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f''
f''
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Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m\ge n-s$$m=n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.

Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m\ge n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.

Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m=n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.
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Gamow
Gamow
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Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m=n-s$$m\ge n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.

Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m=n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.

Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m\ge n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.
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f''
f''
  • 33.8k
  • 4
  • 121
  • 165

Suppose that there is a winning strategy for the prisoners with $m<n-s$.

For an arbitrary sequence of steak allocations such that every prisoner eats steak, some $m$ prisoners will eventually vote for option 1. If we swap the meals of the $n-m$ other prisoners around, the $m$ prisoners cannot tell the difference, so they must vote for option 1 anyway. But $n-m>s$, so we can swap the meals of those $n-m$ such that one never eats steak.

This shows that a winning strategy is impossible for $m<n-s$.

If $m=n-s$, we can use the same strategy that @matega described for the original problem. The $s$ prisoners who receive steak on the first day always vote for option 2, and the other $m$ vote for option 1 if they have received steak at least once.