Rob and his cohorts are correct. Once you determine that the die can be numbered to create a rock, paper, scissors game, every 8 year old boy knows the winner is the one who chooses second. Thus Oppenheimer wins.
Because there exists a neutral game that allows Oppenheimer to never lose. Since he can choose this one, the worst that can happen to him is that there is always a draw. In the worst case, he gets even.
Proving that someone is favoured only requires to show the existence of a non losing strategy.
“In the worst case, he gets even.”
But Ken didn’t just ask whether Oppenheimer can “get even.” He asked whether Oppenheimer (or Einstein) would actually be favored–i. e., better than even odds.
Francois - I am not following you.
If you can set the die up so that by choosing second, you will always possess a die that will win against the first chosen die more often than it will lose, why would you set the die up so that by choosing second, you will possess a die that will only win against the first chosen die as often as it loses?
In other words, why set up a rock, rock, rock game when you can set up a rock, paper, scissors game?
You have a game where the first player can ensure a draw regardless of what the other does. Hence you show that you’re favoured when being the first player.
My answer is the lazy answer, but it does answer Ken’s question. Rob did the work to identify an actual non-transitive relation that shows Oppenheimer can win.
Ken asks: in that game is someone favoured.
The lazy answer: yes, Oppie is favoured because even if he plays dumb and chooses a trivial neutral partition, he will have a draw. In other words, it doesn’t even matter what Einstein does.
The answer requiring more work: Oppie can actually have a strictly dominant strategy where he wins regardless of what Einstein does.
However, answer 1 does answer Ken’s question. It pays to be the first player (which is the actual question).
Would you also say if Oppenheimer and Einstein play tic-tac-toe and Oppenheimer goes first, he is “favored to win”? The only way you could even begin to argue that would be by supposing that Einstein might make a mistake in his response to Oppenheimer’s moves. But Ken’s post asked us to “assume that Oppenheimer and Einstein employ the smartest possible strategies,” so we’re supposed to rule out the possibility of either making a mistake.
Wrong analogy. With tic-tac-toe, you can be the first player and still lose.
From the moment Oppie plays the first move of that game, with a neutral strategy, he can’t lose anymore.
Read the differences between dominant and strictly dominant strategies in 2-player games.
“Oppie is favoured because even if he plays dumb and chooses a trivial neutral partition, he will have a draw.”
If the best Oppenheimer could do would be to choose a neutral partition, then the game would in fact be a draw (since the players are not allowed to make mistakes, per Ken’s specifications). Therefore NEITHER PLAYER would be favored, if that were the case.
OK, Rob, you’re right.
“With tic-tac-toe, you can be the first player and still lose.”
Not if you add in the specification in Ken’s problem, namely, that both players play correctly.
Either you or I watch too much Star Trek.
You realize that this whole thread would have lost its point if everyone didn’t already recognize that you’re Slowtwitch’s resident math genius. 
I’m far from a math genius…however I do understand the difference between a strict order and an order 
.
If you guys are done arguing, I want to know if I was right??? No advantage to either as the rules were laid out…
No, sorry. ; There’s an advantage to Oppenheimer, because there are certain ways that he can number the dice so that he can work it to his advantage no matter which die Einstein chooses.