Professor Egghead participates today in a local game show. There are six jewels (a diamond, an emerald, a moonstone, a ruby, a sapphire, and a topaz), and there are six boxes that carry the numbers 1, 2, 3, 4, 5, 6. Each of the boxes contains exactly one of the jewels. The assignment of jewels to boxes is totally random (and of course unknown to the professor), and each assignment is equally likely.

According to the rules of the game show, Egghead announces three guesses for each of the jewels:

  • I guess that the diamond is in one of the three boxes D1, D2, D3.
  • I guess that the emerald is in one of the three boxes E1, E2, E3.
  • I guess that the moonstone is in one of the three boxes M1, M2, M3.
  • I guess that the ruby is in one of the three boxes R1, R2, R3.
  • I guess that the sapphire is in one of the three boxes S1, S2, S3.
  • I guess that the topaz is in one of the three boxes T1, T2, T3.

After the guesses have been announced, the game show host opens all six boxes. The professor wins a fantastic prize, if each of his six guesses turns out to be correct.

What strategy should the professor apply?
What is his winning probability under an optimal strategy?

  • $\begingroup$ is my answer correct? $\endgroup$ Commented Feb 23, 2015 at 15:31

2 Answers 2


probability of winning is

(3/6)(3/5)(2/4)(2/3)(1/2)(1/1) = 5%

another way to calculate the probability is

(3/6)(2/5)(1/4) = 5%

the strategy is to

group 3 jewels into 3 same boxes, 3 other jewels into the 3 other boxes.

  • 1
    $\begingroup$ How did you prove the optimality of this? I wrote a program which brute forces all possibilities and this does work, but do you have a smarter proof? $\endgroup$ Commented Feb 22, 2015 at 22:37
  • $\begingroup$ no, I have no formal proof. but choosing these boxes ensures that each guess is not interfering too much with the next and previous guesses, which would be a losing strategy (guessing too many times on the same box) $\endgroup$ Commented Feb 23, 2015 at 0:52

I believe that the answer of user3453281 for the winning probability is correct, however I will try to give some more details on how to calculate it:

When assuming uniform probability, then the first guess will always have a $3/6=1/2$ change to be correct, independent of your guess (assuming that you can't guess the same box twice for the same jewel). Because the problem is symmetric, then for simplicity sake lets call the initial guess 1, 2 and 3. If you are correct, then for the following guesses those three boxes will have a lower probability of containing another jewel, namely only two of those three boxes can contain another jewel.
Lets assume that a participant wants every consecutive guess to have the highest probability of being correct. This means that the second guess has to be 4, 5 and 6, which has a probability of being correct of $\frac{3}{3+2/3\cdot3}=3/5$. This equalizes the probability of all boxes, thus the participant can chose any three boxes for the third guess, followed by the forth guess containing exactly the other three boxes, with respective probabilities, $\frac{2/3\cdot3}{2/3\cdot6}=1/2$ and $\frac{2/3\cdot3}{1/3\cdot3+2/3\cdot3}=2/3$. This forth guess equalizes the probability of all boxes again, meaning that the fifth guess can be three boxes, followed by the sixth guess containing exactly the other three boxes, with respective probabilities, $\frac{1/3\cdot3}{1/3\cdot6}=1/2$ and $\frac{1/3\cdot3}{0/3\cdot3+1/3\cdot3}=1/1$. The change of all these six guesses being correct thus has a probability of $1/2\cdot3/5\cdot1/2\cdot2/3\cdot1/2\cdot1/1=1/20$ or 5%.

The best strategy:

In the previous section I already mentioned which strategy should be used, however this leaves some room for variation. When considering that every box and jewel has identical properties (concerning this game show) and the order in which you would place every guess does not affect your final probability of being correct, then the only constraint to the best strategy should be that each box should part of exactly three guesses.


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