Timeline for Five Front Battle
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Feb 23, 2010 at 23:40 | comment | added | David E Speyer | Jason, you are wrong. As zeb says in his original post, if you play a strategy which projects to Uniform(0,2/5), then you will lose against my strategy of playing a random permutation of (0,0,3/10,3/10,2/5). For simplicity, I'll analyse the case that your strategy projects to the uniform measure on [0,2/5]^2 for every pair of fronts. Then the probabilities that you win on 2, 3 and 4 fronts are 9/16, 6/16 and 1/16 respectively. Your expected margin of victory is -(9/16)+(6/16)+3*(1/16)=0, so you break even by Colonel Blotto rules. But you lose 9/16ths of the time by the rules of this game. | |
| Feb 23, 2010 at 22:16 | history | edited | Jason Bandlow | CC BY-SA 2.5 |
confirmed that referenced paper contained a solution
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| Feb 16, 2010 at 3:23 | comment | added | zeb | This is a different game - in the Colonel Blotto game, you are trying to maximize the number of fronts won, while in this game you only care about whether you win more fronts than the other guy. The Colonel Blotto game is much easier than this one... you don't have to worry about correlations between the fronts. | |
| Feb 16, 2010 at 1:58 | history | answered | Jason Bandlow | CC BY-SA 2.5 |