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Old 10-31-2006, 01:12 AM
PairTheBoard PairTheBoard is offline
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Default Re: Ro-Sham-Bo-Fu Game Theory

[ QUOTE ]
No, I'm always using 2 bonus points.

My question is what happens after the six rounds. Do you just win (+1) or lose (-1)? Or does the number of points you scored matter? For example, suppose you could make a deal with the other player to always pick the superior non-tainted object, so you each got the bonus points each round. You would each get +12, which is a good result. Does that make sense, or would it just result in both of you tying and getting 0 outcome for the game.

[/ QUOTE ]

Essentially it's +1 or -1. Actually you are betting a predetermined amount with your opponent. The one with the most points after 6 rounds wins the bet. There are always 3 points put up for grabs each round, along with the 2 free bonus points for picking a NonTainted Object. If the round ties, the 3 points carries over to the next round and you play for 6. If tied the next round is for 9. So there are extra considerations later in the match depending on how much you are ahead or behind.

For example, if you are tied going into the last round it really doesn't matter that there are bonus points. Whoever wins the throw wins. Another example would be if you are 4 points behind with two rounds to go. With Rock Tainted and knowing your opponent will pick scissors you are better off picking scissors than rock, thereby letting the 3 points ride to the last round. That way your opponent can't hide behind scissors in the last round.

So you are really trying to maximize the difference between the points you score and the points your opponent scores. If Rock is tainted, picking Paper vs your opponent's Rock pick gives you a 5 point lead. That's the best result. So, what's the optimum first round strategy?

To avoid confusion I suggest we consistently make Rock the Tainted Object and refer to frequencies for Rock, Paper, and Scissors under this assumption.

PairTheBoard
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