I want to play a game where one person thinks of a number and the other tries to guess it, but I want to play this game by myself.

The puzzle is to come up with a way to play this game.

Rules (have been edited. I'm very sorry to the people who answered before I edited, your answers are good too!):

  1. Need to guess a number 1-10.

  2. There must exist one "correct" number that, if you guess it, will necessarily be correct, as in, it needs to make a difference which number you guess.

  3. Each guess, I should be able to figure out if it's "right" or "wrong", without getting any other information.

  4. I'll keep guessing until I guess right.

  5. I need to be able to play this game entirely by myself. No other people, no computer (pencil and paper is allowed).

  • $\begingroup$ How about a coin or a dice telling you if it is 'right' or 'wrong'? $\endgroup$
    – elias
    Apr 3, 2017 at 14:10
  • $\begingroup$ @elias That's a good idea but I think it would break rule (1) that there must be one correct number, because a die or coin could tell you "correct" multiple times. $\endgroup$
    – Owen
    Apr 3, 2017 at 14:11
  • $\begingroup$ Maybe I got it wrong, but because of rule (3) you keep guessing until you got it right. Hence you won't be told for two different numbers, that they're both correct. It might happen, that all numbers are told to be 'wrong', however. $\endgroup$
    – elias
    Apr 3, 2017 at 14:17
  • $\begingroup$ must it be exactly 1-10? $\endgroup$
    – Sconibulus
    Apr 3, 2017 at 14:18
  • $\begingroup$ @elias OK, I guess you are right. The rule I had in mind is that there must actually "be" a correct number (pre-determined), but I guess I did not do a good job writing the rules. $\endgroup$
    – Owen
    Apr 3, 2017 at 14:24

13 Answers 13


Write the numbers 1-10 on 10 "answer" sheets of paper in different locations on the paper (one number per page).

Make ten corresponding "guess" sheets of paper that have holes (scissors or ripping) where one number would be, and label those with what number it would show (so if you place the 7 guess sheet over the 7 answer sheet you'll see a 7, but all other answer sheets you would see nothing).

Shuffle the answer sheets face down and take one out without looking at it.

Slide the guess sheet for the number you want to guess and then flip it over. If you see the number you're done!

Elsewise, flip back over and repeat with another guess sheet.

Seems fun.

You can also take advantage of symmetry to use less than 20 sheets of paper.

  • $\begingroup$ Great idea! I think this matches all the rules. $\endgroup$
    – Owen
    Apr 3, 2017 at 15:05
  • $\begingroup$ this is a unique idea, if not a bit bulky $\endgroup$
    – RozzA
    Apr 3, 2017 at 21:41
  • 1
    $\begingroup$ Also known as, "create a random number generator with only paper and pencil." Nicely done. $\endgroup$
    – jpmc26
    Apr 5, 2017 at 2:21
  • 2
    $\begingroup$ This seems to me like it only works a limited number of times since you will start to recognise the answer sheets. $\endgroup$ Apr 5, 2017 at 6:35

Make 10 identical pieces of paper. Make sure they're nice and opaque. (Card would be better than paper, if allowed.) Write "RIGHT" on one side of one of them, "WRONG" on one side of each other one.

Turn them so you can't see the writing. Shuffle them. (Another reason why card would be better than paper.) Lay them out in a line. You can put labels 1-10 beside them, if you like.

Now to guess a number you turn over the card in that position.

(This is similar to jousle's answer, but seems rather simpler.)

  • 5
    $\begingroup$ To make it even simpler, use a deck of cards a have a predetermined card (like ace of spades) be the 'right' card. $\endgroup$
    – Kruga
    Apr 4, 2017 at 8:15
  • 1
    $\begingroup$ This feels like it'd be significantly easier to do that the current accepted answer. +1 $\endgroup$
    – Joe
    Apr 5, 2017 at 15:44
  • $\begingroup$ @Kruga as explained in Ian's answer about guessing golf $\endgroup$
    – RozzA
    Apr 10, 2017 at 19:20

Use amidakuji with ten columns numbered 1-10. The top row is the guessing row. You choose a number from the top and trace it down to the bottom. If you reach 1, your guess is right. Otherwise it's wrong and you gained no additional information on what the correct number is.

