Three friends enter a quiz-show as a team, the final round of the show is all or nothing, they only get their prize if they can win this round.

Three boxes labelled 1, 2 and 3 are placed in a room, each box has either a white or a black stone in it, with a 50% probability for each.

The friends enter the room in turn, the first friend get to look in box 2 and 3, and then has to guess the colour of the stone in box 1. Similarly the second friend get to look in box 1 and 3, and then has to guess the colour of the stone in box 2. And finally the third friend get to look in box 1 and 2 before guessing the colour of the stone in box 3.

The friends have no means of communicating what they have seen in the boxes, and they are not told what guess those who came before them made.

The friends win if they collectively get at least two guesses right.

Find an optimal strategy for the friends to use.

Variation 1

What is the optimal strategy if the stones are black with 70% probability and white with 30% probability?

Variation 2

What is the optimal strategy if the friends are allowed to switch the two stones they get to see?

  • $\begingroup$ As for variation 1: Do the participants KNOW the distribution is 70/30? $\endgroup$ Aug 28, 2014 at 13:04
  • $\begingroup$ As for varation 2: Did the participants know the concept beforehand, so they could've come up with a strategy? $\endgroup$ Aug 28, 2014 at 13:20
  • $\begingroup$ @TimCouwelier Yes and yes. $\endgroup$ Aug 28, 2014 at 15:37

5 Answers 5


No matter what they do each of the friends will have exactly 50% probability of answering correctly, this could lead some people to the incorrect conclusion that their probability of victory is 50% no matter what they do.

The trick is that the friends can to some extend coordinate their guesses, so that the probabilities of guessing correctly are not independent.

A solution should try to distribute the correct guesses so that none are "wasted" on three correct guesses or a single correct guess. The theoretical optimal is that all three guesses are incorrect with 25% probability and exactly two of the guesses are correct with 75% probability.

There are a few facts to be deduced about any such optimal strategy:

  • None of the friends may be able to deduce that they will lose by looking at the two other stones. Since a loss scenario involves 3 incorrect guesses it would tell the guesser that he is about to guess wrong by using the agreed upon system, thus he has gained knowledge of the colour of the stone he hasn't seen, which is impossible.
  • All three stones must be a factor in the strategy of at least one of the friends. If one of the stones didn't have any influence on the guessing pattern then the guesser of that stone would be able to deduce the success of both his friends, thus knowing about a loss, which as mentioned above is impossible.

This knowledge can be used to quickly discard some options when searching for answers.

There are actually quite a lot of ways to achieve this, one of them is to let all of the friends guess to opposite colour of the colour of the stone in the box labelled one less than the box they are guessing on. And the first friend should guess the opposite of what is in box 3. This method will fail when all the three stones are of the same colour, and give two correct answers in all other cases, thus 75% chance of victory.


Let $R_1,R_2$ and $R_3$ be the three rooms and let $A, B ,C$ denote the three friends. I represent the white balls by the digit $1$ and the black ones by $0$.

They could adopt the strategy as summarized in the table below where I have indicated all the possibilities where the $plus$ and $dot$ symbol represent the Boolean operators OR and AND respectively.The overhead operator indicates NOT operation.

In this strategy $C$ always guesses $1$ (White ball), $A$ and $B$ guess based on what $R_3$(room 3) has. Their guesses are determined by simple Boolean algebra.

So for eg in the first case i.e in a) first A looks in $R_3$ and on seeing a white ball he performs the operation $R_2 .R_3$ meaning $1.1 = 1$(as per Boolean algebra). $A$ guesses correctly. Next $B$ will perform the NAND operation of $R_1$ and $R_3$ which gives zero.Here $B$ makes an incorrect guess and finally C will simply say white every-time(correct for this case). So they get 2 out of 3 correct. Rest of the table is self-explanatory.

enter image description here

The red ones indicate the cases which would fail. So they can win 6 out of 8 times. Also this method works for the case when all balls have the same colour.

For Variation 1

Here the strategy should be to guess all those combinations correctly which has more combined probability of occurrence. The table below shows one such strategy where $C$ always guess $0$ (Black ball). $A$ guess $1$ if $R_2 = 1$ otherwise $0$. $B$ guess $1$ if $R_1 = R_3$ otherwise $0$.

enter image description here

This way they manage to guess the 6 out of 8 which has a total probability of 0.874.

