The Weakest Player Should Waste the Turn

Three players take turns in a knockout contest. On your turn you may name one opponent and remove them with probability equal to your own hit rate, or you may deliberately aim at nothing. You land one hit in ten. The second player lands three in ten, the third six in ten. The turn cycles in that order over whoever is still in, and the last player standing wins. You move first. Whom should you aim at?

The honest answer is neither of them. Your best opening move is to waste the turn on purpose, which is worth $965/4736 = 20.376\%$. Aiming at the strongest player is worth $89365/442816 = 20.181\%$, and aiming at the middle player is worth $81865/442816 = 18.487\%$. All three of those numbers are exact rationals, and the surprise is not that passing wins but how little it wins by.

First, the one-on-one

Everything in the three-player phase eventually collapses into a duel, so the duel has to be priced before anything else. Two players alternate, the first shoots with hit rate $p$ and the second with rate $q$. The first mover wins immediately with probability $p$. Failing that, both must miss for the position to return to exactly where it started, which happens with probability $(1-p)(1-q)$, and then the first mover is on strike again. Summing over the number of complete rounds of misses gives a geometric series:

The denominator has a reading worth keeping. $p + q - pq$ is the probability that at least one of the two lands a hit in a given round, so equation (1) says the first mover wins with the share of that event that belongs to him. The series converges because $(1-p)(1-q) < 1$ whenever either player can hit at all.

Put your rate and the middle player's into it. Entering that duel on strike is worth

and entering the same duel second is worth the complement of your opponent's first-mover value, which is $1 - 30/37 = 7/37$, about 19 percent. Same two players, same rates, and eight percentage points of difference decided by nothing but who shoots first.

Why a landed hit is a liability

Here is the step the whole problem turns on, and it is about bookkeeping rather than about probability. Removing a player also removes their place in the cycle, so a hit changes who moves next.

Suppose you take out the six-in-ten player on your own turn. The turn passes to the next living player after you, which is the three-in-ten, so it opens the duel and you are worth $7/37$. Now suppose you pass and the three-in-ten takes out the six-in-ten on its turn instead. The next living player after the three-in-ten is you, because the six-in-ten is gone, so you open the duel and you are worth $10/37$. The same elimination, performed by someone else, is worth $3/37$ more to you.

Solving the three-player phase

With all three alive and you to move, a round in which nobody connects returns the position to exactly the same state, so the value of your strategy satisfies an equation in itself. Assume for now that the two strong players aim at each other, which the next section proves rather than assumes. If you pass, then the three-in-ten shoots at the six-in-ten and connects with probability $3/10$, leaving you first in the duel against the three-in-ten, worth $10/37$. Otherwise the six-in-ten shoots back at the three-in-ten and connects with probability $6/10$, leaving you first against a six-in-ten, worth $W(1/10, 6/10) = 5/32$. If both miss you are back where you began. Writing $x$ for the value of passing:

The other two options carry the same tail. You fire, you connect with probability $1/10$, and if you miss the position hands over to the identical pass-round above. Removing the six-in-ten gives you the duel entered second against the three-in-ten, worth $7/37$. Removing the three-in-ten gives you the duel entered second against the six-in-ten, worth $1 - W(6/10, 1/10) = 1/16$, which is a dreadful place to be. So:

As decimals, $20.376\%$, $20.181\%$ and $18.487\%$. The ordering is what the video had room to say. The margins are what it did not.

The assumption that is usually left hanging

Every derivation above leaned on the two strong players aiming at each other. Textbook treatments of the problem tend to wave at that as obvious. It can be settled instead. There are three moves available to each of the three players, so twenty-seven pure stationary strategy profiles in total, and every one of them can be solved in exact rational arithmetic as a linear fixed point of the same shape as equation (3). Exactly one profile is a Nash equilibrium: you fire into the air, the three-in-ten aims at the six-in-ten, and the six-in-ten aims at the three-in-ten. The convenient assumption turns out to be a conclusion.

Restricting yourself to stationary strategies also looks like a restriction, and is not one. With the other two players fixed and stationary, the position you face is a finite Markov decision process over a handful of states, and such a process always admits an optimal stationary policy. So the maximum of the three numbers above is the true optimum, and no scheme that passes twice and then shoots beats $20.376\%$.

Where the result stops being true

Two caveats matter more than the headline. The first is the size of the margin. Aiming at the strongest player trails passing by $0.195$percentage points, which is second best rather than wrong. A candidate who names the six-in-ten has understood the structure of the problem and lost two parts in a thousand of win probability. The gap is fragile in the other direction too: hold your rate and the middle player's fixed, push the strongest player's rate up past about $0.689$, and removing him becomes the better move, because a shooter that dangerous is worth more deleted than the turn order is worth kept.

The second is that none of this is a law about weak players. The conclusion is a consequence of the stated turn order, in which the one-in-ten shoots, then the three-in-ten, then the six-in-ten. Reverse the two strong players so the six-in-ten moves before the three-in-ten and the whole bookkeeping of who inherits the strike changes with it, and the best opening move can change too. The general statement is the one about turn order, not the one about weakness: eliminating a player edits the cycle, and if the edit puts you on the wrong side of a duel, the elimination costs more than it gains.

This family of problems is usually called a truel. The interesting instances are those where surviving into a duel matters more than reducing the number of opponents, and this one sits close to that boundary.

Sources and further reading

• The three-player standoff and its history — Truel
• The equilibrium concept used to justify the two strong players targeting each other — Nash equilibrium
• The state structure behind equations (3) and (5) — Absorbing Markov chain
• Why an optimal stationary policy exists — Markov decision process
• The sum in equation (1) — Geometric series

Every number here was checked three ways before publication. The duel formula in equation (1) was rebuilt by summing its series term by term rather than quoted. All twenty-seven pure stationary profiles were solved in exact rationals, which is what identifies the single equilibrium. And two million simulated games under that equilibrium returned $0.203376$ against the exact $0.203758$, which is the agreement you would expect from that many trials.