Timeline for Games that never begin
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13, 2012 at 21:02 | comment | added | Per Alexandersson | What a strange game, the only way to win is not to play. | |
| Nov 13, 2012 at 13:45 | comment | added | Joel David Hamkins | Incidently, I want now to disagree with the terminology you've adopted four comments above. I think the terminology should be that one says that Alice or Bob wins a particular play, depending on whether that play is in the payoff set or not. A strategy is a winning strategy for a particular player, if all plays of the game that conform with that strategy is in the corresponding payoff set. On this terminology, both players can have winning strategies, but only when there is no play that conforms with both of them. | |
| Nov 13, 2012 at 13:41 | comment | added | Joel David Hamkins | David, I posted an update to my answer, which has a theorem (from AC) that there is a non-determined game with respect to rational engaged strategies. | |
| Nov 12, 2012 at 21:36 | comment | added | David Feldman | So I still can't yet rule out that all such games are determined, right? | |
| Nov 12, 2012 at 21:11 | comment | added | Joel David Hamkins | Yes, that's right. So Alice alone wins the game, even though it seems that Bob's always-play-$3$ strategy is a good one. | |
| Nov 12, 2012 at 20:38 | comment | added | David Feldman | When two strategies have no common run, Alice wins, since the empty set is a subset of the payoff. So my clarified definition rules out having winning strategies for both, right? | |
| Nov 12, 2012 at 19:14 | answer | added | Noah Schweber | timeline score: 4 | |
| Nov 12, 2012 at 19:03 | comment | added | Joel David Hamkins | Come to think of it, for this game Bob's always-play-$3$ strategy is also both rational and engaged, since it does produce a play for many strategies of Alice, and all of these are wins for Bob. So once again, both Alice and Bob have rational, engaged winning strategies. But these two strategies have no common play. | |
| Nov 12, 2012 at 18:40 | comment | added | Joel David Hamkins | Regarding your fixes, consider the game with payoff set for Alice consisting of all sequences with only finitely many nonzero numbers. Bob would seem to have a winning strategy with the "always play $3$" strategy, among others, but this is actually not correct, since we may let Alice play the strategy which plays $0$, as long as the history was almost all $0$, and otherwise adds $1$ to her previous move. This strategy is strongly rational and also engaged, since it has a play when Bob plays almost all $0$s. But the only plays that accord with it are almost all $0$ and hence wins for Alice. | |
| Nov 12, 2012 at 15:49 | history | edited | David Feldman | CC BY-SA 3.0 |
added 1441 characters in body
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| Nov 12, 2012 at 11:53 | answer | added | Tony Huynh | timeline score: 8 | |
| Nov 12, 2012 at 11:11 | answer | added | Joel David Hamkins | timeline score: 8 | |
| Nov 12, 2012 at 7:04 | history | asked | David Feldman | CC BY-SA 3.0 |