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
PMF = (n choose k) * p^k * (1-p)^(n-k)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.
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.
The saying probably assumes that each chronometer has a certain small probability of malfunctioning, resulting in a significant error (basically a fat-tailed error distribution). With three chronometers, you can use a robust estimator of the true value (consensus value or median). With two, there's no robust estimator and if you use the mean, you have twice the probability of being significantly wrong (though only by half as much).
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.
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.
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.
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 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
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.
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.
My reasoning is that a clock is either right or wrong.
The average of a right and a wrong clock is wrong. Half as wrong as the wrong one, but still wrong.
If this is a good mental model for dealing with clock malfunctions depends on the failure modes of the clocks.
This is not how continuous probabilities work. The probability that a clock is exactly right is zero; hence there is always some error in a measurement of time. Adding additional clocks will always cause the error to be less or equal to the maximum error.
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.
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.
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...
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.
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...
I imagined it like:
F T (Alice)
F xx ????????
xx ????????
T ?? vvvvvvvv
?? vvvvvvvv
^ ?? vvvvvvvv
B ?? vvvvvvvv
o ?? vvvvvvvv
b ?? vvvvvvvv
v ?? vvvvvvvv
?? vvvvvvvv
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.
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.
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.
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 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.
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.
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.
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.
I'm having trouble with this one due to a lack of experience, but if there is no consensus between the two parties, my assumption would be that you trust neither and ask again. Why is that not the case in a split-brain scenario here? Do you /have/ to make an immediate decision?
You could see that as a lack of detail of the problem as posed. Alternatively it's a breakdown when applying it as an analogy.
Time critical scenarios are one possibility.
In a safety critical scenario intermittent sensor failure might be possible but keep in mind that consistent failure is too.
A jury scenario is presumably one of consistent failure. There's no reason to expect that an intentional liar would change his answer upon being asked again.
This gives me an idea of how to implement isEven() and isOdd() probabilistically.
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.
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.
What if, in the event of a tie (just Alice & Bob), we always decide to trust Alice. Would that not improve our probability of guessing the tie correctly, i.e. back to 80% success?
no in fact that's the proof that adding Bob doesn't help. If Alice & Bob disagree, then since both are correct with same probability it doesn't matter if you pick alice or bob. So WLOG you choose to trust Alice. But now when Alice and Bob agree, that means that you also trust Alice's output [as in you're only right if Alice is right (which is same as when Bob's right since Bob and Alice match)]. So in both cases you are right when Alice is right, i.e. you trust Alice and that means you don't even care about Bob's output.
I think why it feels odd is that most people intuitively answer a different question. If you had to bet on an outcome then Alice and bob agreeing gives you more information. But here you're not dealing with that question, you're either right and wrong; and whether or not Alice & Bob agree, you're effectively "wagering the same" in both cases (where your wager is 0.8, the probability [or expectation] that one is correct).
Although revisiting this, you have to be a bit careful about the argument.
Basically what you're doing is breaking down p(correct) = p(correct & agree) + p(correct & disagree) where former is 0.8*0.8 and latter is 0.8*0.2. Explicitly computing the conditional probability however makes calculating more difficult: p(correct | agree)*p(agree) + p(correct | disagree)*p(disagree). This is something like (16/17) * (0.8*0.8 + 0.2*0.2) + 0.5 * (0.8*0.2*2) which is not easy to arrive at intuitively unless you grind through the calculation.
So _conditioned_ on them agreeing you are right ~94% while conditioned on them disagreeing it's a coin-toss (because when they disagree exactly one is right, and it's equally likely to be alice or bob). Interesting case where the unconditional probability is actually more intuitive and easier than the conditional.
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.
Thanks. I came up with this Python simulation that matches your 68%:
import random
def lying_flippers(num_flips=1_000_000):
"""
- Bob flips a coin and tells Alice the result but lies 20% of the
time.
- Alice tells me Bob's result but also lies 20% of the time.
- If I trust Bob, I know I'll be correct 80% of the time.
- If I trust Alice, how often will I be correct (assuming I don't
know Bob's result)?
