Simpler, quick game theory question

[Magic_Man]

Quick question related to the RPS game proposed in an earlier thread. Consider this game:

Win with rock: 3 points
Lose with rock: 0 points
Tie with rock: 0 points

Win with scissors: 5 points
Lose with scissors: 2 points
Tie with scissors: 2 points

Win with paper: 5 points
Lose with paper: 2 points
Tie with paper: 2 points


So here's my question. My probabilities are R,P,S. If I'm calculating my opponent's EV for playing rock, do I use this?:

EV = 0*R + (3-2)*P + (3-2)*S = P + S

or do I use this?:

EV = 0*R + 3*P + 3*S = 3P + 3S

One of these is taking into account point differential, and one of them is considering only points earned. I assume that in different situations, I might use one or the other. Can someone enlighten me?

~MagicMan

[Utah]

[ QUOTE ]
So here's my question. My probabilities are R,P,S. If I'm calculating my opponent's EV for playing rock, do I use this?

[/ QUOTE ]Why would you calculate your opponent's EV for playing rock as you are indifferent? Also, why would he calculate it either as he needs to figure out his best overall strategy, which is of course equal to yours given the game.

Since this is a zero sum game with a winner all you care about is the differential. So, for the purposes of your question it would be 0*R + S*1 + P*-5

[Magic_Man]

[ QUOTE ]
[ QUOTE ]
So here's my question. My probabilities are R,P,S. If I'm calculating my opponent's EV for playing rock, do I use this?

[/ QUOTE ]Why would you calculate your opponent's EV for playing rock as you are indifferent? Also, why would he calculate it either as he needs to figure out his best overall strategy, which is of course equal to yours given the game.

Since this is a zero sum game with a winner all you care about is the differential. So, for the purposes of your question it would be 0*R + S*1 + P*-5

[/ QUOTE ]

Right, it's too late to edit now but it should be P*-5, not P*1 as I wrote. I assumed that the way to calculate the best strategy was to find their EV (in differential) for playing R,P, or S given that I am playing a strategy with probabilities a,b,c, and then set all their EVs to be equal, so that it didn't matter what they did. This was the solution given in a previous thread. So, in this case, if my probabilities are R,P,S, my opponents EV's are:

R: 0*R - 5*P + S
P: 5*R + 0*P - 3*S
S:-1*R + 3*P + 0*S

1) S - 5P = 5R - 3S
2) 5R -3S = 3P - R
3) R+P+S = 1

combining 2&3:
5R-3(1-R-P) = 3P - R
5R - 3 + 3R + 3P = 3P - R
9R = 3

4)R = 1/3

combining 1,3,4:
(1-R-P) - 5P = 5R - 3(1-R-P)
2/3 - P - 5P = 5/3 - 3 + 1 + 3P
9P = 1

5) P = 1/9

combining 3,4,5:
S = 1-1/3 - 1/9

6) S = 5/9

So it seems that my best strategy is:
R = 1/3
P = 1/9
S = 5/9

Using that method, no matter what my opponent does, the point differential will be 0. Did I do this correctly, or is my method completely wrong?

~MagicMan

[Utah]

[ QUOTE ]
Using that method, no matter what my opponent does, the point differential will be 0. Did I do this correctly, or is my method completely wrong?

[/ QUOTE ]I did the calculations are your answer is correct. You would expect it to be zero given neither player has an edge and given that both players have exactly the same payouts and strategy options.

[oneeye13]

the avalanche always wins... there is no strategy to counter an infinite series of rocks

[FortunaMaximus]

[ QUOTE ]
the avalanche always wins... there is no strategy to counter an infinite series of rocks

[/ QUOTE ]

Ah, but you're incorrect there. There is the infinite firehose to counter that.