It would be more straightforward to remove the permutations and just display the combinations and the symmetry between heads and tails. And solve it analytically Eg:
if p is the probability that the NPC is correct
P(A|AAAA) = p^4
P(A|BBBB) = (1-p)^4
Anyway, the apparent strangeness of the tie case comes from the fact that the binomial PMF is symmetric with respect to n (the number of participants) and n-k.
PMF = (n choose k) * p^k * (1-p)^(n-k)
So when k = n/2, the symmetry means that the likelihood is identical under p and 1-p, so we're not gaining any information. This is a really good illustration of that; interesting post! (edit: apparently i suck at formatting)
Sailors in the past had a similar adage: “Never go to sea with two chronometers; take one or three.” They relied on precise clocks to calculate longitude.
It's possible to navigate without being able to measure your longitude. Like if you're looking for an island, you should first navigate to the correct latitude and then sail along that latitude until you hit the island. The route is longer, obviously. But that's what you should do if your chronometers disagree.
I think it is different in the continuous case though, because you can average two (reasonably accurate) chronometers and get a better measurement. But we can't average true and false, at least not in the context of this problem definition.
But the chronometers are will sync with each other if you don't store them apart, which would result correlated noise that an average won't fix.
You can definitely average two relatively accurate chronometers but you if you only have two it’s difficult to tell if one is way fast or way slow.
In a perfect world they drift less than a minute per day and you’re relatively close to the time with an average or just by picking one and knowing that you don’t have massive time skew.
I believe this saying was first made about compasses which also had mechanical failures. Having three lets you know which one failed. The same goes for mechanical watches, which can fail in inconsistent ways, slow one day and fast the next is problematic the same goes for a compass that is wildly off, how do you know which one of the two is off?
> In a perfect world they drift less than a minute per day...
A minute per day would be far too much drift for navigation, wouldn't it?
From Wikipedia [1]:
> For every four seconds that the time source is in error, the east–west position may be off by up to just over one nautical mile as the angular speed of Earth is latitude dependent.
That makes me think a minute might be your budget for an entire voyage? But I don't know much about navigation. And it is beside the point, your argument isn't changed if we put in a different constant, so I only mention out of interest.
> Having three lets you know which one failed.
I guess I hadn't considered when it stops for a minute and then continues ticking steadily, and you would want to discard the measurement from the faulty watch.
But if I just bring one watch as the expression councils, isn't that even worse? I don't even know it malfunctioned and if it failed entirely I don't have any reference for the time at the port.
My interpretation had been that you look back and forth between the watches unable to make a decision, which doesn't matter if you always split the difference, but I see your point.
> A minute per day would be far too much drift for navigation, wouldn't it?
Even that was much better than the dead-reckoning they had to do in bluewater before working chronometers were invented. Your ship's "position" would be a triangle that might have sides ten miles long at lower latitudes.
I’ve never heard the bring one or three, I’ve always just heard three. I think that exact saying implies that if you have two and one isn’t working out you’ll go crazy but if you have one you’ll be oblivious until it’s too late.
A well serviced rolex in 2026 with laser cut gears drifts +/- 15sec per day.
One with hand filed gears is going to be +/- a minute on a good day, and that’s what early navigation was using. I have watches with hand filed gears and they can be a bit rough.
Prior to that, it was dead reckoning, dragging a string every now and again to calculate speed and heading and the current and then guesstimating your location on a twice daily basis.
Those two wildly inaccurate systems mapped most of the world for us.
Reading the comment thread here made me realize the idea seems to be that having 2 just means double the probability of one of them failing in some undetectable way. The resulting error magnitude is reduced by half, but the probability of that error is doubled. So it gains you nothing to expected value to have 2. Unlike with 3, where the probability of undetectable failure and the error rate from partial failure are both reduced by the ability to make comparative measurements (eg pick the middle number not the average)
Though not without significant errors, the most amusing to me being that islands had a tendency to multiply because different maps would be combined and the cartographer would mistake the same island on two maps as being separate islands due to errors. A weird case of aliasing I suppose.
