There are a total of 200 players assigned to either A or B at a certain venue. There are actually 100 of each, but the players don't know that. They don't know which one they are. Of course, there is no way to exchange information with others.

If you are logically certain that you are A, you leave immediately; B stays. Each player has a machine that displays the number of other A and B players remaining excluding the machine-holder in the venue every five minutes. The machine keeps track of rounds it shows the rest of the A and B players and it starts as round zero.

The game will end when there are no A's.

Assume that all the players have sufficient logical thinking ability. How many rounds will it be, in the shortest case, before there are no more A's in the venue, i.e., all the machines display that there are no A's?

  • $\begingroup$ This is now identical to the standard blue eyes problem. $\endgroup$
    – Ben Barden
    Commented Aug 4, 2021 at 13:57
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    $\begingroup$ WIthout the guru from the blue eyes problem, there is nothing to kick start the process. If there were only one A, they would not be able to deduce they were an A without a guru pronouncement. Without that base case, no further deductions can be made. $\endgroup$ Commented Aug 4, 2021 at 14:33
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    $\begingroup$ That's an extremely major change to the problem which might (I haven't thought it all through) completely invalidate the existing answer. I'm tempted to roll it back... $\endgroup$
    – bobble
    Commented Aug 5, 2021 at 1:39
  • $\begingroup$ @JaapScherphuis , I disagree with you . Imagine that there is only one A. His machine shows A's count to be 0. Yet, he notices that the game hasn't ended . This will tell him that he is A. Similar logic can be extended to the case of 2 As etc. $\endgroup$ Commented Dec 11, 2022 at 17:32
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    $\begingroup$ @HemantAgarwal There were edits after my comment was made. $\endgroup$ Commented Dec 11, 2022 at 18:55

1 Answer 1


It seems to me that no one ever leaves.

In the original blue eyes puzzle an oracle shows up and gives some information to the islanders that they didn't have before. Specifically the oracle gives them all information that would allow a lone blue eyed islander to leave which starts the induction chain that eventually lets every blue eyed islander leave.

Our players don't have any information that would allow a single A player to leave which means that none of them will leave.

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    $\begingroup$ I think you're right, since the group does not have the knowledge that "there is at least one A here". If the scenario included a bell that rang once all A's had left, I think it would be possible, since you could make deductions based on whether the game had ended or not. But as it is, a blue-eyed person who sees only brown-eyed people will never leave unless they know for certain that there are blue-eyed people remaining, and deduce that they must be that person. $\endgroup$ Commented Aug 4, 2021 at 15:27

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