Ideally we would want them to not include the combination $(1,1,1)$ which has the minimum combined probability and you would want to guess $(0,0,0)$ combination for sure. But I don't think that's possible. Either you can have both $(1,1,1)$ and $(0,0,0)$ or neither of them in your correct 6. I'm not sure of this,maybe there is some other way.

Edit : After reading the comments I realized that i was wrong about $(1,1,1)$ and $(0,0,0)$ combination.

But I think that this is the optimal solution which gives the maximum total probability of occurrence for 6/8 because there is only one other combination which gives a higher value (0.91) i.e. when you exclude the $(1,1,1)$ + any other with two $1's$ viz ((a,b) , (a,c) or (a,e)). But if you choose to exclude any of these two then you would surely fail.

Like for eg. if you try to exclude (a,b) then you need to satisfy the following:-

enter image description here

Firstly the two failed one's should necessarily have all guesses incorrect. With this constraint we run into contradiction when two same initial configuration demand exactly opposite result(see blue coloured in table). Similar contradiction happen for the other two cases. So by contradiction we cant have the 0.91 solution so the next highest has to be the optimal solution.

  • $\begingroup$ Pretty good so far, though I do wonder how you managed to make the logical expressions so complicated. It is actually possible to have just one of 111 and 000 in a 6/8 solution. In any case, can you prove that your Variation 1 solution is the optimal? $\endgroup$ Aug 29, 2014 at 14:18
  • $\begingroup$ I agree that there's no need to invoke Logic gates into this. Actually i started thinking of this like some circuit with input,output using logic gates but later realized that it was an unnecessary complication. Anyways I have edited my answer to show that my solution should be optimal. $\endgroup$
    – Hubble07
    Aug 29, 2014 at 17:34
  • $\begingroup$ It is not that there is anything wrong with using formal logic, it just that I'd choose some shorter forms, like A=R2, B=R1⊕R3 in your first solution. Your proof is nice and simple. $\endgroup$ Aug 29, 2014 at 19:05

Update: Added a simplification of Hubble07's answer for variations 0 and 1.

Variation 0 (original question)

The goal is to get 2/3 correct guesses as often as possible. To do that, A and B should ensure that exactly one of them is correct if C is correct, and that either both or neither of them is correct if C is incorrect. One possible strategy is Hubble07's strategy:

  • C always guesses the same color.
  • If C will be correct, A guesses the stones A and B will match, and B guesses that they will not match.
  • If C will be incorrect, A and B both guess that stones A and B will match.

Since stones A and B will either match or not match, this guarantees exactly 2 correct guesses if C is correct. If C is incorrect, there will be 2 correct guesses if stones A and B match, and 0 correct guesses if they do not.

They only fail if C is incorrect and stones A and B do not match. Each of these has a 50% chance, so they fail 25% of the time, and succeed 75% of the time.

There are many strategies guaranteeing 75% chance of success. This is the same strategy as put forth by Hubble07, but in simplified form.

Variation 1

The probabilities have shifted, but this strategy remains the best.

Wtih a 70% chance of each stone being black and 30% chance of it being white, the possible outcomes range in probability from $2.7\%$ of $www$ up to $24.3\%$ of $bbb$.

If C is going to always guess the same color, that color should obviously be black. Given that if C is correct, A and B will ensure a win, that is $70\%$ chance of winning right there.

If C's guess is incorrect, A and B should both guess that stones A and B will match or both guess that they will not match. There is a $.7\times .7 + .3\times .3 = 58\%$ chance that stones A and B will match, and a $.7\times .3 + .3\times .7 = 42\%$ chance that they will not match. Matching is more likely, so they should both guess match if C is incorrect.

This leads to a total of $70\% + 30\%\times 58\% = 87.4\%$ chance that they will win with this strategy.

As I mentioned, since this is exactly the same as Hubble07's strategy, it yields the same chance of success as Hubble07 indicated.

Variation 2: (Below is all original content).

For the case where the guessers can rearrange the stones they see before leaving the room.