The book “Longitude” is fascinating, and discusses the challenges prior to chronometers (many people died), as well as the rewards offered for a precise chronometer, the attempts, etc.
You seem to agree that two is not a good number. Better bring four then, so that you're not left with only two after your mishap.
Or bring only two, but step on one immediately, to get rid of the cursed pair situation, and also to get the clumsiness out of the way early. Old sailor's trick.
You will have an error rate of less than or equal to 1%. You can't average two measurements and get a result with a higher error rate than the worst of the original measurements had.
You wouldn't be well served by averaging a measurement with a 1% error and a measurement with a 90% error, but you will have still have less than or equal to 90% error in the result.
If the errors are correlated, you could end up with a 1% error still. The degenerate case of this is averaging a measurement with itself. This is something clocks are especially prone to; if you do not inertially isolate them, they will sync up [1]. But that still doesn't result in a greater error.
You could introduce more error if you encountered precision issues. Eg, you used `(A+B)/2` instead of `A/2 + B/2`; because floating point has less precision for higher numbers, the former will introduce more rounding error. But that's not a function of the clocks, that's a numerics bug. (And this is normally encountered when averaging many measurements rather than two.)
There are different ways to define error but this is true whether you consider it to be MSE or variance.
The result in the original article only applies when there are discrete choices. For stuff you can actually average, more is always better.
Oh, and even with discrete choices (like heads vs tails), if you had to give a distribution and not just a single highest likelihood outcome, and we'd judge you by the cross-entry, then going from one to two is an improvement. And going from odd n to the next even n is an improvement in general in this setting.
This principle is also highly relevant in safety critical systems for using redundant sensors. Just adding a second sensor is often not enough. Because if they disagree, which one do you trust.
Bob isn't giving you any actionable information. If Alice and Bob agree, you're more confident than you were before, but you're still going to be trusting Alice. If they disagree you're down to 50% confidence, but you still might as well trust Alice.
To complete the circle, now that we have winnowed the space down to these options, we would normalize them and end up with 0.16 / (0.16 + 0.16) = 0.5 = 50% in both cases.
The reason I'm not putting % signs on there is that, until we normalize, those are measures and not probabilities. What that means is that an events which has a 16% chance of happening in the entire universe of possibility has a "area" or "volume" (the strictly correct term being measure) of 0.16. Once we zoom in to a smaller subset of events, it no longer has a probability of 16% but the measure remains unchanged.
In this previous comment I gave a longer explanation of the intuition behind measure theory and linked to some resources on YouTube.
F T (Alice)
F xx ????????
xx ????????
T ?? vvvvvvvv
?? vvvvvvvv
^ ?? vvvvvvvv
B ?? vvvvvvvv
o ?? vvvvvvvv
b ?? vvvvvvvv
v ?? vvvvvvvv
?? vvvvvvvv
(where "F" describes cases where the specified person tells you a Falsehood, and "T" labels the cases of that person telling you the Truth)
In the check-mark (v) region, you get the right answer regardless; they are both being truthful, and of course you trust them when they agree. Similarly you get the wrong answer regardless in the x region.
In the ? region you are no better than a coin flip, regardless of your strategy. If you unconditionally trust Alice then you win on the right-hand side, and lose on the left-hand side; and whatever Bob says is irrelevant. The situation for unconditionally trusting Bob is symmetrical (of course it is; they both act according to the same rules, on the same information). If you choose any other strategy, you still have a 50-50 chance, since Alice and Bob disagree and there is no reason to choose one over the other.
Since your odds don't change with your strategy in any of those regions of the probability space, they don't change overall.
In voting parity matters. In some cases, effects are interesting - like in some cases if the mafia/werewolf game, when adding a citizen/villager decreses their winning chance by sqrt(pi/2), vide https://arxiv.org/abs/1009.1031
Let's pretend Alice tells the truth 100% of the time, and Bob is still at 20%. It's more easy to intuit that Bob's contribution is only noise.
Slide Alice's accuracy down to 99% and, again, if you don't trust Alice, you're no better off trusting Bob.