  • $\begingroup$ Very simple and looks quite fun $\endgroup$
    – SztupY
    Apr 3, 2017 at 18:18
  • 2
    $\begingroup$ this just blew my mind $\endgroup$
    – RozzA
    Apr 3, 2017 at 21:34
  • $\begingroup$ except this will fail, long term - the same person creating the entire diagram could result in eventual (subconscious?) cheating $\endgroup$
    – RozzA
    Apr 3, 2017 at 21:48
  • $\begingroup$ @RozzA Possible. Of course, the more legs you draw, the better. If your board has 100 legs, maybe you won't remember all of them. $\endgroup$
    – BaSzAt
    Apr 4, 2017 at 6:34


Get a gun with a 10 bullet chamber. Put one bullet in the chamber and roll it. Now, guess a number between 1-10. This number should be the number of times you need to pull the trigger to shoot the bullet out. Now, keep shooting until your number comes up. If the bullet is fired before that, you lose and you roll the chamber once again and keep guessing until you win.


Use a toy and not one of these . :-)
enter image description here

  • 2
    $\begingroup$ except that resets the number to guess: I can just keep guessing 1 and rolling the chamber til I get it, and it might take dozens of tries - or I could guess 1 thru 10 and still not win, due to the target number changing each 'roll' $\endgroup$
    – RozzA
    Apr 3, 2017 at 21:28
  • 4
    $\begingroup$ it would make more sense to write a number on each chamber, and rolling the chamber to the number you pick and firing, then doing the same again with another number $\endgroup$ Apr 4, 2017 at 1:28
  • $\begingroup$ @RozzA - Winning/Losing is a part of game. You can lose everytime in a Roulette too ;) $\endgroup$
    – Techidiot
    Apr 4, 2017 at 4:16
  • $\begingroup$ @Destrucrible Lemon - Yes. Can be done. But it works like this as well. $\endgroup$
    – Techidiot
    Apr 4, 2017 at 4:18
  • $\begingroup$ @Techidiot unless the rules were changed after your answer, this fails #4 $\endgroup$
    – RozzA
    Apr 10, 2017 at 19:19

Because of rule #3, you will always get the correct number eventually. This means that the game is actually more akin to "guessing golf".

  1. Take a deck of cards.
  2. Pull out the 1-10 of spades and 1-10 of hearts (or your favourite suits)
  3. Shuffle the spades and spread them out face down. Choose one and set it aside.
  4. Take the hearts and arrange them with your guesses left to right.
  5. Flip over the chosen spade.
  6. Score the number of cards to the left of the matching heart.


Hearts     1 6 4 7 9 2 3 5 8 10         Chosen spade: 7
           x x x ^--- score 3
  • $\begingroup$ this is the most complete, non-cheatable design - and it comes with a bonus scoring system nicely done! $\endgroup$
    – RozzA
    Apr 3, 2017 at 21:51

I can imagine a method which uses modular arithmetic. My answer does not address all the details.

We would need a method that decides if a given (large) number is divisible by 11 or not, but does not tell you the remainder if it is not divisible.

Once we had this, we should produce a large random number $k$. If it is divisible by 11, throw it away, generate another one. Repeat, if necessary.

A number $n$ in the range 1-10 is 'correct', if and only if $nk-1$ is divisible by 11. This happens iff $n$ is a modular inverse of $k$ mod 11, hence there is a single solution in the range 1-10.

The multiplication and subtraction can be done with paper and pencil, but I still have to find an easy 'divisibility by 11'-test which does not give away the remainder.

