Comments by "" (@PunmasterSTP) on "Game Theory 101 (#34): Rock Paper Scissors" video.
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@chrisnewtownnsw Ah, I think I see what you're saying.
I still think that the <1/3, 0, 2/3> strategy (for <rock, paper, scissors>) still breaks even. Let's break it down by cases. Every 1 out of 3 times when you play rock, there is a 1/3 chance your opponent plays rock and it's a draw, 1/3 chance your opponent plays paper and you lose, and a 1/3 chance your opponent plays scissors and you win. So you break even there. Now, let's think about when you play scissors. No matter how frequently you play it (2/3 of the time, 1/3 of the time, or all of the time, etc.), there's always a 1/3 chance your opponent will play scissors and it will be a draw. There's also a 1/3 chance your opponent will play rock and you'll lose, and a 1/3 chance your opponent will play paper and you'll win. Again, you'd expect to break even, even though you're playing scissors twice as often as rock.
Because <1/3, 1/3, 1/3>, <1/3, 1/3, 1/3> is a Nash equilibrium, you can't make yourself better off by deviating from <1/3, 1/3, 1/3>. And in this case, you'd break even, no matter what you did.
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@chrisnewtownnsw I think I see what you are saying. If it's alright, let's go back to <1/3, 1/3, 1/3> for your opponent's strategy and then <1/3, 0, 2/3> for yours (i.e. 1/3 rock and 2/3 scissors).
We can calculate the probabilities of each each different game occurring. Using the law of multiplication, we'd expect the game <rock, rock> (your opponent plays rock and you play rock) to happen 1/3 * 1/3 = 1/9 of the time. That's a draw, so it doesn't count. The game <rock, paper> occurs none of the time, since you never play paper. The game <rock, scissors> happens 1/3 * 2/3 = 2/9 of the time. We can start adding up your expected returns, and from that round it would be 2/9 * (-1) = -2/9, since you are losing $1 (or whatever units) 2/9 of the time. Does that sound alright?
Next, let's think about your opponent playing paper. <paper, rock> happens 1/3 * 1/3 = 1/9 of the time, and you lose, so you pick up a 1/9 * (-1) = -1/9 from there. <paper, paper> doesn't happen, but <paper, scissors> happens 1/3 * 2/3 = 2/9 of the time. From that, you'd get 2/9 * (1) = +2/9, since you are winning.
Lastly, <scissors, rock> happens 1/3 * 1/3 of the time, so you win 1/9 * (1) = 1/9. <scissors, paper> never happens and <scissors, scissors> happens 1/3 * 2/3 = 2/9 of the time, but it's a draw.
Totaling everything up, your expected earnings are (-2/9) + (-1/9) + (2/9) + 1/9 = 0. As before, both players playing <1/3, 1/3, 1/3> represents a Nash equilibrium, so no one person can become better off by changing their strategy.
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@chrisnewtownnsw I also see what you're saying, and I definitely respect it and I'm enjoying this conversation!
I don't think you can necessarily "read it from left to right". As in, you can't predict what your opponent will play next, even if you knew overall that they're trying to play each strategy exactly 1/3 of the time. Each new round is essentially a toss-up, and no matter what you do, you could never expect more than to break even.
Say that you choose rock for the next round. There's a 1/3 chance your opponent will choose scissors and you'll win, but there's also the same chance they'll choose paper and you''' loose. The other 1/3 of the time you'll break even. The same logic holds if you play paper or scissors as well.
If your opponent did not play randomly*, but rather cycled between rock, paper and scissors every three turns, then you could *absolutely win every time (by playing paper, scissors and rock, respectively).
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