Interestingly, this also happens as a feature of them being independent. If Bob told the truth 20% of the time that Alice told a lie, or if Bob simply copied Alice's response 20% of the time and otherwise told the truth, then the maths are different.
I think there's an annoying thing where by saying "hey, here's this neat problem, what's the answer" I've made you much more likely to actually get the answer!
What I really wanted to do was transfer the experience of writing a simulation for a related problem, observing this result, assuming I had a bug in my code, and then being delighted when I did the math. But unfortunately I don't know how to transfer that experience over the internet :(
(to be clear, I'm totally happy you wrote out the probabilities and got it right! Just expressing something I was thinking about back when I wrote this blog)
I, erroneously, thought that "when Alice and Bob agree there's a 96% chance of them being correct, then surely you can leverage this to get above the 80% chance. What if we trust them both when they agree and trust Alice when they disagree?" Did some (erroneous) napkin math and went to write a simulation.
As I was writing the simulation I realized my error. I finished the simulation anyway, just because, and it has the expected 80% result on both of them.
My error: when we trust "both" we're also trusting Alice, which means that my case was exactly the same as just trusting Alice.
PS as I was writing the simulation I did a small sanity test of 9 rolls: I rolled heads 9 times in a row (so I tried it again with 100 million and it was a ~50-50 split). There goes my chance of winning the lottery!
The betting-voting distinction is interesting and was on my mind while I was reading it.
So much of this breaks down when the binary nature of the variables involved becomes continuous or at least nonbinary.
It's an example of a more general interest of mine, how structural characteristics of an inferential scenario affect the value of information that is received.
I could also see this being relevant to diagnostic scenarios hypothetically.
It depends on what you are doing with the guess. If it is just a question of how frequently you are right or wrong the second person doesn't help. But if you are, for example, betting on your guess the second person improves your odds of coming out ahead significantly, since you can put down a higher wager when they agree than when they disagree.
This kinda reminds me of error correction, and where at some level you can have detectable but not correctable error conditions. Adding Bob is just like adding a parity bit: can give you a good indication someone lied, but won't fix anything. Adding Charlie gives you the crudest ECC form, a repetition code (though for storing one bit, I don't think you can do better?)
I guess if you only need to store one bit you could store either 0 or 11 and on average use less than two bits (for bit flips only), or 111 if you have to also worry about losing/duplicating bits.
The triangulation effect with 3+ observers is fascinating, but there may be a weirder extension: what if the "third observer" isn't another person but the relationship coherence between two people?
Instead of three independent signals, you'd evaluate: given how Alice and Bob usually interact, does their agreement/disagreement pattern here tell you something? (E.g., if they're habitual contrarians, their agreement is the signal, not their disagreement.)
Take it further: human + LLM collaboration, where you measure the ongoing conversational dynamics—tone shifts, productive vs. circular disagreement, what gets bypassed, how contradictions are handled. The quality of the collaborative process itself becomes your truth signal.
You're not just aggregating independent observations anymore; you're reading the substrate of the interaction. The conversational structure as diagnostic.
I misread the article too, but it states that with 3 or 4 on the jury the probability has increased to 0.9, instead of 0.8 with 1 or 2. It's just that an incremental extra jury member from odd to even never increases the information, even when it's 9999 to 10000.
In that case following Alice's input is still the best strategy, but you'll be worse off: you'd only be right if both tell the truth, at 80%80%=64%, or both lie, at 20%20%=4%, for a total of 68%.
In the general case of n intermediate occasional liars, the odds of the final result being accurate goes to 50% as n grows large, which makes sense, as it will have no correlation anymore to the initial input.
>>> To establish this, let’s write a simple simulation. We’ll flip a coin a million times, ask our friends what they saw, and observe the results.
Either they agree, or they disagree.
If they agree, they're either both telling the truth or both lying. All you can do is go with what they agreed on. In this case, picking what they agreed on is the same as picking what one of them said (say, Alice).
If they disagree, then one is telling the truth and one is lying and you have no way to tell which. So just pick one, and it makes no difference if you pick the same one every time (say, Alice).
So you end up just listening to Alice all the time anyway.
