What would be a dominant strategy in the Traveler's dilemma, if there is a $2 reward for having a lower number, but no $2 penalty for having a higher number? Would the final Nash equilibrium be (0,0)?
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I am no expert so please someone correct me if I'm wrong (and assuming you find out the right answer, let me know). Here's my take:
Let's assume that there's a cap at $100 that one can guess. In the original game (where there is +2 for lower number and -2 for higher number) we start out 100,100. Realizing 1 can get more money if he undercuts 2, he offers 99,100 making the payoffs 101,98. Realizing 2 increase his payoff if he undercuts 1, he offers 99,98 making the payoffs 97,100. This goes on and on until both players offer the minimum. What makes this happen is the constant incentive to undercut the other player, driving the reward down to the minimum.
BUT, let's take a look at the modified game with no -2. Both start at 100,100. Player 1 undercuts 99,100 making payoffs 101,100. Should 2 undercut 1? If he does, that makes it 99,98 with payoffs 99,100. That's the same payoff that he had before. If you keep going, you'll see this continue with the pattern of each player having -2 each payoff, yet the effect is staggered (meaning it will be 99,100...99,98...97,98...97,96...) This is why I'd say the nash equilibrium is 99,100 payoff 101,100, because 2 has no incentive to deviate because he'd get the exact same payoff (unless he gets utility from 1 having less money).
I hope this makes sense and please let me know if I made any errors/what the right answer is.
I should've added they make their moves simultaneous, and the other person (who didn't undercut) receives the undercut price.
So, player 1 undercuts player 2, so payoff is 101,99 because player 1 chooses 99, and 99100, so he gets +2, so player 1 has 101, while player 2 stuck with 99 (the low number chosen by player 1). Should 2 now undercut 1? Well, yes. He can improve by undercutting at 98 such that he gets 98 + 2 = 100 and player 1 gets 98, so we have 98,100.
I don't actually know what the answer but would that logic follow through to the point 0,0?
Oh okay. I guess the fact the other player receives the undercut price is a pretty important factor (and potentially much more than a $2 penalty) that I was unfamiliar with. In that case, your answer sounds correct.
bump for any econ majors out there
I don't believe that the incentive structure that initiates the backwards-induction down to (0,0) has been altered by removing the penalty. The function that drives down each bid is still intact -- Player A's payout is optimized by bidding 1 below Player B. The inverse is true for Player B, and both are aware of each other's strategy. This is true at an opening gambit of 100, or all the way down to 1.
A Nash equilibrium is only satisfied if no marginal gain is available by a change in strategy. Since both players marginal gain is always improved by bidding X - 1, where X is the other player, then the recursive nature of this relationship inevitably drives it to (0,0). Both players are assumed to be mechanistic actors so there is no escape from it.
That's considered the paradox of rationality which some believe undermine the merit of game theory. Truly mechanistic game-theoretical reasoning can lead to absurd results that are sub-optimal to those achieved by pure intuition.
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