This is an interesting puzzle. I catch myself fighting the hypothetical a lot.
I think it hinges on what would be the right move if you saw a six, and the market also had six as the favored option. In that situation, it would be appropriate to bet on the six which would move it past the 50% equilibrium, because you have the information from the market and the information from the die. I think maybe your equilibrium price can only exist if there is only one participant currently offering all of those bets, and they saw a six (so it’s not really a true market yet, or there is only one informed participant and many uninformed). In that case, you having seen a six would imply a probability of higher than 50% that it is the weighted side. Given that thinking, if you see that prediction market favoring a different number (“3”), you should indeed bet against it, because very little information is contained in the market (one die throw worth).
The market showing a 50% price for one number and 10% for the rest is an unstable equilibrium. If you started out with a market-maker with no information offering 1⁄6 for all sides and there were many participants who only saw a single die, the betting would increase on the correct side. At each price that favors that side, every person who again sees that side would have both the guess from the market and the information from their roll, then they would use that information to estimate a slightly greater probability and the price would shift further in the correct direction. It would blow past 50% probability without even pausing.
Those don’t seem like very satisfactory answers though.