This riddle was inspired when I watched the latest (as of this post) episode of Curb Your Enthusiasm (S10E4), in which Larry is faced with a dilemma that he never really solved. The dilemma is as follows:

Larry is ordering a private airplane to fly to Mexico for him and his friends: Jeff, Susie, Cheryl, Leon, and Donna. For safety and fuel efficiency reasons, the pilot needs to know the weights of the passengers and their luggage. Getting the weights of luggage is no big deal. However, to get the weights of the passengers there's one problem:

None of Larry's friends will allow anybody to know their weight. To each person, their weight is a very private matter that nobody can know, not even Larry, their partner, or the pilot. So they refuse to tell Larry (or anybody) their weight.

Assume you can easily solve this with the tools in your own home (e.g. no industrial-sized scale). Assume everybody knows their own weight.

How does Larry give the pilot the information he needs, while respecting everybody's wish to keep their weight private?

I have a solution in mind. However, any solution that works just as well should be just as valid. Watching the episode won't help you because Larry doesn't solve this.

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    $\begingroup$ rot13(Jrvtu gur crbcyr + yhttntr, jrvtu gur yhttntr frcnengryl, gura fhogenpg?) I sure hope not... $\endgroup$ – Avi Feb 13 '20 at 19:19
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    $\begingroup$ @Avi Sorry, maybe I should add that the solution is practically doable, without anything fancy (like an industrial sized scale). Imagine you can solve this easily with the tools in your own home $\endgroup$ – Bridgeburners Feb 13 '20 at 19:22
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    $\begingroup$ Do you need to know (a) only the summed weight of all passengers and their luggages, (b) the weight of each passenger added to their luggage, (c) the summed weight of the passengers and the summed weight of their luggage, or (d) the individual weights of each passenger and each luggage? $\endgroup$ – Orntt Feb 14 '20 at 3:45
  • $\begingroup$ It seems fairly clear that the total weight of all cargo is required, and that one of two section of that weight is trivial to obtain, but the second involves greater difficulty, hence the requirement of a solution: let Lary obtain the sum of all personal weights without needing any individual to share exactly their own weight to anybody else. $\endgroup$ – Nij Feb 14 '20 at 3:55
  • $\begingroup$ Note that, if each person is particularly paranoid about it, you would need to ensure that there's no way that the others in the group could conspire to reveal the person's true value. This is no simple task, as anyone could choose to put "zero" into their claimed value, and so if everyone except one puts in zero, it reveals the remaining person's weight. $\endgroup$ – Glen O Feb 14 '20 at 4:11

Everyone weighs themselves, and writes down numbers on a series of slips of paper (they choose how many) so that the numbers on those papers add up to their weight. All of the slips of paper are tossed into a hat, shaken up, and then added up at the end. If there are any handwriting-matching worries, you can split the papers up between people for an initial add, and then add the results of those aggregators together.

  • $\begingroup$ So foolproof. +1. $\endgroup$ – Nautilus Feb 14 '20 at 20:20
  • $\begingroup$ My solution was actually equivalent to Deusovi's, but I accepted this because it's much harder for people to conspire to reveal others' weights. $\endgroup$ – Bridgeburners Feb 16 '20 at 17:22

One way to do this is:

Larry whispers a random number to one person. They add their weight to it and whisper to the next person. The next person adds their weight to it and whispers to the next person. This continues until the last person, who whispers the total back to Larry. Larry then subtracts his original random number, and gets the total weight (which can be announced publicly).

Nobody has enough information to determine anyone else's weight, because all they know is a number that is "a random number, plus a bunch of other weights".

(However, two people could work together -- for example, the last person knows the random number, so they and the first person in line could work together to determine Larry's weight.)

This could be modified to fix one minor concern:

If the random number is chosen poorly, or with a known distribution, it's possible to get probabilistic information about other people's weights. To fix this, establish an upper bound on the total weight (say, in total they don't weigh more than a million pounds). Then, the random number is chosen uniformly from 0 to that upper bound, and all arithmetic is done modulo this upper bound. This removes any sort of probabilistic information about weights.

