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An Adventurer is standing on the first of eight sequential platforms, ready to advance. But a Booby-trapper is positioning a trap to catch them in a game played over several rounds.

Each round, the Adventurer secretly chooses either 1, 2, or 3 and the Booby-trapper secretly picks a platform. Then, the choices are revealed and the Adventurer advances a number of platforms equal to their chosen number. Then, if the Adventurer is standing on the platform that the Booby-trapper picked for this round, the game is over and the Adventurer wins nothing. Then, if the Adventurer is standing on the 6th, 7th, or 8th platform, the game is over and the Booby-trapper pays the Adventurer $100. Then, if the game is not yet over, a new round begins.

How much should the Adventurer be willing to pay the Booby-trapper to play this game?

Assume that both players are perfectly rational gamblers trying to maximize their money, have access to random number generators, and are aware that both players know all of this.

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  • $\begingroup$ For example, if in the first round A chooses 2 and B picks the 4th platform, A advances to the 3rd platform and the game continues into the second round. Then if A chooses 1 and B picks the 5th platform, A will advance to the 4th platform. Since this was not the B's pick for this round, the game continues into the third round. If A chooses 2 and B picks the 6th platform, A will advance to the 6th platform. Since this was B's pick, the game is over and A wins nothing. $\endgroup$
    – taef
    Nov 5, 2022 at 18:33
  • $\begingroup$ The idea behind this game is not original. I saw it on a game show around 2002 (and am guessing it was taped in the 1980s) but I don't remember the name of the game show. $\endgroup$
    – taef
    Nov 5, 2022 at 18:33
  • $\begingroup$ The show might have been Pitfall!, but the eight platforms are all that carries over: Pitfall!'s pits are pre-placed and the platforms are crossed one at a time. $\endgroup$ Nov 6, 2022 at 8:33

1 Answer 1

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Suppose the Adventurer is standing on a particular platform. There is some reward that the Adventure can expect to win on average if he manages to successfully advance 1, 2, or 3 platforms: call these $a$, $b$, and $c$, respectively.

Some basic calculation on the resulting payoff matrix shows that the Adventurer's best strategy is to move 1, 2, or 3 spaces in the proportion $bc:ac:ab$, the Booby-Trapper's best strategy is trap the spaces 1, 2, or 3 ahead in the proportion $ab+ac-bc:ab+bc-ac:ac+bc-ab$, and the resulting value of the game is $\frac{2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for the Adventurer.
Thus, if we normalize the \$100 reward to 1, the Adventurer's expectation is $\frac{2}{3}$ on platform 5, $\frac{4}{7}$ on platform 4, $\frac{8}{17}$ on platform 3, $\frac{16}{43}$ on platform 2, and finally $\frac{32}{105}$ at platform 1 - a payout just shy of \$30.48.

Addendum:

This process generalizes: Whenever there are $n$ winning platforms and up to $n$ spaces for the Adventurer, I conjecture that $E(P_k) = \frac{n-1}{\sum\limits_{i=1}^{n}\frac{1}{P_{k+i}}}$, and have shown that this holds for $n\leq 4$. I'm sure it can be shown that the matrices that show up at larger sizes have the requisite determinants for this pattern to continue, but I can't prove it yet.
If there are more winning platforms than the Adventurer can reach, they can be ignored in calculations. If there are fewer, then you calculate as if the Adventurer can only reach those platforms that have been given values so far: e.g. if there are 2 winning platforms and the Adventurer can move 4 spaces, then you calculate first with the 2-platform recurrence, then the 3-platform recurrence, before performing all further calculations with the 4-platform formula.

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