This is a proposed solution for a Jane Street Puzzle January 2020 variation where the game is split into rounds \(r\) and the \({sum}_{r} \neq 11\) is implemented only on each round. For example, if Nate starts with a \(7\) then \({sum}_{0} = 7\). When Alter follows with a \(3\), then \({sum}_{0} = 10\). After the next Nate move, we have \(r = 1\) and the sum is reset.

We’ll first approach the strategy that is described by Nate as unfair. We’ll define a choice as \(c_{p,t}\) where \(p\) is a person with \(p \in (A, N)\) for Alter and Nate respectivelly and \(t\) the timestep (round) of play, with \(t \in [0, n]\). We’ll also define a playing position as \({pos}_{p,t}\) which is the number where a player starts from on each round. For example, \({pos}_{N,0} = 0\).

Let’s analyze the winning strategy from a reverse time perspective: For Nate to lose he has to be forced \(c_{N,n-1} \in [89,98]\). This will allow Alter to send the sum to \(99\) (\({pos}_{N,n} = 99\)) and force Nate’s \(c_{N, n} \in [100,109]\). This is a core idea of the Nate’s winning strategy. In fact, it is his only strategy and we’ll explain why later. Let’s first do a simple analysis of the chain of events that leads to \(c_{N,n-1} \in [89,98]\):

For this to be guaranteed to happen, Alter needs to force \({pos}_{N,n-1} = 88\), meaning that \(c_{N,n-2} \in [78,87]\) needs to happen. As such, we recognize a series of events that create Alter’s strategy:

\(t\) \(c_ {N, t-1}\) \({pos} _{N,t}\)
\(9\) \([100,101]\)  
\(8\) \([98,89]\) \(99\)
\(7\) \([87,78]\) \(88\)
\(6\) \([76,67]\) \(77\)
\(5\) \([65,56]\) \(66\)
\(4\) \([54,45]\) \(55\)
\(3\) \([43,34]\) \(44\)
\(3\) \([32,23]\) \(33\)
\(2\) \([21,12]\) \(22\)
\(1\) \([1,10]\) \(11\)

The issue for Alter here is that he needs to force \({pos} _{N,t}\) into the values of the table. Otherwise, Nate will have the opportunity of doing so and the tables are switched. For example, if after \(c_ {N, 0}\), Alter chooses \({pos} _{N,1} \neq 11\) and thus \({pos} _{N,1} \geq 12\), allowing at least \(c_ {N, 1} \in [13, 22]\). In that way \({pos} _{A,2} = 22\) by Nate and Alter has effectively taken Nate’s position in the winning strategy. This is the equivalent of Alter forfeiting a round \([1]\).

The interesting part is that, this is exactly what’s happening when \({pos} _{N,t} \neq 11\) (which is equivalent to \({c} _{N,t} + {c}_{A,t} \neq 11\)). If Nate forces \({pos} _{A,0} = 10\) \([2]\) then, at \(t=1\), Alter cannot force \({pos} _{N,1} = 11\), because of the aforementioned restriction. So forcibly \(c_ {N, 1} \in [13, 22]\). Now a new timestep/ round starts and Nate has the possibility of placing \({pos}_{A,1} = 22\) with a current sum of: \({c} _{N,1} + {c}_{A,1} = 10 + {c}_{A,1}\), which is allowed. It follows that the same is true for \(t>1\).

Thus, Nate always wins \([3]\).

Notes:

  • \([1]\) The game is inflexible in the sense that forfeiting a round (= not playing the best move) always leads to losing the game.
  • \([2]\) If Nate chooses something else instead of \({pos} _{A,0} = 10\), then this is the equivalent of Nate forfeiting a round.
  • \([3]\) This is a pretty cool puzzle because it displays our inability to quickly analyze \(n+1, n > 0\) level effects of our actions. That’s why we result to strategy building.
  • \([4]\) Jane Street looks like a pretty cool place to work - Puzzling over Game Theory riddles is fun (it seems they have weekly courses on it also) - The 90% book reimbursement for work-related books, is incredibly appealing.
  • \([5]\) On the same note, one can answer \(3\) here.