Weirdly this reminds me of the Raft consensus protocol: two nodes cancel each other when failing consensus as you can't tell which is the valid one, three gives you better chances, if one fails you have the other two that can get consensus. Of course in the off chance you have two failures you cannot get consensus with the only living node. Adding another two nodes make you robust to two failures.
Now replace fail with lying and you have the exact same problem.
One way to think about this is you have a binomial distribution with p=0.8 and n=number of lying friends. Each time you increase n, you shift the probability mass of the distribution "to the right" but if n is even some of that mass has to land on the "tie" condition.
The second liar often gives you information you never wanted! But that is offset by the excellent information of when the liars agree. Fascinating.
I didnt't math during the thinking pause, but my intuition was a second liar makes it worse (more likey to end up 50-50 situation) and additional liars make it better as you get to reduce noise.
Is there a scenario where the extra liar makes it worse, you would be better yelling lalalallala as they tell you the answer?
It would be more straightforward to remove the permutations and just display the combinations and the symmetry between heads and tails. And solve it analytically Eg: if p is the probability that the NPC is correct
Anyway, the apparent strangeness of the tie case comes from the fact that the binomial PMF is symmetric with respect to n (the number of participants) and n-k. So when k = n/2, the symmetry means that the likelihood is identical under p and 1-p, so we're not gaining any information. This is a really good illustration of that; interesting post! (edit: apparently i suck at formatting)Sailors in the past had a similar adage: “Never go to sea with two chronometers; take one or three.” They relied on precise clocks to calculate longitude.
Sailors had a lot of harmful sayings.
It's possible to navigate without being able to measure your longitude. Like if you're looking for an island, you should first navigate to the correct latitude and then sail along that latitude until you hit the island. The route is longer, obviously. But that's what you should do if your chronometers disagree.
I think it is different in the continuous case though, because you can average two (reasonably accurate) chronometers and get a better measurement. But we can't average true and false, at least not in the context of this problem definition.
But the chronometers are will sync with each other if you don't store them apart, which would result correlated noise that an average won't fix.
You can definitely average two relatively accurate chronometers but you if you only have two it’s difficult to tell if one is way fast or way slow.
In a perfect world they drift less than a minute per day and you’re relatively close to the time with an average or just by picking one and knowing that you don’t have massive time skew.
I believe this saying was first made about compasses which also had mechanical failures. Having three lets you know which one failed. The same goes for mechanical watches, which can fail in inconsistent ways, slow one day and fast the next is problematic the same goes for a compass that is wildly off, how do you know which one of the two is off?
> In a perfect world they drift less than a minute per day...
A minute per day would be far too much drift for navigation, wouldn't it?
From Wikipedia [1]:
> For every four seconds that the time source is in error, the east–west position may be off by up to just over one nautical mile as the angular speed of Earth is latitude dependent.
That makes me think a minute might be your budget for an entire voyage? But I don't know much about navigation. And it is beside the point, your argument isn't changed if we put in a different constant, so I only mention out of interest.
> Having three lets you know which one failed.
I guess I hadn't considered when it stops for a minute and then continues ticking steadily, and you would want to discard the measurement from the faulty watch.
But if I just bring one watch as the expression councils, isn't that even worse? I don't even know it malfunctioned and if it failed entirely I don't have any reference for the time at the port.
My interpretation had been that you look back and forth between the watches unable to make a decision, which doesn't matter if you always split the difference, but I see your point.
[1] https://en.wikipedia.org/wiki/Marine_chronometer
> A minute per day would be far too much drift for navigation, wouldn't it?
Even that was much better than the dead-reckoning they had to do in bluewater before working chronometers were invented. Your ship's "position" would be a triangle that might have sides ten miles long at lower latitudes.
I’ve never heard the bring one or three, I’ve always just heard three. I think that exact saying implies that if you have two and one isn’t working out you’ll go crazy but if you have one you’ll be oblivious until it’s too late.
A well serviced rolex in 2026 with laser cut gears drifts +/- 15sec per day.
One with hand filed gears is going to be +/- a minute on a good day, and that’s what early navigation was using. I have watches with hand filed gears and they can be a bit rough.
