You're in front of a rather odd eight ball pool table with the holes covered up such that the whole table is perfectly flat. It has unusual inner dimensions, 9 and 26, with all balls (1-15) with radius 1 and lighter than usual (except the white ball).

You rack it up and you're just to about to make the first break...

enter image description here

Earlier this day your wealthy tech (he's got a lot of cool technological stuff) friend made a challenge for you.

Friend: "Hey, I've been thinking of an idea for a while. You know that odd pool table I bought with the light balls?"

You: "Yeah? What about it?"

Friend: "Since you're a good puzzler, I've been wanting to make a puzzle for you which I don't really have an answer to. If you manage to do well, you'll get a good prize."

You: "Sure! Let's do this."

Friend: "Alright. I'll go home and make the puzzle, I'll get back to you soon!"

You: "Awesome!"

A couple of hours went by and then he finally brought the puzzle.


  • Breaks only.

  • At most 100 rounds (100 breaks).

  • Before starting, tell me the number of rounds you want to play and choose only one option A-E, where
    option A corresponds to value 20<x≤25,
    option B to 15<x≤20,
    option C to 10<x≤15,
    option D to 5<x≤10, and
    option E to 0<x≤5.

  • Correct value(round 1) gives 1 dollar, correct value(round 2) gives 2 dollars, correct value(round 3) gives 3 dollars,..., correct value(round 100) gives 100 dollars which means maximum winnings are 5050 dollars. The catch is that if you're wrong in any round, you lose all the money you made in previous rounds.

  • The white ball is not included in the puzzle (except when breaking, of course).


Always ∑ <--->(••) c to c → <---> ?

(Note that my high tech equipment will take care of the above.)

Take your time and when you're ready, tell me the option you've decided and the total number of rounds you want to play.

Good luck!

Finally, after some thinking you feel confident enough to give him an answer.


What option do you choose to get as much money as possible? And how many rounds?

Note: Although mathematics is needed for you to solve the puzzle, it isn't complicated.

Hint 1

Assume the balls have equal probability to end up anywhere on the table (completely random). Oh, and one more thing; not really sum

  • 3
    $\begingroup$ What does "Correct value(round 1)" mean? If I say I want to play 3 rounds and choose option C, why can't you simply say "no, that's not the correct value"? An example would help a lot. $\endgroup$ Jul 4, 2022 at 12:51
  • $\begingroup$ @RayButterworth The friend don't have an answer to this puzzle. "...I've been wanting to make a puzzle for you which I don't really have an answer to". Correct value refers to the values of A-E. F ex round 1, if you chose option C (for 3 rounds) and the value is within the range of C you'll get a dollar and continue to round 2. In round 2, if it's still within the range of C, you'll continue to round 3. In round 3, if it's still within the range of 3 you win the money, if it's not - 0 dollars. The friend don't know the optimal answer (Let's just say he's not as good at math as you). $\endgroup$ Jul 4, 2022 at 13:34
  • 3
    $\begingroup$ I'm still totally lost as to what "the value" refers to and what determines whether it is "correct". $\endgroup$ Jul 4, 2022 at 13:43
  • $\begingroup$ @RayButterworth Yes, intentionally done (enigmatic puzzle). No knowledge of pool or anything like that needed. Just mathematical knowledge. If it's still unsolved within some days, I'll make an edit and clarify (in a subtle way) what determines it. $\endgroup$ Jul 4, 2022 at 13:56
  • $\begingroup$ Can you give away if the question mark is an unknown or just part of the sentence? $\endgroup$ Jul 4, 2022 at 16:01

1 Answer 1


I will give this a try but I'm really not sure whether this is anywhere close to being correct

We should choose

Option E for 100 times.

The "rebus" puzzle is

Always some distance of two balls center to center is within the range.

To show that our chosen option is correct

Let us consider the maximum of the minimum distance between two balls, i.e., how far apart can the closest two balls be at most. Let us assume that this distance is at least 5. Then, consider the following picture

Here the black borders illustrate the original table and red circles are original balls. The pink circles are circles of radius 2.5, i.e., diameter 5, so the grey borders would correspond to a table of dimensions 12x29 (1.5 added to each side).

Now, the question is, how many circles of diameter 5 can we fit inside the larger rectangle? The densest packing of circles in the plane is hexagonal packing, which, in theory, would allow for 16.07 circles inside that region, but in a finite rectangle of this size, we cannot get even close to that density. In fact, there doesn't seem to be any way to fit 15 circles of diameter 5 inside the large rectangle, so we can deduce that there must always be at least two balls whose distance is less than 5. Therefore, we will always win with these rules!

Some thoughts

I honestly have no idea, whether this is the intended answer. The part about picking the number of rounds doesn't seem to make sense unless there is some finite 0<p<1 probability of winning in which case the correct number of rounds would be $\lfloor\frac{p+1}{p-1}\rfloor$. So it might be that the actual intended solution does not give a sure win.

The mathematics could very quickly become very complicated, however, since there are really no exact solutions to be found for the average distance distributions for the balls (only that the pairwise distance must be at least one diameter and the maximum distance must be less than between two opposite corners). I'm not even sure there is any non-complicated way of actually proving the 15-ball packing argument used above.

Also, since the width of the table is less than five ball diameters, it is not actually possible to rack the balls in the usual triangle formation though that is not really relevant for the puzzle. It makes me wonder though whether the balls are supposed to have unit diameter instead of unit radius. However, that doesn't seem to affect the packing argument.

  • $\begingroup$ You've interpreted the rebus correctly and your answer is the intended one. I gotta say, I made this puzzle a bit messy (and numbers off). My idea was that the "non complicated way" to prove this was with the pigeon hole principle. You can always show that there exist a distance between two balls that are always within 5 units away. However I don't know how to show if a distance will be in the range of the other options aswell. Just option E. You'll get the tick anyways ofc. $\endgroup$ Jul 22, 2022 at 10:41

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