Robert can win.
One strategy would be to
on each throw, randomly choose between two rocks, three papers, and one scissors. (Secretly roll a 6-sided die to decide.)
The opponent will naturally soon realise what's happening. But it won't help. Here are the possible results from Alicia's point of view:
\ Robert R R P P P S | Total |
Rock | +5 +5 -10 -10 -10 +15 | -5 |
Paper | +10 +10 -5 -5 -5 -10 | -5 |
Scissors | -10 -10 +5 +5 +5 0 | -5 |
So whatever the opponent chooses, the expected value will favour the strategy given above. By the law of large numbers, the average amount gained from a single throw
should gradually approach five sixths of an Euro (pretty close to one US Dollar, actually) as the game goes on.
(Thanks, @Adayah, for pointing out that I hadn't actually included this number in my original answer)
The reason this game isn't symmetrical has to do with one trump beating the other.
The way I went about solving this puzzle was to first notice that the trump asymmetry
must give Robert an advantage, because both players would like to play their trumps often, but Robert's trump getting played a lot makes it unprofitable for Alicia to play her trump. This means that Robert is in an arbitrage position: all he has to do is to bet "mostly paper", and hedge his bets by betting against himself a bit (choosing anything else than paper is bad for Robert, assuming Alicia chooses randomly), so that his advantage "spreads" to all possible plays by Alicia.
This bet hedging is extremely important, because in arbitrage betting, you can occasionally get away without making sure you win at every possible outcome. In a game of repeated RPS, however, there is an intelligent adversary, and in game theoretical calculations, this guarantees that you will always hit the worst possible outcome with any strategy. Therefore, the only thing you need to optimise is the outcome in the worst case scenario. (For more details and info on the subject, see the excellent comments by Gareth and Jaap below.)
The fact that the first suitable betting strategy I stumbled upon happened to spread the advantage evenly (making the worst case equal to the best case), and with minimum loss of advantage, speaks volumes for the excellent design of the puzzle. Thanks, @jafe!