Friday, April 12, 2019

mathematics - Strategy to beat the Casino reversed


Let's consider Strategy to beat the Casino puzzle, but reverse it - fix number of losses and ask a question about amount of rounds:



Two players, A and B, and the Casino play a game. The game consists of $N$ rounds. $N$ can be chosen by players.


Each round involves showing zeros or ones: each player picks 0/1 and also the Casino picks 0/1.


If the Casino and both players A and B show the same number ([0,0,0] or [1,1,1]), then the two players win the round. Otherwise - the Casino wins.



The players have to play all $N$ rounds, but if they lose at least $M=5$ times - the Casino will kill them. Ha-ha.


Player A is going to learn the Casino's choice for all incoming rounds right before the game starts. Unfortunately, the only way of communicating this information to player B is via the numbers chosen by A.


The evening before the game, A and B meet and agree on a common strategy and on amount of rounds to play. What is the biggest $N$ they can pick with no danger to lives?



For example, a trivial strategy would be each odd round to communicate what Casino will chose in the next round. Then they can play $N$=8 rounds, while losing 4 at most.


It's pretty easy to figure out how to win 8 out of $N$=12. But can we do 9 out of $N$=13? If not - prove it, if yes - what's the strat?
Also, can we prove that $N$=14 is not achievable? (Unless it is magically is...)




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