4 ms·

Ok, I guess I need to understand how this is an equilibrium and read more carefully what the rules of this game are exactly. Because from reading it once I unde

by kmonad 11y ago

Ok, I guess I need to understand how this is an equilibrium and read more carefully what the rules of this game are exactly. Because from reading it once I understood that pirates are supposed to decide on a proposed distribution. And according to my understanding, a pirate would never vote YES if offered 0. Therefore, I seem to not see how the situation we are discussing is a viable equilibrium. (I do see how once in this "state" unilateral change of votes means no advantage to any one player, but is this really interesting for this game, i.e. is it a viable outcome of it despite being stable once reached? Maybe I have it all wrong though.)

lmm 11y ago
> I do see how once in this "state" unilateral change of votes means no advantage to any one player Right, so it meets the definition of an equilibrium. > but is this really interesting for this game, i.e. is it a viable outcome of it despite being stable once reached? In "reality" no. But the Nash Equilibrium formalism gives us no way of determining which of the two possible equilibria (this one and the usual solution) will happen in practice. So you need some more powerful idea to solve that (see references to "superrationality" and "trembling hand" elsewhere in this thread), and none of those theories is mathematically complete.
- kmonad 11y ago
  Yep okay. So is what this then shows, really, that a Nash equilibrium is not so meaningful for this game?
  - lmm 11y ago
    It's meaningful, but it's insufficient to completely solve it.
the_af 11y ago
> And according to my understanding, a pirate would never vote YES if offered 0. A pirate will vote YES if offered 0 when the alternative is getting killed in one of the following turns. Pirates always prefer to live first, and only then they prefer to maximize their money.
- kmonad 11y ago
  what "following turns"? edit: to clarify, the only pirate that would have to consider being killed is the one making the offer. thus yes, offering yourself 0 is possible and in fact seems to be a requirement for the (one of the) stable solutions. But if I'm not most senior, I would never accept 0.
  - the_af 11y ago
    The game has turns in theory: each split proposal is one turn. The most senior pirate proposes a split; if it's not accepted, he gets killed and the next pirate proposes a split. That's the next "turn", assuming the first split gets rejected. Theoretically, a pirate could accept a seemingly inconvenient proposal if he evaluates that rejecting it would result in himself getting killed further down the line, when it's his own turn to propose a split. You're right that this never happens in the 5-pirate game as stated (I'm not sure about the general N-pirate, C-coins game). The pirates who are offered 0 coins reject the split, but they don't get to alter the result anyway.

by kmonad 11y ago

lmm 11y ago

> I do see how once in this "state" unilateral change of votes means no advantage to any one player Right, so it meets the definition of an equilibrium. > but is this really interesting for this game, i.e. is it a viable outcome of it despite being stable once reached? In "reality" no. But the Nash Equilibrium formalism gives us no way of determining which of the two possible equilibria (this one and the usual solution) will happen in practice. So you need some more powerful idea to solve that (see references to "superrationality" and "trembling hand" elsewhere in this thread), and none of those theories is mathematically complete.

kmonad 11y ago
Yep okay. So is what this then shows, really, that a Nash equilibrium is not so meaningful for this game?
- lmm 11y ago
  It's meaningful, but it's insufficient to completely solve it.

the_af 11y ago

> And according to my understanding, a pirate would never vote YES if offered 0. A pirate will vote YES if offered 0 when the alternative is getting killed in one of the following turns. Pirates always prefer to live first, and only then they prefer to maximize their money.

kmonad 11y ago
what "following turns"? edit: to clarify, the only pirate that would have to consider being killed is the one making the offer. thus yes, offering yourself 0 is possible and in fact seems to be a requirement for the (one of the) stable solutions. But if I'm not most senior, I would never accept 0.
- the_af 11y ago
  The game has turns in theory: each split proposal is one turn. The most senior pirate proposes a split; if it's not accepted, he gets killed and the next pirate proposes a split. That's the next "turn", assuming the first split gets rejected. Theoretically, a pirate could accept a seemingly inconvenient proposal if he evaluates that rejecting it would result in himself getting killed further down the line, when it's his own turn to propose a split. You're right that this never happens in the 5-pirate game as stated (I'm not sure about the general N-pirate, C-coins game). The pirates who are offered 0 coins reject the split, but they don't get to alter the result anyway.