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    $\begingroup$ Larry could make sure his weight stays hidden this way. But you're right, people can conspire to uncover others' weights. For example, person 1 can tell person 3 the number they whispered to person 2, letting person 3 know person 2's weight. However, person 1 is risking their own weight being learned, if person 3 can convince person 0 (Larry) to tell them the original number $\endgroup$ – Bridgeburners Feb 13 '20 at 19:48
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    $\begingroup$ @Bridgeburners That's a possibility no matter the strategy, though. If the total is public, then everyone but one person can conspire to learn the weight of that last person. And if it's private, same deal but now it works against everyone but Larry. This is true no matter what strategy you use. $\endgroup$ – Deusovi Feb 13 '20 at 20:32
  • $\begingroup$ @Deusovi truth - but there's a difference between "everyone else conspiring together could target one of their number" and "multiple pairs exist where jsut the two could conspire together to target someone". The chance that two people turn against you is a lot higher than the change that everyone else does, working together... especially since the large-group version means that the revealed weights are revealed publicly, thus notably increasing the vulnerability of everyone who involves themselves. $\endgroup$ – Ben Barden Feb 13 '20 at 21:55
  • $\begingroup$ @BenBarden That's a true statement, but the problem didn't specify anything like that -- it seemed to just be about making sure that nobody accidentally learned someone else's weight (or accidentally got enough information to infer it). And as we get more into practical details, this becomes less of a puzzle that can be objectively answered, and more of a problem that can be solved in many different ways. $\endgroup$ – Deusovi Feb 13 '20 at 22:32
  • $\begingroup$ Re: your first comment - not if you make it possible for everybody to keep part of their own weight secret yet still report it to the others, for all possible combinations of people... @Deusovi $\endgroup$ – Nij Feb 14 '20 at 3:40

Each person gets

as many pieces of identical paper as there are people in the group.

Then they will

write a single number on each slip of paper, such that the sum of all the numbers an individual has written is their own weight.

Once that is done, they trade

one of their pieces of paper with one of another person's pieces of paper, exactly once for each other person.

This leaves them with

some unknown fraction of every person's weight, such that it is impossibly to exactly determine the weight of any individual even if every other person was to conspire against them - because that person is still holding one of their own pieces of paper!

So from there, each person

adds up the numbers on the pieces of paper they hold after the trades are completed, and reports that sum to the group, who can then find the final sum for the entire group.

It is possible to break this system by

finding progressive sums throughout the trading process, and tracking the sums as each trade is made, and working backwards from the final known sum to the first known sum, to determine the number on each piece of paper. Combining that information with the source of each piece of paper can be used to calculate the individual weights. But that's fixed by just making the trades transparent and preventing any sums from being calculated until all trades are finished.

This answer is

a modified version of the answer by Ben Bardon. However it eliminates the problem of handwriting analysis by allowing everybody to keep some of their weight secret and still report it publicly.

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    $\begingroup$ Cool solution! This is a common problem, but I never saw this kind of solution before. $\endgroup$ – justhalf Feb 14 '20 at 3:49
  • $\begingroup$ No need to trade imho, we can just put them all in a box and mix. Good answer. $\endgroup$ – John Hamilton Feb 14 '20 at 14:44
  • $\begingroup$ You may need to read the entire answer and why that provision was included, @JohnHamilton $\endgroup$ – Nij Feb 14 '20 at 19:11
  • $\begingroup$ How do you solve the conspiracies? $\endgroup$ – padawan Feb 18 '20 at 18:26
  • $\begingroup$ You can't. There is a conspiracy that breaks every scheme for the sharing of this information. The best that can be done is force any conspiracy to involve every person except one of them, and then make it certain that the single person can detect the conspiracy if they choose to. Which is coincidentally what this system does. $\endgroup$ – Nij Feb 18 '20 at 18:31

Larry gives the others five empty pieces of paper first, then writes +10, +15, +20, -20, and -25 (any distinct non-zero five numbers adding up to 0 will do) on five other pieces of paper, having everyone randomly pick one. Everyone adds that number they drew to their weight, writes the result down on their empty piece of paper, and then dispose of the drawn ones. Then Larry adds them up combined with his own weight and informs the pilot.

  • $\begingroup$ @JMP I know, but I just wanted to avoid "lolz no, it doesn't work" nitpicks. I've been hastily downvoted quite a few times because people thought I was wrong or even cheating for some reason whether or not it was a good answer. $\endgroup$ – Nautilus Feb 13 '20 at 20:18
  • $\begingroup$ This looks promising. But can you be a little more specific? Who writes the original numbers that add to 0? Who adds the numbers together? It's important to know who has what information $\endgroup$ – Bridgeburners Feb 13 '20 at 20:25
  • $\begingroup$ @Bridgeburners Edited my answer to add some clarification. $\endgroup$ – Nautilus Feb 13 '20 at 20:36
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    $\begingroup$ Suppose you're in the group and you draw a piece of paper and it has, written on it, -2,343,916. Would you be comfortable with Larry ending up with a piece of paper that has the sum of that number and your weight? What if it says -10, and you outweigh anyone else by at least 50 pounds? $\endgroup$ – Bridgeburners Feb 13 '20 at 21:29
  • $\begingroup$ This is the safest and most practical solution to date (except for possibly Ben Barden's). $\endgroup$ – COTO Feb 14 '20 at 16:53

If only the total weight of all passengers AND their luggage together is required:

Each person weighs themselves together with their luggage. Then add all the weights up to get the total weight. Passengers do not allow anyone to weigh their luggage individually, so no-one knows the weight of any person.