Prior to that, it was dead reckoning, dragging a string every now and again to calculate speed and heading and the current and then guesstimating your location on a twice daily basis.
Those two wildly inaccurate systems mapped most of the world for us.
Reading the comment thread here made me realize the idea seems to be that having 2 just means double the probability of one of them failing in some undetectable way. The resulting error magnitude is reduced by half, but the probability of that error is doubled. So it gains you nothing to expected value to have 2. Unlike with 3, where the probability of undetectable failure and the error rate from partial failure are both reduced by the ability to make comparative measurements (eg pick the middle number not the average)
> A well serviced rolex in 2026 with laser cut gears drifts +/- 15sec per day.
Modern Rolex (and Omega et al) are more like +/-2s.
You would probably use a chronometer:
"The precision of a COSC-certified chronometer must be between -4 and +6 seconds per day."
https://www.cosc.swiss/cosc-faq
Interesting, thanks for sharing.
Though not without significant errors, the most amusing to me being that islands had a tendency to multiply because different maps would be combined and the cartographer would mistake the same island on two maps as being separate islands due to errors. A weird case of aliasing I suppose.
15s per day sounds EXTREME. Are you even serious? Even the cheapest Chinese hand watch has 50x better precision.
Cheap watches are almost always quartz watches (electronic). Those are much more accurate than even high end mechanical watches.
15s per day is not unusual for a good mechanical watch.
The book “Longitude” is fascinating, and discusses the challenges prior to chronometers (many people died), as well as the rewards offered for a precise chronometer, the attempts, etc.
I'd much rather take three than one... you might step on the one and crush it.
You seem to agree that two is not a good number. Better bring four then, so that you're not left with only two after your mishap.
Or bring only two, but step on one immediately, to get rid of the cursed pair situation, and also to get the clumsiness out of the way early. Old sailor's trick.
I am a sailor in the present and this same rule is why I have 3 digital multimeters on my boat.
Two is one and one is none
Any navigator worth their salt would take rather take more than fewer chronometers, for various and frankly obvious reasons.
This saying must originate with a landlubber...
To spell out the point:
If the chronometer error rate is 1%, averaging two will give you a 2% error rate.
You will have an error rate of less than or equal to 1%. You can't average two measurements and get a result with a higher error rate than the worst of the original measurements had.
You wouldn't be well served by averaging a measurement with a 1% error and a measurement with a 90% error, but you will have still have less than or equal to 90% error in the result.
If the errors are correlated, you could end up with a 1% error still. The degenerate case of this is averaging a measurement with itself. This is something clocks are especially prone to; if you do not inertially isolate them, they will sync up [1]. But that still doesn't result in a greater error.
You could introduce more error if you encountered precision issues. Eg, you used `(A+B)/2` instead of `A/2 + B/2`; because floating point has less precision for higher numbers, the former will introduce more rounding error. But that's not a function of the clocks, that's a numerics bug. (And this is normally encountered when averaging many measurements rather than two.)
There are different ways to define error but this is true whether you consider it to be MSE or variance.
[1] https://www.youtube.com/watch?v=T58lGKREubo
No, not at all.
The result in the original article only applies when there are discrete choices. For stuff you can actually average, more is always better.
Oh, and even with discrete choices (like heads vs tails), if you had to give a distribution and not just a single highest likelihood outcome, and we'd judge you by the cross-entry, then going from one to two is an improvement. And going from odd n to the next even n is an improvement in general in this setting.
This principle is also highly relevant in safety critical systems for using redundant sensors. Just adding a second sensor is often not enough. Because if they disagree, which one do you trust.
One example of this is in airplanes.
Bob isn't giving you any actionable information. If Alice and Bob agree, you're more confident than you were before, but you're still going to be trusting Alice. If they disagree you're down to 50% confidence, but you still might as well trust Alice.
Better than 50% confidence: they only lie 20% of the time, so when they disagree it's still 64% likely to be heads (.8 x .8)
No, it's 50% -- given that e.g. the flip is H, the base probability is both 16% for HT and 16% for TH.
