Rock, Paper, Scissors and Trump

Alicia and Robert are two high-stakes gamblers who are playing a match of rock-paper-scissors. Each winning throw is worth 10 €, paid by the other player. These players have deep pockets, so matches can last a long time and proper strategy is important.

To make things interesting, Alicia and Robert have added a special rule to the game. Each player selects a trump, which is a throw that nets them an extra 5 € every time they play it, regardless of whether they win, lose or draw with that throw. Alicia has selected rock as her trump, and Robert's trump is paper.

Examples:

Alicia throws scissors, Robert throws rock. Robert wins 10 €.
Alicia throws paper, Robert throws paper. Robert wins 5 €.
Alicia throws rock, Robert throws scissors. Alicia wins 15 €.

Can either player win this game in the long run? What strategy should they use?

• @jafe one question. Can I assume that the person I am not rooting for can throw anything and won't use the same strategy as the one I am rooting for?
– R.D
Aug 23 '18 at 7:34
• @R.D Both players play a strategy that maximizes their winnings. If it's the same strategy that the opponent uses, then they play the same strategy.
– Jafe
Aug 23 '18 at 7:37
• where does the extra 5 come from - the other player? so for rock/paper Alicia is down 10?
– JMP
Aug 23 '18 at 7:51
• @JonMarkPerry The other player, yes.
– Jafe
Aug 23 '18 at 7:54
• If Alicia throws Scissors and Robert throws Paper, does Alicia win 5€, or does Alicia win 10€ and Robert win 5€? Aug 23 '18 at 11:59

Looks like

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 |
Alicia +------------------------------+-------+
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.

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!

• Bass's proposed strategy is in fact the optimal strategy for Robert in the following sense: whatever strategy Alicia chooses, Robert comes out ahead by 5€ on average, and for any other strategy Robert chooses Alicia has a strategy that does better than that. Alicia also has an optimum strategy (her ratio is 1:2:3) but all it achieves is losing by an average of 5€ whatever strategy Robert uses. Aug 23 '18 at 12:03
• Relevant magic words to look up if interested -- which I bet Bass already knows, but others reading this may not -- are "Nash equilibrium", "minimax theorem", and "two-player zero-sum game". Aug 23 '18 at 12:06
• I was working on the same thing. The opponent has a similar strategy, but weighted differently, they must follow in order not to lose even more. If either player diverges from their strategy, the bias allows the opponent to take advantage and increase their average gains or decrease their average losses. Aug 23 '18 at 12:08
• Yeah, me too. I'd computed the minimax strategy, came here to post it, and saw that Bass had already done so, so all I could do was show off in the comments :-). Aug 23 '18 at 12:11
• It’s fun also that this game ends up being basically single-round rock-paper-scissors again, the single round being the choice of trump. Aug 24 '18 at 9:44
A/R R S P ±
R +5 +15 -10 +10
S -10 0 +5 -5
P +10 -10 -5 -5
-5 -5 +10 0

This table shows Alicia's return for each game. Reading as negated gives Robert's.

The first choice for A would then be Rock, but R's trump beats this, and once this is known, Rock is a useless choice for A, and to stop R playing trumps, she should choose Scissors.

But this leads to a cat-and-mouse game, with no discernible strategy, so no determinable winner.

• Oohh this makes more sense. My approach was more one dimensional while you went beyond. Nice!
– R.D
Aug 23 '18 at 8:13

I think Robert would win if he keeps throwing his trump.

The cases are

1. Alicia throws Rock (5e) < Rob's paper (15e)

2. Alicia's Scissors(15e) > Rob's paper (20e)

3. Alicia's Paper (15e) = Rob's Paper (25e)

We see that if Robert sticks to paper only then no matter what Alicia chooses, he will be in the lead.

Now if Alicia follows the same and throws her trump i.e rock then Robert will still be in the lead because his trump is better than hers.

Now let's consider the case of Alicia throwing only scissors. If she catches on to Rob's strategy then she will throw only scissors which is the second case.
But Rob will always have a 5e advantage and all he needs to do is switch to rock when he feels there is a chance that Alicia will chose scissors twice in a row.

• Please add a comment if you are going to downvote a post by a new contributor. Aug 23 '18 at 17:51
• Not the down-voter, but if it becomes apparent that Robert always plays his trump of Paper, then Alicia can counter that by always playing Scissors, which would lead to Alicia winning all of Roberts euros. Aug 24 '18 at 14:13
• True but I did say if Alicia threw it twice in a row he just needed to change it. It's not the best strategy I know XD
– R.D
Aug 25 '18 at 8:50
• If Robert reacts mechanically to Alicia playing scissors twice by switching to rock, then Alicia can just play scissors twice and then switch to paper. Any determinstic strategy, like the one you've described, will fail as soon as Alicia figures out what it is. Mar 1 '19 at 9:23

Just like the original RPS,

So, Alicia might think that Robert would use his trump, and so would play scissors, but Robert knows this, and would play rock instead, and you could go endlessly down the rabbit hole, though there is no way to have a 100% chance to beat a random order. But, Robert will play his trump more, but Alicia won’t as much, because she knows that is they both use their trump, Rob will net 10, but Alicia can net 15 if Rob plays scissors to her trump rock.
Therefore Rob will generally play scissors less, and also rock less, because if they both play rock then she nets 5, and his trump more, which would make her more likely to use scissors, and eventually this will contradict itself, because no matter how deep you go into the rabbit hole, you can’t get a strategy that will give you an advantage.

• The original rock-paper-scissors does have an optimum strategy, which is to choose randomly between the three choices. Any other strategy is sub-optimal against a rational opponent. Aug 23 '18 at 15:28
• All your observations are pretty much correct, I think. @Sneftel's helpful comment seems to be exactly the missing bit of game theory trivia that would have tied all the parts together and maybe lead you to another conclusion altogether.
– Bass
Aug 24 '18 at 4:11