robinesque wrote: RRR 0
6/10 results are favorable
Wait, what? Symmetry requires S>P and P>S with the same frequency, assuming non-exploitable strategies. Looking at your list, you have SSS<->PPP, PSS<->PPS, RSS<->RPP, RRS<->RRP and RPS<->RRR each with the same probability within the pairs, and each pair cancels for the expectation value.
Also note that your results do not have the same probability. RPS will occur twice as often as RPP or RSS, as it has 6 out of 3^3 equal options and not just 3.
Neglecting events with P=S (nothing happens), the expected payout is 1, and the winning probability is 0.5 (and 0.5 losing)
Using this numbers, (p(b+1)-1)/b = 0. The best strategy to maximize the expected logarithm of your money is to do nothing (or risk as few as possible).
However, the Kelly strategy is not optimal: You want to beat a lot of opponents, so you need to take more risk in order to have a chance to perform much better than the average.
Imagine you have 1 million other players and play 20 rounds: Chances are good that one player will win all 20 rounds
(as 2^20 ~ 10^6 and draws are very unlikely). In order to have a chance to win this, you would have to bed all your points in nearly all rounds (maybe with the exception of the last two, depending on the results of other players). But with just two players, that is obviously a bad idea. The other one could play carefully and you would lose with very high probability.
Considering #S>#P: The probability of #R=r rocks with n other players is (n choose r)
The following probability of #S=#P for fixed even n-r is (n-r choose (n-r)/2), if not #S=#P both #S>#P and #P>#S have the same probability (1-(n-r choose (n-r)/2))/2.
And for even n-r the probability of #S>#P is just 1/2
Introduce a sum over r, multiply (n choose r) with the appropriate probability from above, and you get the chance of #S>#P. However, I don't see how you want to use it.