Recently I gave a second thought on the solution of 98E and got confused about some technical detail. It occurs to me that there may be something wrong with the problem.↵
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In the [editorial](http://codeforces.net/blog/entry/2323), it claims that the expected utility for "bluffing and move on" should be 1-P(n,m-1) and the reason was "In the same manner we fill other cells of the matrix.". Obviously not a compelling reason. ↵
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I guess the writer's proof was something like this: ↵
Since the opponent chooses to move on. It must lose when I'm not bluffing. So it chooses to behave optimally in the case that I am bluffing which is equivalent to the case that I showed andabandondiscarded this card.↵
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The problem is that they are actually not equivalent because I did notabandondiscard that card. Since f[n][m] may not coincide with 1 — f[m][n], I can bluff about this card even though the opponent knows I'm bluffing. I waste a step and change the state to 1 — f[m][n] instead, and I may actually benefit from it. The cards that known by the opponent but not discarded matters!↵
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So now the claim made in the editorial seems quite suspicious. I'm looking forward to a clear and intuitive explanation.
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In the [editorial](http://codeforces.net/blog/entry/2323), it claims that the expected utility for "bluffing and move on" should be 1-P(n,m-1) and the reason was "In the same manner we fill other cells of the matrix.". Obviously not a compelling reason. ↵
↵
I guess the writer's proof was something like this: ↵
Since the opponent chooses to move on. It must lose when I'm not bluffing. So it chooses to behave optimally in the case that I am bluffing which is equivalent to the case that I showed and
↵
The problem is that they are actually not equivalent because I did not
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So now the claim made in the editorial seems quite suspicious. I'm looking forward to a clear and intuitive explanation.
