Suppose for the sake of contradiction all the knights end up with wine. This also means everyone ends up with exactly one glass. Step 1: No two consecutive knights have red wine. Otherwise, there exists three knights with glasses $WRR$ in order. It is easy to verify that the knight starting with the white glass will end up empty-handed, a contradiction. Step 2: No three consecutive knights have white wine. Otherwise, there exist four knights with glasses $RWWW$ in order. It is easy to verify that the knight starting with the red glass will end up with two glasses, a contradiction. Step 3: A knight with white wine does not sit between two knights with red wine. Otherwise, we have $RWR$ in order, and the leftmost knight starting with the red glass will end up with two glasses, a contradiction. Step 4: The number of knights is divisible by 3. Claim 1 shows that the sequence of wine glasses takes the form $RW^{k_1}RW^{k_2}RW^{k_3} \cdots$. Claim 2 shows $k_i < 3$. Claim 3 shows $k_i > 1$. Hence $k_i = 2$ for all $i$. So the sequence is $RWW$ repeated over and over again. Since 50 is not divisible by 3, we have the final contradiction. I'm curious about this incorrect translation bit; what was the version intended by the proposer?