There is a 100% strategy:

  • A always rearranges so that C has a black stone, if possible.
    • A always guesses whatever color they left in box C.
  • B always rearranges so that C has a black stone, if possible.
    • B always guesses white.
  • C always rearranges so that B has a white stone, if possible.
    • C always guesses black.

Here is a table, showing the initial layout, the layout after A leaves the room, the layout after B leaves, the final layout after C leaves, what the three guesses were, and the tally of results.

$$ \begin{matrix} \underline{\text{Initial}} & \underline{\text{A}} & \underline{\text{B}} & \underline{\text{C (final)}} & \underline{\text{Guessses}} & \underline{\text{Result}}\\ www & www & www & www & wwb & ++- \\ wwb & wwb & wwb & wwb & bwb & -++ \\ wbw & wwb & wwb & wwb & bwb & -++ \\ wbb & wbb & wbb & bwb & bwb & +++ \\ bww & bww & wwb & wwb & wwb & +++ \\ bwb & bwb & bwb & bwb & bwb & +++ \\ bbw & bwb & bwb & bwb & bwb & +++ \\ bbb & bbb & bbb & bbb & bwb & +-+ \end{matrix} $$

As the table shows:

  • C only guesses incorrectly if all the stones are white.
  • B only guesses incorrectly if all the stones are black.
  • A only guesses incorrectly if there are exactly two white stones and A started with one of them.

There are no overlaps between these conditions, so it is always the case that at least two of them guessed correctly.


As @eBusiness notes, by coupling their guesses the friends can achieve 75% success. By choosing the opposite of the 'previous' box, they score every time there is a change in colour between one box and the next. This is useful because there are always either 0 (25%) or 2 (75%) changes.

Variant 1: If the probability of black stones is 70%, however, all friends guessing black succeeds with probability $.7^3+3*.7^2*.3=.784$. The previous approach now only works for the 63% of the time both colours are present.

Variant 2: If they may swap the stones, friend 1 guesses the colour they see in box 2. Friend 2 guesses the colour they see in box 1. Friend 3 then switches these stones they see, making the previous 2 correct with 100% probability.

  • $\begingroup$ In variant 2 they have to guess the colour of the stone currently in the box. $\endgroup$ Oct 19, 2014 at 9:03
  • $\begingroup$ Ah, I see. Is the probability still 50/50? $\endgroup$ Oct 19, 2014 at 9:37
  • $\begingroup$ The probability is 50/50 for variant 2. Also note that Hubble07 has a better solution than you for variant 1. $\endgroup$ Oct 19, 2014 at 11:13

Variation 2: I believed there was a 100% strategy - reached as follows:

A guesses BLACK, and if possible switches boxes 2 and 3 so 2 would hold a black. B guesses BLACK, unless he sees two whites in 1 and 3. If possible and needed, he switches boxes 1 and 3 to put a white in box 3. C guesses WHITE.

The box contents are as follows: enter image description here

Based on the explained logic, here are the guesses (red = mistakes)

enter image description here

There is never more then one mistake, so always at least two correct answers, guaranteeing a 100% win.

PS: also interesting variation: if box switching is allowed, how can they maximize odds of getting all three right. (I found a 3/8 strategy there)

As pointed out below that is NOT correct, as the assumptions for option B aren't interpreted correctly in the table. However, if B assumes BLACK 100% of the time, winning chances are 7/8

  • 1
    $\begingroup$ You wrote "B guesses BLACK, unless he sees two whites in 1 and 3." But I see that two whites appear 3 times (first three cases) in box 1 and 3 for turn 2, so shouldn't B's guess have three whites? $\endgroup$
    – Hubble07
    Aug 29, 2014 at 10:56
  • $\begingroup$ Indeed there is an error in that table, this solution as described only get 75% wins. Concerning your PS, better than 3/8 is possible. $\endgroup$ Aug 29, 2014 at 14:24
  • $\begingroup$ Yea it appears I stuffed up. Initially I was assuming B always guessing black, but that meant there'd be two failures on the first option (so 7/8) rates. My attempted 'fix' for that issue indeed didn't entirely fix it (it made it worse, because I mistakenly went based on what A saw, not B.). I'll edit the reply to that for now, and see if I can come up with a way to get to 100% anyway. $\endgroup$ Aug 29, 2014 at 14:44

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