The Problem of Points: Splitting an Interrupted Bet

The problem. Two players bet $100 each on a fair coin-flip game: first to win 6 flips takes the whole $200 pot. The game is interrupted after Player A has won 5 flips and Player B has won 3. How should the $200 be split fairly? Splitting by flips won so far (5:3) feels intuitive but is wrong — it ignores that the game was already tilted heavily toward A finishing the job.

Why proportional-so-far splitting is wrong

A 5:3 split of $200 gives A $125 and B $75. But that number only reflects flips already played — it says nothing about how likely each player was to have actually won the whole game had play continued. A is one win away from taking everything; B needs three more wins in a row (essentially) to catch up. The fair split has to be based on the probability each player would have gone on to win, not on the score so far.

The correct approach: play out the game hypothetically

The insight, credited to Pascal and Fermat's 1654 correspondence, is to ask: if the game had continued, what's the probability A eventually reaches 6 wins before B does? Since A needs 1 more win and B needs 3 more, the game must finish within at most 3 more flips (after 3 flips, either A has gotten the 1 win it needs, or B has swept all 3 and reached 6). Enumerate the $2^3 = 8$ equally likely outcomes of the next 3 flips (each flip win/loss for A):

A wins the game unless B wins all of the next 3 flips. $P(\text{B wins all 3}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$. So $P(\text{A eventually wins}) = 1 - \frac{1}{8} = \frac{7}{8}$, and $P(\text{B eventually wins}) = \frac{1}{8}$.

The fair split of the $200 pot is therefore in the ratio 7:1 — A gets $\frac{7}{8} \times 200 = 175$, i.e. $175, and B gets $\frac{1}{8} \times 200 = 25$, i.e. $25.

Worked example: a closer score

Suppose instead A has 4 wins and B has 3, first to 6 wins. A needs 2 more, B needs 3 more, so the game resolves within at most 4 more flips. B only wins the whole thing if B wins at least 3 of the next 4 flips (3 or 4 outright, since A would need only 2). Count favorable outcomes for B out of 16 equally likely flip sequences: $\binom{4}{3} + \binom{4}{4} = 4 + 1 = 5$. So $P(\text{B wins}) = \frac{5}{16}$ and $P(\text{A wins}) = \frac{11}{16}$. The pot splits 11:5 — much closer than 7:1, reflecting that this score was genuinely closer to even.

The general technique: expected value under hypothetical continuation

This puzzle is the historical birthplace of expected value: instead of asking "what happened," ask "what would happen on average if we kept going," and split resources in that ratio. The same logic underlies how a poker pot gets split when players agree to stop and run the remaining cards out mathematically (an "equity chop"), and it's the right frame any time a fair division needs to account for future uncertainty rather than just the count of events already resolved.