If the luggage is considered to be of known weight, substitute with some small objects of unknown weight, then weigh all the objects together and subtract from the total weight.

  • $\begingroup$ Beat me to it by about a minute $\endgroup$ – Glen O Feb 14 '20 at 4:19
  • $\begingroup$ So, Larry immediately weighs each small object and calculates each person's individual weight. Unless you're proposing they all weigh as a group, in which case the objects do nothing useful to hide any potential information escape. $\endgroup$ – Nij Feb 14 '20 at 9:38
  • $\begingroup$ Just to say that in real life the total weight is NOT all that the pilot needs to know. $\endgroup$ – DJClayworth Feb 14 '20 at 15:27

My first a bit harder but imo the most efficient way,with guarantee that noone will know anyones weight:

Let everyone fill a water bottle with amount of water that corresponds to their weight, like if someone weights 50 kilo bottle has to be 500ml full and so on(consider that bottles are 2l each so everyone can fill right amount). Then you put something on the bottle that you cannot see the amount of water inside so even if you see someone elses bottle you dont have slight idea how much of water there is. Then there is a big bucket, also dark one with small hole to put water in. Everyone walks one by one and pours their water to the bucket. In the end you just weight water and know how much do all people weight in total.

My second and simple approach to this problem is following:

everyone type on the same piece of paper, better even print it, so its impossible to recognize by handwriting,then throw into the same bottle all the papers and make pilot read all of them and sum it.Quite important here is that the pilot sums it, as if it does one of the guys who goes to plane he first of all will know his own weight and can easier guess other persons weights


If the total weight of the passengers is needed, this would be the method I go with. Let J, S, C, L, and D be the actual weights of each person.

Step 1:

Larry makes an estimation for each person E(J), E(S), E(C), E(L), E(D), writes his estimations down, and hands each estimation to the corresponding person.

Step 2:

Each person writes down the result og their actual weight minus Larry's estimation. For instance, Larry guessed that Susie was 54, but she is actually 52. Then, E(S) - S = +2. She writs +2.

Step 3:

They select one person among them, say Jeff, and hand all the pieces of papers to him (not the estimations, but how off the estimations were).

Step 4:

Jeff sums up the numbers: E(J) + E(S) + E(C) + E(L) + E(D) - (J + S + C + L + D), and tells the result to Larry. Larry adds this number to the total of initial estimations, and this will be the actual sum of their weights.

  • $\begingroup$ To spare people's feelings, maybe it is best not to give estimates but use some random numbers that are plausible weights and tell everybody that this is so. $\endgroup$ – Jaap Scherphuis Feb 14 '20 at 9:15
  • $\begingroup$ How does this prevent Larry and Jeff from conspiring to determine the weight of every other person immediately? $\endgroup$ – Nij Feb 14 '20 at 9:35
  • $\begingroup$ In real life the total weight is not all that the pilot needs to know. $\endgroup$ – DJClayworth Feb 14 '20 at 16:08
  • $\begingroup$ @Nij Larry hands out separate estimates, but he gets a total amount of mistakes. He will never know whose whose weight he guessed true. $\endgroup$ – padawan Feb 14 '20 at 20:13
  • $\begingroup$ I think you have not understood what "Larry and Jeff .. conspiring" means. $\endgroup$ – Nij Feb 14 '20 at 21:43

This is not really in the spirit of the puzzle, but I'll add it for completeness.

A pilot does not actually need to know the weight of every person on board. They need to know a) the total of moments about the centre of gravity - i.e. their weight times the distance from the plane's centre of gravity. b) the total weight.

So the first problem can be solved by assigning each person a seat, and telling them the distance of their seat from the CofG (the pilot will know that). Each person multiplies their weight by that number and writes it on a piece of paper and puts it in a box. The pilot adds up the total of those numbers and uses it in their calculation. (It does work, and I can explain it in more detail if necessary).

As long as the pieces of paper can't be traced back to a specific person nobody ever knows anyone's weight. (The pilot can't tell the difference between a weight of 100lb at a distance of 30in and a weight of 50lb at a distance of 60in).

The total weight is a well known problem with solutions elsewhere.

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    $\begingroup$ All aircraft have a maximum takeoff weight (MTOW), the pilot would need to know that this is not exceeded. There is no information that could be added to the above schematic that wouldn't make it possible to infer the weight of the seat occupant. Spirit of the puzzle aside; in the spirit of airmanship, make no mistake that if there were any doubt, they would be getting on some bloody scales! $\endgroup$ – Toby Wilson Feb 14 '20 at 16:30
  • $\begingroup$ @TobyWilson The pilot also needs to know the CofG (and yes of course they probably would want to weigh everybody). But in theory if the pilot trusts the passengers they could work out what they need from this. I don't think anyone, even the pilot, could infer the weight from the info they are given. Could you explain how you think they could work it out? $\endgroup$ – DJClayworth Feb 14 '20 at 16:45

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