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This requires a very unnatural set-up (a game that has an average payoff that depends on your capital modulo some number).

• tel 16y ago
  With this understanding, the paradox resolves itself: The individual games are losing only under a distribution that differs from that which is actually encountered when playing the compound game. You don't need a modulo-style set up. You just need a dependence of the two games such that a player can predict when to play each game to always be optimal. The ratchet is just one example where the player always avoids the really bad odds in game B that help to make it a losing game. It's exactly the same thing as Simpson's Paradox.
  • JoachimSchipper 16y ago
    Sure, the particular distributions used are a detail; but it still needs a very odd setup, right?
    • tel 16y ago
      Not really. I'd say the necessary elements are: + A choice between two "games", each losing on average. + At least one game has periods of local payoff great enough to overwhelm the long-run losses in the other game + Some flow of information between the games so that playing one game will help you predict the payoff periods of the other game. The example that flies to mind is investment. Game A is to lose value of money you hold on to via inflation. Game B is the generally losing game of day trading. The net game can be profitable as long as you spend time in A learning to accurately predict game B's upswings. Of course, the real information flow from game B to game A is already heavily capitalized making the net game even more difficult.