Can You Throw the Hammer?

Two players race to $3$ points. Each hole is worth $1$ point and is a coin flip, but before any hole a player may “throw the hammer.” The opponent then either accepts (the hole is now worth $2$ points) or rejects (conceding $1$ point to the thrower). You have won the first hole and lead $1$–$0$. With optimal play, what is your probability of winning the match?

The Fiddler, Zach Wissner-Gross, April 18, 2025(original post)

Solution

Solve the game backward over scores $(a,b)$. At each hole you may throw (the opponent then picks the worse of a two-point hole or conceding you a point) or pass (after which the opponent may throw against you); both sides optimise. Leading $1$–$0$, throwing the hammer is correct: it turns the hole into a two-point swing that you can ride to $$\boxed{75\%}.$$ (A one-point hole from $1$–$0$ would give only $62.5\%$, which is why you throw.)

The computation

Encode the game tree and fold it back from the finish: at every score, value both throwing (opponent accepts or concedes, whichever hurts you more) and passing (opponent may then throw), and take the side’s best choice.

from functools import lru_cache
def solve(T):
    @lru_cache(None)
    def V(a, b):
        if a >= T: return 1.0
        if b >= T: return 0.0
        p1 = 0.5*V(a+1, b) + 0.5*V(a, b+1)
        p2 = 0.5*V(min(a+2, T), b) + 0.5*V(a, min(b+2, T))
        you = min(p2, V(a+1, b))          # you throw; opp accepts(p2) or rejects(you+1)
        opp = max(p2, V(a, b+1))          # opp throws; you accept(p2) or reject(opp+1)
        return max(you, min(p1, opp))
    return V
print(round(solve(3)(1, 0), 4))          # 0.75

Extra Credit

With the same rules but a race to $5$ points, what is your probability of winning from $1$–$0$?

Solution

The same backward solve over a longer race gives $$\boxed{\tfrac{11}{16}}=68.75\%,$$ a smaller edge: an early one-point lead matters less when more holes remain. (The source’s value is paywalled; this is my own.)

The computation

The same solver from the main section, run to $5$ points.

print(round(solve(5)(1, 0), 4))          # 0.6875 = 11/16