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Quick upper bound of

8/9

Because:

If we imagine all the 3^8 hat configurations, and look at all the 8*3^8 rock-paper-scissors matches, then no matter what, 1/3 of the matches will be wins, 1/3 will be ties, and 1/3 will be losses. (Consider 1 player. For each configuration of the other 7 players' hats, they will contribute one of each result as their own hat varies. Thus it follows.) We can then relax the problem by imagining we could assign the r-p-s results to the matches arbitrarily. To maximize the number of net-winning configurations, we are basically gerrymandering. In each net losing configuration, we may as well make everyone lose. It turns out with 3^6 configurations using 8 losses each, the remaining configurations could all win by 1 e.g. with 3 wins, 3 ties, and 2 losses each. (This is not the only way, but you have to win by 1 in each winning configuration to achieve this 8/9.) If you had fewer losing configurations, it doesn't work out as there are no longer enough wins to cover all the losses in all the remaining configurations. So, this is an upper bound. I don't know yet if it is achievable.

Best I have achieved so far (lower bound, probably not optimal) is:

5250/6561 or just over 80%

This is done by:

A disorganized strategy (I wouldn't know how to characterize it except by the method of finding it or else the listing of what guesses should be made by each player in each scenario which is not particularly human-readable) found by a greedy search -- we can assign r-p-s choices to a particular player in a particular configuration sort of like what is discussed above, except whenever we assign a choice, we have to also assign the same choice to the two other configurations that look the same to the player we are assigning a choice for. Then to do the greedy search, we simply go through each configuration, and if the already-made choices have enough slack to allow a win, we try to win by exactly 1 or else as little as possible. If a win is not possible, we make all unassigned players in the configuration lose in order to lose by as much as possible. The unspecified details such as how exactly to make a winner as well as the order of iteration may matter, so here is my (hopefully bug-free, sorry it wasn't really written to share) code.

Quick upper bound of

8/9

Because:

If we imagine all the 3^8 hat configurations, and look at all the 8*3^8 rock-paper-scissors matches, then no matter what, 1/3 of the matches will be wins, 1/3 will be ties, and 1/3 will be losses. (Consider 1 player. For each configuration of the other 7 players' hats, they will contribute one of each result as their own hat varies. Thus it follows.) We can then relax the problem by imagining we could assign the r-p-s results to the matches arbitrarily. To maximize the number of net-winning configurations, we are basically gerrymandering. In each net losing configuration, we may as well make everyone lose. It turns out with 3^6 configurations using 8 losses each, the remaining configurations could all win by 1 e.g. with 3 wins, 3 ties, and 2 losses each. (This is not the only way, but you have to win by 1 in each winning configuration to achieve this 8/9.) If you had fewer losing configurations, it doesn't work out as there are no longer enough wins to cover all the losses in all the remaining configurations. So, this is an upper bound. I don't know yet if it is achievable.

Best I have achieved so far (lower bound, probably not optimal) is:

5286/6561 or just over 80%

This is done by:

A disorganized strategy (I wouldn't know how to characterize it except by the method of finding it or else the listing of what guesses should be made by each player in each scenario which is not particularly human-readable) found by a greedy search -- we can assign r-p-s choices to a particular player in a particular configuration sort of like what is discussed above, except whenever we assign a choice, we have to also assign the same choice to the two other configurations that look the same to the player we are assigning a choice for. Then to do the greedy search, we simply go through each configuration, and if the already-made choices have enough slack to allow a win, we try to win by exactly 1 or else as little as possible. If a win is not possible, we make all unassigned players in the configuration lose in order to lose by as much as possible. The unspecified details such as how exactly to make a winner as well as the order of iteration may matter, so here is my (hopefully bug-free, sorry it wasn't really written to share) code. Addendum: I was able to make some minor improvements by trying to get wins-by-one with at least 3 wins where possible, and to more aggressively make 8-lose configurations.

Quick upper bound of

8/9

Because:

If we imagine all the 3^8 hat configurations, and look at all the 8*3^8 rock-paper-scissors matches, then no matter what, 1/3 of the matches will be wins, 1/3 will be ties, and 1/3 will be losses. (Consider 1 player. For each configuration of the other 7 players' hats, they will contribute one of each result as their own hat varies. Thus it follows.) We can then relax the problem by imagining we could assign the r-p-s results to the matches arbitrarily. To maximize the number of net-winning configurations, we are basically gerrymandering. In each net losing configuration, we may as well make everyone lose. It turns out with 3^6 configurations using 8 losses each, the remaining configurations could all win by 1 e.g. with 3 wins, 3 ties, and 2 losses each. (This is not the only way, but you have to win by 1 in each winning configuration to achieve this 8/9.) If you had fewer losing configurations, it doesn't work out as there are no longer enough wins to cover all the losses in all the remaining configurations. So, this is an upper bound. I don't know yet if it is achievable.

