Game One Is Worth 31.25 of Your 100, and Nobody Asked Who Is Better

A holding pays $200 if a team wins four games first, you must take a symmetric position on every game, and committing the whole hundred to game one produces the right payoffs a week too early. Backward induction on the lattice fixes the amount at half the gap between the two successor values, $31.25. The same number is 5/16 of the holding, which is the chance the other six games split three each, and no win probability appears anywhere in the derivation.

A holding pays $\$200$ if a particular team is the first to win four games in a seven-game series, and nothing otherwise. You are required to take a symmetric position on every single game: commit an amount $s$ and your holding moves to $W + s$ if that game goes the team's way and to $W - s$ if it does not. Starting from $\$100$ , how much should ride on game one?

The tempting move is to commit the whole hundred. It has the right flavour, since a symmetric position of $100$ turns $100$ into either $200$ or $0$ , which are the two payoffs the contract can produce. The trouble is that it produces them a week too early. The correct amount is $\$31.25$ , and finding it uses no information whatever about how good either team is.

Values first, positions second

Label a node by the pair $(a, b)$ , the wins and losses accumulated so far. Absorbing values are known outright: $V(4,b) = 200$ and $V(a,4) = 0$ . Everything else follows from the symmetry of the position structure, which forces equal weights on the two successors.

V(a,b) \;=\; \tfrac{1}{2}\,V(a+1,b) \;+\; \tfrac{1}{2}\,V(a,b+1)

(1)

Those one-halves are not an opinion about either team. They are what the symmetric position structure forces: if a commitment of $s$ gains $s$ one way and loses $s$ the other, then the only value assignment under which no schedule of positions creates money out of nothing is the equally weighted one. The true strengths of the teams are irrelevant to the recursion for the same reason that the real drift of a stock is irrelevant to a hedge.

Rolling (1) back from the absorbing edges fills the lattice. At nil-nil the value is $100$ by symmetry. One game up it is $200 \cdot 21/32 = 131.25$ , one game down it is $200 \cdot 11/32 = 68.75$ .

Fig. 1 — The whole lattice, filled in backwards from the absorbing edges. Each interior number is the average of the one to its right and the one below it.

With the values in hand the position at any node is forced. It has to be the number that lands on both successor values at once, so it is half the gap between them:

s(a,b) \;=\; \frac{V(a+1,b) - V(a,b+1)}{2}, \qquad s(0,0) = \frac{131.25 - 68.75}{2} = 31.25 .

(2)

Check it: from $100$ , a win gives $131.25$ and a loss gives $68.75$ , which are exactly the values of the two states you can be in. Nothing is left over and nothing is missing, so the schedule never needs new money.

The same number, counted a completely different way

Where does $21/32$ come from? From one game up, the team needs three more wins before the opponent needs four, so it takes the series exactly when it wins at least three of the six remaining games. Summing the top of the binomial gives $(20 + 15 + 6 + 1)/64 = 42/64 = 21/32$ .

Now subtract the two conditional chances and watch what survives:

\frac{21}{32} - \frac{11}{32} \;=\; \frac{20}{64} \;=\; \frac{\binom{6}{3}}{2^6} \;=\; \frac{5}{16}

(3)

The $\binom{6}{3}$ is the signature of something specific. Game one changes the outcome of the series precisely when the other six games split three each, because only then does game one break a tie. That is what it means for a game to be decisive, and equation (3) says the gap in value between winning and losing game one is the chance that game one turns out to matter.

Fig. 2 — The other six games. The central bar is the decisive count, 20 of 64, and the four right-hand bars are the 42 of 64 that win the series from one game up.

So the answer can be written without touching the lattice at all. The value gap is the payoff times the chance game one decides the series, and the position is half of that:

s(0,0) \;=\; \tfrac{1}{2}\cdot 200 \cdot \Pr(\text{game 1 decides}) \;=\; 100 \cdot \tfrac{5}{16} \;=\; 31.25

(4)

Two derivations that share no step landing on the same number is the strongest evidence available for an answer of this kind.

