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It's not clear to me how much this overlaps with FirstName LastName, but I feel that a good strategy for a large number of players is

safety in numbers i.e. assume that one's hat is the same as the most popular hat in the ones that can be seen. (Tie-breaking rules to follow).

This means that

In any situation where there the most popular hat appears at least two times more than the second most popular hat, we win. e.g. if there are 100 contestants and the hats are 36 paper, 34 scissors and 30 stone then everyone assumes that their hat is paper; they shout "scissors" and the score is 36 wins, 30 losses and 34 draws.

Things do not go our way if

there is a tie for the most popular hat. In this case we always fail e.g. if there are 34 paper, 34 scissors and 32 stone the people with paper hats will assume that they have scissors hats; the people with scissors hats will assume that they have paper hats (the people with stone hats do something not yet specified, but we have already racked up 34 losses and 34 draws).

Well also need to consider

What happens when there is exactly one more of the most popular hat than the second most popular. Here we can use a tiebreaker to say that if anyone views a two-way tie for the most popular hat they assume that they have the "winner" hat (i.e. for a paper-scissors tie assume scissors; a scissors-stone tie assume stone; a stone-paper tie assumee stone) and for a three way tie choose paper.

This would mean that

When there are $n$ paper hats and $(n-1)$ scissors hats and fewer than $(n-1)$ stone hats we lose, but when there are $n$ scissors hats, $(n-1)$ paper hats and fewer than $(n-1)$ stone hats we win. (Similarly winning half of other $n,n-1,n>$ cases and one third of $n,n-1,n-1$ cases).

Mathematically, if we have $m$ players the count of losing cases is

$$3\sum_{m/3<n\le m/2}\frac {m!}{(n!)^2(m-2n)!}+3\sum_{m/3< n<m/2}\frac{m!}{(n-1)!n!(m-2n+1)!} $$ out of $3^m$ possible arrangements of hats. (With an extra $2(3k+1)!/(k!)^2(k+1)!$ term when $m=3n+1$). If you prefer binomial coefficients to factorials, the above expression is $$3\sum_{m/3<n\le m/2}\binom{2n}{n}\binom m{2n}+3\sum_{m/3\le n<m/2}\binom{2n-1}n\binom m{2n-1} $$

My intuition is

that the proportion of losing cases will tend to zero as $m\to\infty$.

Numerical investigations will follow.

It's not clear to me how much this overlaps with FirstName LastName, but I feel that a good strategy for a large number of players is

safety in numbers i.e. assume that one's hat is the same as the most popular hat in the ones that can be seen. (Tie-breaking rules to follow).

This means that

In any situation where there the most popular hat appears at least two times more than the second most popular hat, we win. e.g. if there are 100 contestants and the hats are 36 paper, 34 scissors and 30 stone then everyone assumes that their hat is paper; they shout "scissors" and the score is 36 wins, 30 losses and 34 draws.

Things do not go our way if

there is a tie for the most popular hat. In this case we always fail e.g. if there are 34 paper, 34 scissors and 32 stone the people with paper hats will assume that they have scissors hats; the people with scissors hats will assume that they have paper hats (the people with stone hats do something not yet specified, but we have already racked up 34 losses and 34 draws).

Well also need to consider

What happens when there is exactly one more of the most popular hat than the second most popular. Here we can use a tiebreaker to say that if anyone views a two-way tie for the most popular hat they assume that they have the "winner" hat (i.e. for a paper-scissors tie assume scissors; a scissors-stone tie assume stone; a stone-paper tie assumee stone) and for a three way tie choose paper.

This would mean that

When there are $n$ paper hats and $(n-1)$ scissors hats and fewer than $(n-1)$ stone hats we lose, but when there are $n$ scissors hats, $(n-1)$ paper hats and fewer than $(n-1)$ stone hats we win. (Similarly winning half of other $n,n-1,n>$ cases and one third of $n,n-1,n-1$ cases).

Mathematically, if we have $m$ players the count of losing cases is

$$3\sum_{m/3<n\le m/2}\frac {m!}{(n!)^2(m-2n)!}+3\sum_{m/3< n<m/2}\frac{m!}{(n-1)!n!(m-2n+1)!} $$ out of $3^m$ possible arrangements of hats. (With an extra $2(3k+1)!/(k!)^2(k+1)!$ term when $m=3n+1$). If you prefer binomial coefficients to factorials, the above expression is $$3\sum_{m/3<n\le m/2}\binom{2n}{n}\binom m{2n}+3\sum_{m/3\le n<m/2}\binom{2n-1}n\binom m{2n-1} $$

My intuition is

that the proportion of losing cases will tend to zero as $m\to\infty$.

