Two Six-Packs of Beer on a Game Theory Wager

[Orpheus]

I think you unconsciously included a cheating proposition in your mental depiction of the scenario. You say Player 2 (with perfect knowledge and execution) cannot be beaten by Player 1. I say: that may be true if Player 1 is a mere mortal, but a Player 1 with perfect knowledge and execution would have a pretty nice advantage over a 'mortal' Player 2, too.

I can think of *many* counterexamples to your premise if BOTH players have perfect knowledge and play perfectly.

How about a 'perfect duel'? The first person to shoot wins. [literally a perfect -ahem- 'execution']. You can be Player 2; tell me how perfect knowledge gets you a tie. The perfect strategy would consist of NOT letting yourself be Player 2!

It's quite common for the party that acts first to have a distinct advantage in games, in business, etc.

Also, if both parties have perfect knowledge and execution, many common games can go either way.

Take, for example: Texas Hold'Em. If both have perfect knowledge and execution, Player 2 will still lose if s/he doesn't have the hands. In fact, because there will be no bluffing or uncertainty, every hand will be folded, and the outcome will be a simple addition of how many winning hands each player will be dealt on the BB (net: 1 SB from SB) vs. SB (net: 2 Sb from BB) vs any other position (net: 3 SB from SB and BB).

"Perfect knowledge" is NOT inconsistent with randomness. Randomness means that the exact distribution of (cards, dice, cute willing babes) is beyond your control. You can still have perfect knowledge of the random outcome is still possible in two distinct forms: a) each player may know all that can be known (i.e. both know the hands dealt, but not undealt hands); or b) both have perfect knowledge of the entire sequence of hands -- dealt and undealt. Either way, "Perfect Knowledge plus Perfect Execution" do not equal "perfect control over the outcome" in TX HE.

There's another issue: If player 2 cannot lose, than either the game is biased against player 1 (who can lose) or it is moot (because Player 1 ALSO cannot lose) We cannot judge if this is true unless WE have perfect knowledge of the game. Mathematicians have been debating for centuries whether chess, for example, might be solvable -- possesing a perfect strategy for black or white. WE do not have perfect knowledge or play, so we cannot decide.

Or imagine, for example, a game where players take turns playing pieces on a N-grid (for the sake of simplicity, imagine they place their token on the intersections, as in Go or Pente, rather than the squares defined by the intersections, as in Chess). After placing each token, the player gets a score that is equal to the the sum of the distances to every remaining empty intersection (i.e. the final score will be the number of remaining empty squares x the average distance)

Our perfect (algebraic) knowledge of this game tells us that, if N is odd, the first player always wins, because placing his first token in the center spot represents an insuperable advantage. Player 2 must make an inferior 2nd move, and every later spot s/he chooses will already have been passed over as "known inferior, or at best equal) by Player 1 [Also, Player 1 will have one more empty square on each turn to gain points with: each of his/her tokens fills a square, preventing player 2 from getting points for its distance)

You could say that this is a biased game, because Player 1 has a guaranteed winning strategy, but in YOUR example, you presumed that Player 2 had a never-lose strategy. Isn't THAT either biased (if he always wins) or moot (if he always ties)? However, to players with bad math skills (inperfect knowledge), many such games seem reasonable. We can't prove there is no universal winning strategy for black ore white, so we continue to play. Someday a computer may "solve the game" and it'll be as meaningless as tic-tac-toe to anyone who can memorize the solution.

Therefore: there exist zero sum games with an insurmountable first player advantage, if both players are otherwise equal (e.g. a duel or the token game above); and games where the outcome can be perfectly known and played, but either player may win (e.g with a random element, as in TXHE). The chess example disproves your believe that your postulate is a known premise of game theory: mathematicians have argued if there is a "solution" to chess for centuries, yet if you postulate were a known truth, any math grad student could prove "if there is a universal winning strategy to chess, it must mean a victory for Black (Player 2)" However, the general belief is tbat it would more likely be a victory for White (Player 1) because chess is a game where the first move is a significant advantage (like the duel or disk game)

You owe someone some beer.

[laja]

he owes me some beer ;]

luckily I had complete information before I made the bet and consulted my game theory teacher^^;;