What the general statement is

Position as a discrete derivative

For a payoff that depends only on the sequence of outcomes, the position at a node is the difference quotient of the value function across the next outcome. Equation (2) is that difference quotient with the step normalised to one, and it is the discrete counterpart of a hedge ratio: how much the value moves per unit of the thing that moves it.

The value process is a martingale under the symmetric weights, which is the reason the schedule is self-financing. Rearranging (1) gives $\tfrac{1}{2}(V_{\text{up}} - V) + \tfrac{1}{2}(V_{\text{down}} - V) = 0$ : the expected change in value across any game is zero, so no node ever requires a top-up. The pair of conditional probabilities does the bookkeeping, $21/32 + 11/32 = 1$ and their average is the $1/2$ the root started from.

The construction was checked by walking every one of the seventy completed first-to-four series in exact fraction arithmetic. Terminal wealth is exactly $200$ when the team wins and exactly $0$ when it does not, on all seventy. The holding never goes negative, the minimum along any path is $0$ , and no position ever exceeds the holding that backs it.

What the win probabilities do and do not change

Suppose the team really wins each game with probability $0.7$ . The schedule in (2) does not change by a cent, and it still lands on exactly $200$ or exactly $0$ . Simulating two hundred thousand series at true per-game chances of $0.70$ , $0.50$ and $0.30$ gives zero mismatches at every one, with the team taking the series $87.4\%$ , $50.0\%$ and $12.7\%$ of the time.

What the true probabilities change is the value. A team that wins $87.4\%$ of its series is holding something worth about $0.874 \times 200 \approx 175$ in expectation, not $100$ . The distinction is the same one that separates a price from a forecast: the replicating schedule is determined by the structure of the positions available, while the expected payoff depends on the world. Only the first is what the question asks for.

It is worth seeing a naive alternative fail. Spreading the hundred evenly, at $100/7 \approx 14.29$ per game, reaches only $157.14$ after four straight wins, which is short of $200$ . An even schedule cannot work because the chance of being the decisive game is not constant across the lattice: it is $5/16$ at the start and $1$ at three-three, where the whole holding has to be committed.

Where exact replication breaks

The construction rests on assumptions that are easy to leave unstated. Each of them names a way the exact replication can fail.

The positions must be symmetric and available in any size, in either direction, on every game. Take away the freedom to size freely, by a cap or a minimum, and the schedule can no longer hit both successor values, so the terminal payoff is only approximated. Take away symmetry, for instance if committing $s$ returns $0.95s$ on a win while still costing $s$ on a loss, and the value recursion in (1) picks up a strictly negative drift; the payoff is then unreachable from a starting holding of $100$ , and the honest starting figure is lower.

The series must also stop. Absorption at four wins is what makes the lattice finite and the backward induction well posed. Replace it with an unbounded race and the deciding probabilities no longer come from a fixed binomial, and the value function has to be built as a limit rather than filled in.

Games are treated as separate outcomes with no carry-over. If a result changed the terms of later games, the state $(a,b)$ would no longer be enough to determine the value, and the whole two-index lattice would have to be replaced by something that remembers the order in which the wins arrived.

Sources and further reading

Wikipedia: Backward induction and Binomial options pricing model, which is equation (1) with a discount factor attached.
Wikipedia: Martingale and Central binomial coefficient, for the two objects that meet in equation (3).
Wikipedia: Banzhaf power index, where the probability that a single vote is decisive plays the role that the deciding game plays here.
John Cox, Stephen Ross and Mark Rubinstein, “Option Pricing: A Simplified Approach”, Journal of Financial Economics7 (1979), 229–263, for the lattice argument in its original setting.

The deciding count above was rebuilt from first principles rather than quoted: enumerating all sixty-four arrangements of the other six games and counting the twenty that finish three each gives the $5/16$ directly, and the curve of values was verified to satisfy the martingale identity exactly at every non-absorbing node.