There’s an interesting tactical problem that I’ve run into in Hanabi games a few times recently.
It is Alice’s turn. Bob’s turn will follow, then Charlie’s turn, and then Dan’s turn. There is no fifth player. (Hence, four-card hands.)
The current score is 27. Bob holds two coloured 5s, both marked. Charlie holds the 1 black and three non-black cards, and none of Charlie’s cards are marked. There is exactly one token available, and no misfires have occurred. There is exactly one card remaining in the draw deck. No relevant Flamboyants are available. (Either Flamboyants are not in use, or the two remaining Flamboyants are those that give tokens.)
Charlie is an extremely intelligent and experienced player, and can be assumed to figure out very complex chains of logic, but the players have not set a convention to deal with this specific scenario. All have demonstrated ability to both give and follow finesses and bluffs, including advanced variations such as reverses, long finesses (and long partially-reversed finesses), trash bluffs, and even hidden finesses. They have not established any of the conventions that are more unusual on BGA.
How does Alice clue in order to give the best possible chance of achieving a score of 30 at the end of the game?
Note that if Alice plays or discards, Bob will get only one more turn, so cannot play both coloured 5s. Alice must clue, or the probability of achieving 30 drops to 0. So, positional discard hints are impossible. Likewise, Bob and Charlie must both play. Nobody cares what Dan does.
Now, the answer is trivial if Charlie holds only a single 1. But what if Charlie holds x 1r 1k x? x 1r x 1k? In these scenarios, cluing 1s would give at best a 50/50 chance of Charlie picking the correct card to play.
I have my own answer to this problem, and I have used it successfully at least once, but at least one other time “Charlie” expressed confusion and did not play the correct card. I will let others ponder and perhaps attempt to solve this before I explain my approach. (Maybe someone will have a better idea than I did.)
It is Alice’s turn. Bob’s turn will follow, then Charlie’s turn, and then Dan’s turn. There is no fifth player. (Hence, four-card hands.)
The current score is 27. Bob holds two coloured 5s, both marked. Charlie holds the 1 black and three non-black cards, and none of Charlie’s cards are marked. There is exactly one token available, and no misfires have occurred. There is exactly one card remaining in the draw deck. No relevant Flamboyants are available. (Either Flamboyants are not in use, or the two remaining Flamboyants are those that give tokens.)
Charlie is an extremely intelligent and experienced player, and can be assumed to figure out very complex chains of logic, but the players have not set a convention to deal with this specific scenario. All have demonstrated ability to both give and follow finesses and bluffs, including advanced variations such as reverses, long finesses (and long partially-reversed finesses), trash bluffs, and even hidden finesses. They have not established any of the conventions that are more unusual on BGA.
How does Alice clue in order to give the best possible chance of achieving a score of 30 at the end of the game?
Note that if Alice plays or discards, Bob will get only one more turn, so cannot play both coloured 5s. Alice must clue, or the probability of achieving 30 drops to 0. So, positional discard hints are impossible. Likewise, Bob and Charlie must both play. Nobody cares what Dan does.
Now, the answer is trivial if Charlie holds only a single 1. But what if Charlie holds x 1r 1k x? x 1r x 1k? In these scenarios, cluing 1s would give at best a 50/50 chance of Charlie picking the correct card to play.
I have my own answer to this problem, and I have used it successfully at least once, but at least one other time “Charlie” expressed confusion and did not play the correct card. I will let others ponder and perhaps attempt to solve this before I explain my approach. (Maybe someone will have a better idea than I did.)