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Pirates gold coins12/31/2023 P2 splits of 3? P4 will get 1 P4 will makes sure P3 gets one, P3 follow its rule 1 by splitting of 1 from the group of 3 -> division 5111 P2 splits of 2? P4 will get 1 if they split of 1 or 3, since P4 cannot get 2 and thus splits the group of 2, so P4 will split of 2 -> division 2222 (best option for P2) P2 splits of 1? P4 will get 1 P4 will makes sure P3 gets one, P3 follow its rule 1 by splitting of 1 -> division 5111 Now P39 has the same power relative to P38 and it is best to maximize P40s (and their own) winnings by telling P38 the same, then P38 will tell this to 37 etc. P40 Has the last action, thus a lot of influence over P39.ġ/ It will tell P39 that they will get the same amount, unless the largest stack is odd and splitting that in half gives P40 more.Ģ/ It will also say that it will do exactly what P39 wants, as long as this does not violate 1/ The first pirate can only guarantee 50 coins. All P1 can ensure is one of those 123-coin piles (or better) once 1967 coins are still in a single pile on P21's turn. P1 can't be sure they will do this, though certainly encouraging P2 through P20 to guarantee single-coins for 22 through 40 is likely to make them feel like upsetting others. If the pirates in the latter half are surly and grumpy, knowing they will only get one coin, they may split their piles less evenly, creating single-coin or piles. 6 of these get split let's say the 124 does, for 2 62's and 5 61/62 pairs, leaving 10 123s. P22 (who is getting 1 coin no matter what) will at best split the first into 492 and 492, and P23 the other into 492 and 493. It's in P21's interest to split this pile into 984 and 985. P21 will not want to pull off a single coin - the 19 pirates picking piles after P21 are getting singles, but P21 is going to get the smallest thing left after dividing up the 1967 coins still in the big pile. If anyone after you splits what's left, you'll be right in line after me for the second biggest."Īssuming this line of logic holds up, eventually we will be at P21 and there will be 20 single coins on the table. You're best served by just pulling off 1. "Listen, P2, I know you want to split that big 'un so we can both get half, but the pirates after you will just keep splitting it. The first pirate, to ensure the most coins possible, needs to explain this to each choosing pirate in turn. Everyone else would like more piles of 1, since this makes the piles they are choosing from more likely to have more than 50 coins. The last pirate would like 40 piles of either 49 or 50, so that they will not be frozen out. So the first pirate definitely splits into and 1986 and 1. The first pirate would love to end up with one pile of 1947 coins and 40 piles of 1 coin. Everyone is aware that there is no alliance formation anywhere, and that all other pirates are intelligent enough. $\left\lceil\frac$ won't bother about how the earlier pirates have performed their division, but rather focus on maximizing her guaranteed wealth from the current coin-set she has. If there is an alliance formation (there can be several such cases), then in the worst case the rest of the pirates form an alliance and go against $P_1$.
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