Best I have achieved so far (lower bound, probably not optimal) is:

5250/6561 or almost 80%

This is done by:

A disorganized strategy (I wouldn't know how to characterize it except by the method of finding it or else the listing of what guesses should be made by each player in each scenario which is not particularly human-readable) found by a greedy search -- we can assign r-p-s choices to a particular player in a particular configuration sort of like what is discussed above, except whenever we assign a choice, we have to also assign the same choice to the two other configurations that look the same to the player we are assigning a choice for. Then to do the greedy search, we simply go through each configuration, and if the already-made choices have enough slack to allow a win, we try to win by exactly 1 or else as little as possible. If a win is not possible, we make all unassigned players in the configuration lose in order to lose by as much as possible. The unspecified details such as how exactly to make a winner as well as the order of iteration may matter, so here is my (hopefully bug-free, sorry it wasn't really written to share) code.

Quick upper bound of

8/9

Because:

If we imagine all the 3^8 hat configurations, and look at all the 8*3^8 rock-paper-scissors matches, then no matter what, 1/3 of the matches will be wins, 1/3 will be ties, and 1/3 will be losses. (Consider 1 player. For each configuration of the other 7 players' hats, they will contribute one of each result as their own hat varies. Thus it follows.) We can then relax the problem by imagining we could assign the r-p-s results to the matches arbitrarily. To maximize the number of net-winning configurations, we are basically gerrymandering. In each net losing configuration, we may as well make everyone lose. It turns out with 3^6 configurations using 8 losses each, the remaining configurations could all win by 1 e.g. with 3 wins, 3 ties, and 2 losses each. (This is not the only way, but you have to win by 1 in each winning configuration to achieve this 8/9.) If you had fewer losing configurations, it doesn't work out as there are no longer enough wins to cover all the losses in all the remaining configurations. So, this is an upper bound. I don't know yet if it is achievable.

Best I have achieved so far (lower bound, probably not optimal) is:

5250/6561 or just over 80%

This is done by:

A disorganized strategy (I wouldn't know how to characterize it except by the method of finding it or else the listing of what guesses should be made by each player in each scenario which is not particularly human-readable) found by a greedy search -- we can assign r-p-s choices to a particular player in a particular configuration sort of like what is discussed above, except whenever we assign a choice, we have to also assign the same choice to the two other configurations that look the same to the player we are assigning a choice for. Then to do the greedy search, we simply go through each configuration, and if the already-made choices have enough slack to allow a win, we try to win by exactly 1 or else as little as possible. If a win is not possible, we make all unassigned players in the configuration lose in order to lose by as much as possible. The unspecified details such as how exactly to make a winner as well as the order of iteration may matter, so here is my (hopefully bug-free, sorry it wasn't really written to share) code.

Quick upper bound of

8/9

Because:

If we imagine all the 3^8 hat configurations, and look at all the 8*3^8 rock-paper-scissors matches, then no matter what, 1/3 of the matches will be wins, 1/3 will be ties, and 1/3 will be losses. (Consider 1 player. For each configuration of the other 7 players' hats, they will contribute one of each result as their own hat varies. Thus it follows.) We can then relax the problem by imagining we could assign the r-p-s results to the matches arbitrarily. To maximize the number of net-winning configurations, we are basically gerrymandering. In each net losing configuration, we may as well make everyone lose. It turns out with 3^6 configurations using 8 losses each, the remaining configurations could all win by 1 e.g. with 3 wins, 3 ties, and 2 losses each. (This is not the only way, but you have to win by 1 in each winning configuration to achieve this 8/9.) If you had fewer losing configurations, it doesn't work out as there are no longer enough wins to cover all the losses in all the remaining configurations. So, this is an upper bound. I don't know yet if it is achievable.

Quick upper bound of

8/9

Because:

If we imagine all the 3^8 hat configurations, and look at all the 8*3^8 rock-paper-scissors matches, then no matter what, 1/3 of the matches will be wins, 1/3 will be ties, and 1/3 will be losses. (Consider 1 player. For each configuration of the other 7 players' hats, they will contribute one of each result as their own hat varies. Thus it follows.) We can then relax the problem by imagining we could assign the r-p-s results to the matches arbitrarily. To maximize the number of net-winning configurations, we are basically gerrymandering. In each net losing configuration, we may as well make everyone lose. It turns out with 3^6 configurations using 8 losses each, the remaining configurations could all win by 1 e.g. with 3 wins, 3 ties, and 2 losses each. (This is not the only way, but you have to win by 1 in each winning configuration to achieve this 8/9.) If you had fewer losing configurations, it doesn't work out as there are no longer enough wins to cover all the losses in all the remaining configurations. So, this is an upper bound. I don't know yet if it is achievable.

Best I have achieved so far (lower bound, probably not optimal) is:

5250/6561 or almost 80%

This is done by:

A disorganized strategy (I wouldn't know how to characterize it except by the method of finding it or else the listing of what guesses should be made by each player in each scenario which is not particularly human-readable) found by a greedy search -- we can assign r-p-s choices to a particular player in a particular configuration sort of like what is discussed above, except whenever we assign a choice, we have to also assign the same choice to the two other configurations that look the same to the player we are assigning a choice for. Then to do the greedy search, we simply go through each configuration, and if the already-made choices have enough slack to allow a win, we try to win by exactly 1 or else as little as possible. If a win is not possible, we make all unassigned players in the configuration lose in order to lose by as much as possible. The unspecified details such as how exactly to make a winner as well as the order of iteration may matter, so here is my (hopefully bug-free, sorry it wasn't really written to share) code.