To complete the circle, now that we have winnowed the space down to these options, we would normalize them and end up with 0.16 / (0.16 + 0.16) = 0.5 = 50% in both cases.
The reason I'm not putting % signs on there is that, until we normalize, those are measures and not probabilities. What that means is that an events which has a 16% chance of happening in the entire universe of possibility has a "area" or "volume" (the strictly correct term being measure) of 0.16. Once we zoom in to a smaller subset of events, it no longer has a probability of 16% but the measure remains unchanged.
In this previous comment I gave a longer explanation of the intuition behind measure theory and linked to some resources on YouTube.
https://news.ycombinator.com/item?id=35796740
I imagined it like:
(where "F" describes cases where the specified person tells you a Falsehood, and "T" labels the cases of that person telling you the Truth)In the check-mark (v) region, you get the right answer regardless; they are both being truthful, and of course you trust them when they agree. Similarly you get the wrong answer regardless in the x region.
In the ? region you are no better than a coin flip, regardless of your strategy. If you unconditionally trust Alice then you win on the right-hand side, and lose on the left-hand side; and whatever Bob says is irrelevant. The situation for unconditionally trusting Bob is symmetrical (of course it is; they both act according to the same rules, on the same information). If you choose any other strategy, you still have a 50-50 chance, since Alice and Bob disagree and there is no reason to choose one over the other.
Since your odds don't change with your strategy in any of those regions of the probability space, they don't change overall.
In voting parity matters. In some cases, effects are interesting - like in some cases if the mafia/werewolf game, when adding a citizen/villager decreses their winning chance by sqrt(pi/2), vide https://arxiv.org/abs/1009.1031
When it comes to the wisdom of crowds, see https://egtheory.wordpress.com/2014/01/30/two-heads-are-bett...
Let's pretend Alice tells the truth 100% of the time, and Bob is still at 20%. It's more easy to intuit that Bob's contribution is only noise.
Slide Alice's accuracy down to 99% and, again, if you don't trust Alice, you're no better off trusting Bob.
Interestingly, this also happens as a feature of them being independent. If Bob told the truth 20% of the time that Alice told a lie, or if Bob simply copied Alice's response 20% of the time and otherwise told the truth, then the maths are different.
I paused and wrote out all the probabilities and saw no way to improve beyond 80% - I scrolled down hoping to be proven wrong!
(I'm the author)
I think there's an annoying thing where by saying "hey, here's this neat problem, what's the answer" I've made you much more likely to actually get the answer!
What I really wanted to do was transfer the experience of writing a simulation for a related problem, observing this result, assuming I had a bug in my code, and then being delighted when I did the math. But unfortunately I don't know how to transfer that experience over the internet :(
(to be clear, I'm totally happy you wrote out the probabilities and got it right! Just expressing something I was thinking about back when I wrote this blog)
I, erroneously, thought that "when Alice and Bob agree there's a 96% chance of them being correct, then surely you can leverage this to get above the 80% chance. What if we trust them both when they agree and trust Alice when they disagree?" Did some (erroneous) napkin math and went to write a simulation.
As I was writing the simulation I realized my error. I finished the simulation anyway, just because, and it has the expected 80% result on both of them.
My error: when we trust "both" we're also trusting Alice, which means that my case was exactly the same as just trusting Alice.
PS as I was writing the simulation I did a small sanity test of 9 rolls: I rolled heads 9 times in a row (so I tried it again with 100 million and it was a ~50-50 split). There goes my chance of winning the lottery!
The betting-voting distinction is interesting and was on my mind while I was reading it.
So much of this breaks down when the binary nature of the variables involved becomes continuous or at least nonbinary.
It's an example of a more general interest of mine, how structural characteristics of an inferential scenario affect the value of information that is received.
I could also see this being relevant to diagnostic scenarios hypothetically.
It depends on what you are doing with the guess. If it is just a question of how frequently you are right or wrong the second person doesn't help. But if you are, for example, betting on your guess the second person improves your odds of coming out ahead significantly, since you can put down a higher wager when they agree than when they disagree.
The post discusses this exact point near the end.
This gives me an idea of how to implement isEven() and isOdd() probabilistically.