"""
# Invert flip 20% of the time.
def maybe_flip_flip(flip: bool):
if random.random() < 0.2:
return not flip
return flip
def sum_correct(actual, altered):
return sum(1 if a == b else 0 for (b, a) in zip(actual, altered))
half_num_flips = num_flips // 2
twenty_percent = int(num_flips * 0.2)
actual_flips = [random.choice((True, False)) for _ in range(num_flips)]
num_heads = sum(actual_flips)
num_tails = num_flips - num_heads
print(f"Heads = {num_heads} Tails = {num_tails}")
bob_flips = [maybe_flip_flip(flip) for flip in actual_flips]
alice_flips = [maybe_flip_flip(flip) for flip in bob_flips]
bob_num_correct = sum_correct(actual_flips, bob_flips)
bob_percent_correct = bob_num_correct / num_flips
alice_num_correct = sum_correct(actual_flips, alice_flips)
alice_percent_correct = alice_num_correct / num_flips
# Trusting Bob should lead to being correct ~80% of the time.
# This is just a verification of the model since we already know the answer.
print(f"Trust Bob -> {bob_percent_correct:.1%}")
# Trusting Alice should lead to being correct ?% of the time.
# This model produces 68%.
print(f"Trust Alice -> {alice_percent_correct:.1%}")
print()>>> 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.
As expected by Shannon's information theory (error correction, entropy, channel capacity).
Shannon's information theory asssumes optimal coding, but doesn't say what the optimal coding is.
Anyway if a single observer who lies 20% of the time gives you 4 out of 5 bits correct, but you don't know which ones...
And N such observers, where N>2, gives you a very good way of getting more information (best-of-3 voting etc), to the limit, at infinite observers, of a perfect channel...
then interpolating for N=2, there is more information here than for N=1. It just needs more advanced coding to exploit.
I don't have the math for this, but a colleague does, and after spending a few minutes on Matlab came up with about 0.278 bits/flip of channel capacity (Shannon) for the single observer, and I think around 0.451 bits/flip of channel capacity for the dual observers. That's the theoretical capacity with optimal coding. Whatever coding schemes need to be employed to get there, i.e. what redundancy to add to the bit stream... that's the hard part.
Here i was thinking of transformer architecture
i thought of something completely different
No one commenting on the inaccuracy (or at least imprecision) of the Python output cited in the article??
"A:T, B:T - chances - H 6.0% | T 94.0% | occurs 34.0% of the time"
By the simplest of math for unrelated events, the chance of both A & B lying about the coin is 20% of 20%, or .2 * .2 = 0.04, or 4.0% ...
The "Let's prove it" section contains the correct analysis, including that our chance of being correct is 80% with two friends.
The code output for three players is similarly flawed, and the analysis slight misstates our chance of being correct as 90.0% (correctly: 89.6%).
Or am I missing something about the intent or output of the Python simulation?
This is an interesting question!
But no, the python output is correct (although I do round the values). It's counterintuitive but these are two different questions:
1. What are the odds that both players lie? (4%)
2. Given that both players say tails, what are the odds that the coin is heads (~6%)
The answer for question (2) is 0.02 / 0.34 = 6%
One way of expressing this is Bayes Rule: we want P(both say tails | coin is heads):
* we can compute this as (P(coin is heads | both say tails) * P(coin is heads)) / P(both say tails)
* P(coin is heads | both say tails) = 0.04 (both must lie)
* P(coin is heads) = 0.5
* P(both say tails) = 0.04 * 0.5 + 0.64 * 0.5 = 0.34
I think that might not be convincing to you, so we can also just look at the results for a hypothetical simulation with 2000 flips:
* of those 2000 flips, 1000 are tails
* 640 times both players tell the truth
* 40 times both players lie
* 680 times (640 + 40) both players *agree*
* 320 times the players disagree
40 / 680 ~= 6%
We go from 4% to 6% because the denominator changes. For the "how often do they both lie" case, our denominator is "all of our coin flips." For the "given that they both said tails, what are the odds that the coin is heads" case, our denominator is "all of the cases where they agreed" - a substantially smaller denominator!
The three players example is just me rounding 89.6% to 90% to make the output shorter (all examples are rounded to two digits, otherwise I found that the output was too large to fit on many screens without horizontal scrolling).
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.
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?
Better to stay away from lying friends, no?
then no one would have any friends
LOL very true
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