A little sagemath gives

a loss rate of 41.6% when $m=8$, a loss rate of 14.7% when $m=100$, and a loss rate of 4.6% when $m=1000$.

It's not clear to me how much this overlaps with FirstName LastName, but I feel that a good strategy for a large number of players is

safety in numbers i.e. assume that one's hat is the same as the most popular hat in the ones that can be seen. (Tie-breaking rules to follow).

This means that

In any situation where there the most popular hat appears at least two times more than the second most popular hat, we win. e.g. if there are 100 contestants and the hats are 36 paper, 34 scissors and 30 stone then everyone assumes that their hat is paper; they shout "scissors" and the score is 36 wins, 30 losses and 34 draws.

Things do not go our way if

there is a tie for the most popular hat. In this case we always fail e.g. if there are 34 paper, 34 scissors and 32 stone the people with paper hats will assume that they have scissors hats; the people with scissors hats will assume that they have paper hats (the people with stone hats do something not yet specified, but we have already racked up 34 losses and 34 draws).

Well also need to consider

What happens when there is exactly one more of the most popular hat than the second most popular. Here we can use a tiebreaker to say that if anyone views a two-way tie for the most popular hat they assume that they have the "winner" hat (i.e. for a paper-scissors tie assume scissors; a scissors-stone tie assume stone; a stone-paper tie assumee stone) and for a three way tie choose paper.

This would mean that

When there are $n$ paper hats and $(n-1)$ scissors hats and fewer than $(n-1)$ stone hats we lose, but when there are $n$ scissors hats, $(n-1)$ paper hats and fewer than $(n-1)$ stone hats we win. (Similarly winning half of other $n,n-1,n>$ cases and one third of $n,n-1,n-1$ cases).

Mathematically, if we have $m$ players the count of losing cases is

$$3\sum_{m/3<n\le m/2}\frac {m!}{(n!)^2(m-2n)!}+3\sum_{m/3\le n<m/2}\frac{m!}{(n+1)!n!(m-2n-1)!} $$ out of $3^m$ possible arrangements of hats. (With an extra $2(3k+1)!/(k!)^2(k+1)!$ term when $m=3n+1$). If you prefer binomial coefficients to factorials, the above expression is $$3\sum_{m/3<n\le m/2}\binom{2n}{n}\binom m{2n}+3\sum_{m/3\le n<m/2}\binom{2n+1}n\binom m{2n+1} $$

My intuition is

that the proportion of losing cases will tend to zero as $m\to\infty$.

Numerical investigations will follow.

It's not clear to me how much this overlaps with FirstName LastName, but I feel that a good strategy for a large number of players is

safety in numbers i.e. assume that one's hat is the same as the most popular hat in the ones that can be seen. (Tie-breaking rules to follow).

This means that

In any situation where there the most popular hat appears at least two times more than the second most popular hat, we win. e.g. if there are 100 contestants and the hats are 36 paper, 34 scissors and 30 stone then everyone assumes that their hat is paper; they shout "scissors" and the score is 36 wins, 30 losses and 34 draws.

Things do not go our way if

there is a tie for the most popular hat. In this case we always fail e.g. if there are 34 paper, 34 scissors and 32 stone the people with paper hats will assume that they have scissors hats; the people with scissors hats will assume that they have paper hats (the people with stone hats do something not yet specified, but we have already racked up 34 losses and 34 draws).

Well also need to consider

What happens when there is exactly one more of the most popular hat than the second most popular. Here we can use a tiebreaker to say that if anyone views a two-way tie for the most popular hat they assume that they have the "winner" hat (i.e. for a paper-scissors tie assume scissors; a scissors-stone tie assume stone; a stone-paper tie assumee stone) and for a three way tie choose paper.

This would mean that

When there are $n$ paper hats and $(n-1)$ scissors hats and fewer than $(n-1)$ stone hats we lose, but when there are $n$ scissors hats, $(n-1)$ paper hats and fewer than $(n-1)$ stone hats we win. (Similarly winning half of other $n,n-1,n>$ cases and one third of $n,n-1,n-1$ cases).

Mathematically, if we have $m$ players the count of losing cases is

$$3\sum_{m/3<n\le m/2}\frac {m!}{(n!)^2(m-2n)!}+3\sum_{m/3< n<m/2}\frac{m!}{(n-1)!n!(m-2n+1)!} $$ out of $3^m$ possible arrangements of hats. (With an extra $2(3k+1)!/(k!)^2(k+1)!$ term when $m=3n+1$). If you prefer binomial coefficients to factorials, the above expression is $$3\sum_{m/3<n\le m/2}\binom{2n}{n}\binom m{2n}+3\sum_{m/3\le n<m/2}\binom{2n-1}n\binom m{2n-1} $$

My intuition is

that the proportion of losing cases will tend to zero as $m\to\infty$.

