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Three identical, mute triplets, $F$, $S$, and $T$, each of whom is a perfect logician, have been kidnapped. Their captor begins by instructing them that in a room that they will shortly enter, are six wine goblets, three of which contain normal wine, and three of which are poisoned. The poison is such that once even a drop is drunk, ten seconds later, they will die; but is of course otherwise indistinguishable from the wine. They are told that each person must drink from one glass; they may choose to drink all of it, or just take a sip (and their survival does not depend how much they drink).

To make things more interesting, the captor takes $F$ into the room, and truthfully points out which three are poisoned. They leave the room. The platter on which the six wine glasses are on then spins an unknown amount.

1. $F$ enters the room, selects a glass, and drinks all of its contents. You know that $F$ fully drunk the glass, while the others do not.




2. After enough time for the poison to take effect (if $F$ had been poisoned), $S$ enters the room, sees if $F$ is alive or dead, selects a glass, and either drinks all of or takes a sip from it.




3. After enough time for the poison to take effect (if $S$ had been poisoned), $T$ enters the room, sees if $F$ and $S$ are alive or dead, selects a glass, and either drinks all of or takes a sip from it.

Fifteen seconds later, exactly one person lies dead on the floor.

The state of the platter is shown below, where a white circle represents an empty glass, and a purple circle shows a full (or sipped-from) glass. Note how the glasses are in groups of two around the edge, these are indented to be equally spaced.

Puzzle: Which triplet, $F$, $S$, or $T$, was poisoned?

There's no lateral-thinking tag, it's a purely logical puzzle. So, you can assume that:

• no communication occurred


• the glasses are all identical


• the glasses' relative positions do not shift during the rotation


• each person has perfect logical capabilities and memory


• none of them are suicidal


• only $F$ saw the glasses


• you cannot tell the difference between a full and sipped-from glass


• when $T$ enters he is unsure if it was $F$ or $S$ who sipped and who emptied


• the glass were left in the positions they were taken from


• each person does not care if the others survive; they have no priorities other than surviving themselves

If in doubt, feel free to ask.

Puzzle created by me; I have the solution.

I have to go now, so I'll just post this partial answer. Back soon.

First of all,

If $F$ had died, it would imply that the three poisoned glasses do not all share the same position in their respective couples (if all of them were the left one of the couple, just take one from the right and you will live). Therefore there would be, say, two "right" poisonous glasses and one "left". Let us call "$+$" a glass that has a probability of $2/3$ of being good and $1/3$ of being poisonous.

In that case,

To maximize his chances, $F$ would have chosen a $+$-glass, and, being unlucky, would have chosen the only wrong one. Therefore, if $F$ is dead and has chosen a "left" glass, then $S$ only needs to empty any other left glass, and $T$ will empty the last one, and they will both live.

Now first conclusion:

Since the empty glasses are not both on the same side (left/right), we know that $F$ is not dead.