We will need to use a mixture of common sense and conditional
probability to solve this problem, as there is no obvious
systematic point to get started with this anaysis. The key equation
we need is:
$$
P(A> B) = P(A> B|A=1)P(A=1)+\dots +P(A> B|A=6)P(A=6)
$$
An obvious starting point would be to consider the case where dice
B has only one number on it, 3, say.
For A to have a greater than 50% chance to beat B, A must have 4
higher numbers, Say 4444xx.
For B to have greater than 50% chance to beat C, C must have 4
lower numbers, say xx2222. Can C beat A? If C = 662222 and A =
444411 then P(C beats A) = 1/3 x 1 + 2/3 x 1/3 = 5/9
To find dice with equal probabilities after a bit of fiddling
around we settle on the answer 5/9. The trick is to adjust down any
dice which are too strong and up any dice which are too weak. There
is some flexibility to adjust numbers without altering the
probabilities for any other wins/losses.
An answer is
A
5
5
4
4
1
1
B
4
4
3
3
3
2
C
6
6
2
2
2
2
Explicitly, we have
$$
\begin{eqnarray}
P(A> B) &=& P(A> B|A=5)\times \frac{1}{3}+P(A>
B|A=4)\times \frac{1}{3}+P(A> B|A=1)\times \frac{1}{3}\\
&=&1\times \frac{1}{3}+ \frac{2}{3}\times
\frac{1}{3}+0\times\frac{1}{3}\\
&=& \frac{4}{9}
\end{eqnarray}
$$