I think I managed to come up with a method that tells if a given number $t$ is divisible by 11 or not, still it does not tell its remainder in the latter case. (It works as a divisibility test for other prime numbers as well.) It has finite steps with probability 1.

Generate a random number $r$.
Calculate the product $p=tr$.
With a usual method (division, or the difference between the sum of even and odd placed digits, as linked by @notboughtdirtyesterday) check if the product is divisible by 11.
If $p$ is not divisible by 11, then for sure $t$ is not divisible by 11 as well, we are done. We might know the remainder of $p$, but that does not tell anything about $t$.
If $p$ is divisible by 11, then either $t$, $r$ or both are divisible by 11.
Use the usual method to check if $r$ is divisible with 11.
If $r$ is not divisible by 11, then $t$ is divisible by 11, we are done. The remainder of $r$ does not tell anything about $t$ in this case either.
If $r$ is divisible by 11, start again the whole procedure with another randomly chosen $r$.

I'd suggest using 6-7 digit long $r$s, which are not giving away their remainder just by looking at them. At least not for someone who is not a number-juggler.

  • 1
    $\begingroup$ math.hmc.edu/funfacts/ffiles/10013.5.shtml might be of interest, it deals with the simple trick of 11 divisibility. $\endgroup$
    – user19641
    Apr 3, 2017 at 17:01
  • $\begingroup$ @notstoreboughtdirt but also as a side effect it gives the remainder, which easily gives away the correct answer, hence should be kept secret. $\endgroup$
    – elias
    Apr 4, 2017 at 11:08
  • $\begingroup$ I like this answer because, in theory, it is possible to play entirely within one's own head without any external physical objects. Assuming a method of division by 11 without revealing the remainder exists, which it likely does (it may be complicated, but some humans may still be able to handle the complexity). $\endgroup$
    – Keavon
    Apr 5, 2017 at 8:33
  • $\begingroup$ Thanks, @Keavon, in the meantime I found a method to do it. I will edit it in my answer in the next few days. $\endgroup$
    – elias
    Apr 5, 2017 at 14:04

It's not a very interesting game, but here's how such a game could be played.

Player 1: "I am thinking of a number, 1 to 10."

Player 1: "Is it 10?"

flip a coin -

  heads is Yes, tails is No.
  if Yes, you've guessed it, and you're done.
  otherwise ...

Player 1: "Is it 9?" coin flip
Player 1: "Is it 8?" coin flip

...and so on.

Play continues until you have guessed right.

If you get to "Is it 2?" and still get No,
you know the answer must be 1.

  • 2
    $\begingroup$ This doesn't enforce exactly one correct, you may get through all numbers without finding a correct one. $\endgroup$
    – Sconibulus
    Apr 3, 2017 at 14:26
  • 1
    $\begingroup$ @Sconibulus Not true. (You may have missed my edit, which fixed that.) $\endgroup$
    – Rubio
    Apr 3, 2017 at 14:27
  • 3
    $\begingroup$ This is so wrong. You are missing the entire probability concept. If you are playing normal guessing game the chance to guess the number with the first try is 1/10, the second one will be 1/9 and so on. In your modified version is 1/2 for every try regardless of number of tries already made. I hope you see what is the problem. $\endgroup$
    – Mojo Risin
    Apr 3, 2017 at 17:37
  • 2
    $\begingroup$ @MojoRisin There's no requirement that the game be fair, just that it can be played to find a single correct answer. It's not a good game, but it appears to be a valid one. $\endgroup$
    – Rubio
    Apr 3, 2017 at 17:45

It pretty much boils down to turning it into a game of chance we can't control, so there are countless ways.

You can stick a pencil in the ground. Then make a guess, close your eyes and start moving the tip randomly. Open your eyes after a certain time and fold a piece of paper into several identical pieces whose width is to be used as our "unit". Measure the length of the shadow in mod 10 and add 1 to the result to tell if you've guessed right or wrong. Round the length if necessary.