Numerical investigations will follow.

It's not clear to me how much this overlaps with FirstName LastName, but I feel that a good strategy is

safety in numbers i.e. assume that one's hat is the same as the most popular hat in the ones that can be seen. (Tie-breaking rules to follow).

This means that

In any situation where there the most popular hat appears at least two times more than the second most popular hat, we win. e.g. if there are 100 contestants and the hats are 36 paper, 34 scissors and 30 stone then everyone assumes that their hat is paper; they shout "scissors" and the score is 36 wins, 30 losses and 34 draws.

Things do not go our way if

there is a tie for the most popular hat. In this case we always fail e.g. if there are 34 paper, 34 scissors and 32 stone the people with paper hats will assume that they have scissors hats; the people with scissors hats will assume that they have paper hats (the people with stone hats do something not yet specified, but we have already racked up 34 losses and 34 draws).

Well also need to consider

What happens when there is exactly one more of the most popular hat than the second most popular. Here we can use a tiebreaker to say that if anyone views a two-way tie for the most popular hat they assume that they have the "winner" hat (i.e. for a paper-scissors tie assume scissors; a scissors-stone tie assume stone; a stone-paper tie assumee stone) and for a three way tie choose paper.

This would mean that

When there are $(n+1)$ paper hats and $n$ scissors hats and fewer than $n$ stone hats we lose, but when there are $(n+1)$ scissors hats, $n$ paper hats and fewer than $n$ stone hats we win. (Similarly winning half of other $(n+1),n,n>$ cases and one third of $n+1,n,n$ cases).

Mathematically, if we have $m$ players the count of losing cases is

$$3\sum_{m/3<n\le m/2}\frac {m!}{(n!)^2(m-2n)!}+3\sum_{m/3\le n<m/2}\frac{m!}{(n+1)!n!(m-2n-1)!} $$ out of $3^m$ possible arrangements of hats. (With an extra $2(3k+1)!/(k!)^2(k+1)!$ term when $m=3n+1$). If you prefer binomial coefficients to factorials, the above expression is $$3\sum_{m/3<n\le m/2}\binom{2n}{n}\binom m{2n}+3\sum_{m/3\le n<m/2}\binom{2n+1}n\binom m{2n+1} $$

My intuition is

that the proportion of losing cases will tend to zero as $m\to\infty$.

Numerical investigations will follow.

It's not clear to me how much this overlaps with FirstName LastName, but I feel that a good strategy for a large number of players is

safety in numbers i.e. assume that one's hat is the same as the most popular hat in the ones that can be seen. (Tie-breaking rules to follow).

This means that

In any situation where there the most popular hat appears at least two times more than the second most popular hat, we win. e.g. if there are 100 contestants and the hats are 36 paper, 34 scissors and 30 stone then everyone assumes that their hat is paper; they shout "scissors" and the score is 36 wins, 30 losses and 34 draws.

Things do not go our way if

there is a tie for the most popular hat. In this case we always fail e.g. if there are 34 paper, 34 scissors and 32 stone the people with paper hats will assume that they have scissors hats; the people with scissors hats will assume that they have paper hats (the people with stone hats do something not yet specified, but we have already racked up 34 losses and 34 draws).

Well also need to consider

What happens when there is exactly one more of the most popular hat than the second most popular. Here we can use a tiebreaker to say that if anyone views a two-way tie for the most popular hat they assume that they have the "winner" hat (i.e. for a paper-scissors tie assume scissors; a scissors-stone tie assume stone; a stone-paper tie assumee stone) and for a three way tie choose paper.

This would mean that

When there are $n$ paper hats and $(n-1)$ scissors hats and fewer than $(n-1)$ stone hats we lose, but when there are $n$ scissors hats, $(n-1)$ paper hats and fewer than $(n-1)$ stone hats we win. (Similarly winning half of other $n,n-1,n>$ cases and one third of $n,n-1,n-1$ cases).

Mathematically, if we have $m$ players the count of losing cases is

$$3\sum_{m/3<n\le m/2}\frac {m!}{(n!)^2(m-2n)!}+3\sum_{m/3\le n<m/2}\frac{m!}{(n+1)!n!(m-2n-1)!} $$ out of $3^m$ possible arrangements of hats. (With an extra $2(3k+1)!/(k!)^2(k+1)!$ term when $m=3n+1$). If you prefer binomial coefficients to factorials, the above expression is $$3\sum_{m/3<n\le m/2}\binom{2n}{n}\binom m{2n}+3\sum_{m/3\le n<m/2}\binom{2n+1}n\binom m{2n+1} $$

My intuition is

that the proportion of losing cases will tend to zero as $m\to\infty$.

Numerical investigations will follow.