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.
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.
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!
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.
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.
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.
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.
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.)
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...
My lived childhood is old enough to be someone's "research."
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.
gus_massa•1mo ago
> 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 :)