Some great questions, will answer them tonight. About to get on a plane back to Portland for a few days. For now, I'll tackle this one (note to Hundrye: I'm responding to this in a way that it will be comprehensible to as many people as possible, apologies for excessive length).
Quote:
Originally Posted by Hundrye
pool:
player 1 has 3 points, player 2 2 points, player 3 1 point.
last match, team A v team B picking right gives 1000 points. team A has 60%, team B 40%, what would you say nash equilibrium is? if you actually know how to calculate this, how does one calculate it? (if the answer comes down to x picks x 100% of time obv disregard last part).
To start, let's consider the same question except without Player 3. Just one on one, 1vs2, trying to decide who to pick. Throughout this post, I'll use (x, y) to mean "The player picks team A x% of the time, and team B y% of the time".
Player 1: (60, 40)
Player 2: (40, 60)
In this case, this is a NASH equilibrium. Neither player can profitably deviate. You can solve this by saying, "what strategy of player 1 would make player 2 indifferent between picking A and B?". Then do the reverse. This is what you come up with. If Player 1 picks A any more often, Player 2 is correct just to pick B 100% of the time. If Player 1 picks B any more often, Player 2 is correct to just pick A 100% of the time. As it is, it's an equilibrium, with Player 2 playing this distribution to make sure that Player 1 can't profitably deviate, either.
Now, let's add your Player 3, and see how he affects things. The first thing to notice is that Player 2 does not give a flying **** about Player 3. There's no scenario in which Player 3 could possibly take a victory away from Player 2. So Player 2's strategy is going to be still completely dependent on what Player 1 does. If Player 1 goes (60, 40), it doesn't matter what Player 2 does (but (40, 60) makes it an equilibrium). If Player 1 plays any higher percentage of A, Player 2 will play (0, 100), any higher percentage of B, it'll be (100, 0). You get the drill by now.
Player 1, however, isn't so lucky. His chances of winning are dependent on not matching the other two, so he absolutely cares what Player 3 does. He can play (60, 40), which would lead to the following:
Player 1: (60, 40)
Player 2: (40, 60)
Player 3: (100, 0)
This corresponds to an 61.6% winrate: 36% (win by picking A and A winning) + 9.6% (win by picking A and A losing, but everyone else picking A) + 16% (win by picking B and B winning) = 61.6%.
Nash Equilibrium? Well...I don't think so. The problem is that Player 1 can “cheat” and profitably deviate. Let's look again at the expectation from playing A and playing B against Player 2's and Player 3's current strategies:
Expectation of Player 1 playing A = (Chance of A winning) + (Chance of A losing but everybody picking A) = 60% + 16%.
So we win 76% of the time by picking A.
Expectation of Player 1 playing B = (Chance of B winning) + (Chance of B winning put everybody picking B) = 40% + 0%
So we win 40% of the time by picking B.
Wow!!! We win so much more by picking A! So this isn't an equilibrium, right? If we can profitably deviate by playing A more (100% A leads to a winrate of that whopping 76% against these two strategies), well then it can't be.
The thing is, though, if Player 1 plays A any more, Player 2 will instantly switch to (0, 100). Here's what we'll be looking at then:
Player 1: (100, 0)
Player 2: (0, 100)
Player 3: Is **** out of luck, but let's say (100, 0).
Player 1 wins 60% of the time. Even though that's a lower percentage, this actually IS a Nash Equilibrium – no individual player can profitably deviate. Player 2 wins 40% of the time (great for him), Player 3 wins 0% of the time but there's nothing he can do about it, and given Player 2 and Player 3's strategies, Player 1 is making his best move, as well.
The cool thing about this problem (and many other 3+ player games), though, is that there isn't REALLY nothing Player 3 can do. While it's true that no strategy adjustment can do better against 1 and 2's current strategies, let's look at what happens if these strategies are played:
Player 1: ?
Player 2: (0, 100)
Player 3: (0, 100)
Now Player 1 should just pick B and win every time! Because of this, Player 3 can induce Player 1 to change his strategy, a change which will give Player 3 a chance of winning again.
Anyway, you're getting the idea that actual gameplay is a lot more nuanced than NASH equilibrium would make it seem. The fact that Player 2 doesn't care what Player 3 does, but Player 1 cares a ton, and all 3 are independent agents, makes it so "pure" equilibria (you can tell I've only taken one class when I start to make terms up) aren't really present this time.
The best static strategy Player 1 can do (trusting Player 2 and Player 3 will play the static strategy that is optimal for them) is (60, 40). Player 2 will play (40, 60), Player 3 will play (100, 0), and Player 1 will win 61.6% of the time. It's not a NASH equilibrium, though, because Player 1 can cheat and do better (by playing more A). It's just that the other players will adjust against the cheating in a way that leads to an equilibrium that is worse than 61.6% for A.
Pretty cool problem to think about game theory wise. Everybody obviously feel free to correct errors I may have made/suggest alternate solutions, I've been on the road all weekend so have only been thinking about this during taxi rides, state of the relationship conversations, etc.
Last edited by mersenneary; 07-18-2011 at 08:06 PM.