Solution 1 May 29, 2007


Does there exist two unfair 6-sided dice labeled with numbers 1..6 each such that probability of their sum being j is a number in \left(\frac2{33},\frac4{33}\right) for each 2\leq j\leq 12?


This is impossible. Let the probabilities for the sides of the first die be p_1,\, p_2,\ldots,\, p_6 and similarily let q_1,\ldots,\,q_6 be those for the second die.

The probability for the dice showing a sum of 2 is p_1q_1>\frac2{33}. This implies q_1>\frac2{33p_1}. Similarily, p_6q_6>\frac2{33} is the probability of a sum of 12, and we have q_6>\frac2{33p_6}. For the probability of a sum of 7 we have

\frac4{33}>\sum_{i=1}^6p_iq_{7-i} \geq p_1q_6+p_6q_1>\frac2{33}\left(\frac{p_1}{p_6}+\frac{p_6}{p_1}\right)
2 > \frac{p_1}{p_6}+\frac{p_6}{p_1}

This is a contradiction; multiplying both sides by p_1p_6\, and bringing all terms to one side gives

0 > (p_1-p_6)^2,\,

which is false. This completes the proof.