To play this game solo, one must emulate two states by themselves.

   /¯¯\___________________________________________ /¯¯\
   \__/¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ \__/ 

Current Guess ———Distributed Apparatus——— Correct Guess

That is, the State of the Current Guess & the State of the Correct Guess

You also have to emulate an apparatus between them that can check whether the current guess matches, without revealing its state. This requires partitioning the data.

Given you only have paper and pencil, I'd say the ideal solution involves origami.

Ones makes 10 identical origami boxes, and labels their tops, from 1 to 10.

Then the 'Correct Guess State' is emulated by closing you eyes, shuffling up the boxes, and placing a small rolled up piece of paper in one of the boxes. You close it, and close all the others, with your eyes still closed. Given you mix up the boxes sufficiently, before opening your eyes, you should then be able to play out the game rather simply by guessing a number, opening the box and seeing if you were correct or not.

This method could be played over and over.

Now, I am sure you can see how the same game could be played by simply ripping up a paper into 10 labeled pieces, marking one of their backs (so it doesn't bleed through) and placing the pieces all label up. Checking would occur by flipping a piece over to see if it contained the mark. This of course, could only be played a limited number of times before new pieces would have to be made. Perhaps varying the symbol/color written on the back of a piece of paper would elongate lifetime of the pieces.

Both these methods employ the same concepts of maintaining partitioned states(data), which I may also add, as food for thought, is the secret to designing parallelization into a system.

"Partitioning the data is the key to parallelizing the program"

— Joe Armstrong, co-founder of Erlang




It doesn't matter when the correct number is decided as long as you don't get any extra information. So how about this:

Every time you guess a number, flip a coin (or use any other random source) to determine whether it's correct. But don't do as Rubio says. You should design the process so that the chance any number is the answer doesn't depend on your order of guessing. If that number is correct, you win. Otherwise, you write it down so you know it's incorrect.

Here's a way to determine whether a number is the answer with a coin. Suppose there are $n$ numbers left. Flip the coin $\lceil \log_2 n \rceil$ times. Treat the result as a binary number. If it's $0$, the first number is the answer. Else if it's less than $n$, the first number is not the answer. Else, repeat.

I don't have enough reputation, so I can't add a comment to Rubio's answer.

  • Pick a number between 1 and 10 inclusive

  • Write the number down on a piece of paper

  • Hide the number and write down on another piece of paper where you hid it.

  • Forget the number you thought of - you might need help with this bit

  • Try and guess your number - you only get one chance so be careful!


Buy a Macbook, go to terminal, type: "c=$((1 + $RANDOM % 10))" - without the double qoutes. Then guess a number and type: "echo $c". If you get really bored which I find hard to imagine... type: "sudo shutdown -r now"

  • $\begingroup$ This violates rule #5. It also only allows one guess before telling you the answer, violating rules #3 and #4. $\endgroup$ Apr 6, 2017 at 12:36

The best way to encode a number from 1 to 10 without using any extra materials will be to use your fingers, you have exactly 10 of them. In this case we can simplify the game to the following - each time you make a guess lift the corresponding finger and check if there is a marker under it.

So far so good but this will only work if you can hide something or at least mark somehow one of your fingers without knowing exactly which one. If you are absolutely forbidden to use extra materials you can use your hair (it depends how many games you want to play :D). Place the marker (the hair) on the table and then close your eyes. Stick your fingers together and move them randomly above the marker. Place your hands down in such way the the marker will be under one of your fingers but without knowing which one.

And you are set. Pick a number and simply lift up your finger, if you find the maker your guess is correct.

  • 4
    $\begingroup$ For some reason I became bald after playing this, any explanation? $\endgroup$
    – EKons
    Apr 3, 2017 at 17:26
  • 2
    $\begingroup$ Not everyone has 10 fingers, though. $\endgroup$ Apr 3, 2017 at 17:40

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