A game of kettle I just got inspired by a situation in my office:
Let us assume there are $n$ players. Each player must in some time go to the kitchen and prepare some tea. Each player loses $t_1$ to fill the kettle and $t_2$ to boil it.
Each player needs to select from two strategies:


*

*Fill the kettle after preparing the tea and set the kettle

*Leave the kettle empty


For simplicity we assume:


*

*The order of players entering the kitchen is random

*Each player uses all the water in the kettle

*If previous player filled the kettle, then the water is always "almost boiling" for the next player (the amount of time lost to boil it is negligible)


So - if every player selects the first strategy, every player will lose only $t_1$ per tea. If however only one player chooses it, he will lose $2t_1 + t_2$ and other players will gain $\frac{t_1}{n}$
My question: Is the game analogous to this already analyzed? If so, what is its name and where can I find the analysis of it?
 A: Interesting problem! The game theoretic answer, however, is quite simple if we assume that players do not go to the kitchen twice in a row: you don't fill in the water for the others.
We assume that a player wants to minimize his loss (in time, as you describe). Now, if someone already put in water for you, the time you lose if you don't put in water for your successor vs time lost if you do is
$$t_2<t_1+t_2.$$
Similarly, if nobody put in water for you, the time lost if you don't vs if you do is
$$t_1+t_2<2t_1+t_2.$$
That is, no matter what the others do, you individually always do better by not helping the others. This is called a dominant strategy.
If we relax the assumption that you never go twice in a row, this might change, because you might benefit from your own action if you go again. The expected time (over the next two times you go to the kitchen) if you put in the water (assuming your predecessor put in water for you) is
$$t_1+1/n(t_1),$$
compared to the time lost when you don't put in water:
$$1/n(t_1+t_2).$$
Then putting in water after yourself is optimal if
$$t_1+1/n(t_1)<1/n(t_2+t_1)\iff t_1<t_2/n.$$
In particular, it is optimal if there are few people ($n$ small), because then it is very likely you benefit from your own action, and if $t_1<<t_2$ (i.e., boiling takes very long compared to filling in water). If the inequality holds, then putting in water after yourself can be a Nash equilibrium.
