes/no answer.
So a yes might rule out 75% of remaining options (for example) which provides 2 bits of information.
Despite summing to 1, the exact values of P(true) and P(false) are dependent on the options which have previously been discounted. Then those variables get multiplied by the amount of information gained by either answer.
A result which conveys 2 bits of information should occur with 25% expected probability. Because that’s what we mean by “two bits of information.”
The number of bits of information you gained is -log₂ (m/n).
If you ask a question which always eliminates half of the options, you will always gain -log₂ (1/2) = 1 bit of information.
If you go with the dumber approach of taking a moonshot, you can potentially gain more than that, but in expectation you'll gain less.
If your question is a 25-75 split, you have a 25% chance of gaining -log₂ (1/4) = 2 bits, and a 75% chance of gaining -log₂ (3/4) = 0.415 bits. On average, this strategy will gain you (0.25)(2) + (0.75)(0.415) = 0.8113 bits, which is less than 1 bit.
The farther away you get from 50-50, the more bits you can potentially gain in a single question, but - also - the lower the number of bits you expect to gain becomes. You can never do better than an expectation of 1 bit for a trial with 2 outcomes.
(All of this is in the article; see footnote 3 and its associated paragraph.)
The article explicitly calls out the expectational maximum of one bit:
>> You'll also notice that you're never presented with a question that gives you more than 1 expected information, which is backed up by the above graph never going higher than 1.
So it's strange that it then goes on to list an example of a hypothetical (undescribed, since the scenario is impossible) yes/no question with an expected gain of 1.6 bits.
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.
> 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.
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.
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".)
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
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.
A lot more upbeat than "Alice in Borderland".
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?
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.
It’s mostly about how to elicit the information from the contestants. Once you have the probabilities from them, it seems relatively straightforward.
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 :)
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!
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
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.