This kinda reminds me of error correction, and where at some level you can have detectable but not correctable error conditions. Adding Bob is just like adding a parity bit: can give you a good indication someone lied, but won't fix anything. Adding Charlie gives you the crudest ECC form, a repetition code (though for storing one bit, I don't think you can do better?)
I guess if you only need to store one bit you could store either 0 or 11 and on average use less than two bits (for bit flips only), or 111 if you have to also worry about losing/duplicating bits.
The triangulation effect with 3+ observers is fascinating, but there may be a weirder extension: what if the "third observer" isn't another person but the relationship coherence between two people?
Instead of three independent signals, you'd evaluate: given how Alice and Bob usually interact, does their agreement/disagreement pattern here tell you something? (E.g., if they're habitual contrarians, their agreement is the signal, not their disagreement.)
Take it further: human + LLM collaboration, where you measure the ongoing conversational dynamics—tone shifts, productive vs. circular disagreement, what gets bypassed, how contradictions are handled. The quality of the collaborative process itself becomes your truth signal.
You're not just aggregating independent observations anymore; you're reading the substrate of the interaction. The conversational structure as diagnostic.
Surely the extrapolation is wrong?
at 4 heads, just randomly select a jury of 3. and you're back on track.
at a million heads, just sum up all their guesses, divide by one million, and then check the over/under of 0.50
For 4 vs 3, you’re saying the same thing.
He wrote:
If our number N of friends is odd, our chances of guessing correctly don’t improve when we move to N+1 friends.
I misread the article too, but it states that with 3 or 4 on the jury the probability has increased to 0.9, instead of 0.8 with 1 or 2. It's just that an incremental extra jury member from odd to even never increases the information, even when it's 9999 to 10000.
As expected by Shannon's information theory (error correction, entropy, channel capacity).
What happens if Bob lies to Alice 20% of the time and Alice lies to me 20% of the time but I only get input from Alice?
In that case following Alice's input is still the best strategy, but you'll be worse off: you'd only be right if both tell the truth, at 80%80%=64%, or both lie, at 20%20%=4%, for a total of 68%.
In the general case of n intermediate occasional liars, the odds of the final result being accurate goes to 50% as n grows large, which makes sense, as it will have no correlation anymore to the initial input.
>>> To establish this, let’s write a simple simulation. We’ll flip a coin a million times, ask our friends what they saw, and observe the results.
Either they agree, or they disagree.
If they agree, they're either both telling the truth or both lying. All you can do is go with what they agreed on. In this case, picking what they agreed on is the same as picking what one of them said (say, Alice).
If they disagree, then one is telling the truth and one is lying and you have no way to tell which. So just pick one, and it makes no difference if you pick the same one every time (say, Alice).
So you end up just listening to Alice all the time anyway.
Weirdly this reminds me of the Raft consensus protocol: two nodes cancel each other when failing consensus as you can't tell which is the valid one, three gives you better chances, if one fails you have the other two that can get consensus. Of course in the off chance you have two failures you cannot get consensus with the only living node. Adding another two nodes make you robust to two failures.
Now replace fail with lying and you have the exact same problem.
Here i was thinking of transformer architecture
i thought of something completely different
One way to think about this is you have a binomial distribution with p=0.8 and n=number of lying friends. Each time you increase n, you shift the probability mass of the distribution "to the right" but if n is even some of that mass has to land on the "tie" condition.
I wrote a quick colab to help visualize this, adds a little intuition for what's happening: https://colab.research.google.com/drive/1EytLeBfAoOAanVNFnWQ...
Why not unconditionally trust Bob?
You can, but trivially that strategy is also no better than unconditionally trusting Alice.
Better to stay away from lying friends, no?
The second liar often gives you information you never wanted! But that is offset by the excellent information of when the liars agree. Fascinating.
I didnt't math during the thinking pause, but my intuition was a second liar makes it worse (more likey to end up 50-50 situation) and additional liars make it better as you get to reduce noise.
Is there a scenario where the extra liar makes it worse, you would be better yelling lalalallala as they tell you the answer?