Edge cases

Too few participants

If zero or one person enters, the round is void. Everyone is refunded; no fees are taken.

Everyone has the same error

If every participant has the exact same (e.g. everyone predicted the same price), the round is void. Everyone is refunded.

Your error ties the round's median

The median cutoff is exclusive: you need an error strictly below the round’s median error to win. If your error exactly equals the median, you lose along with everyone at or above it, you don’t get a tiebreaker.

Best-coalition exception: majority tied at the median

If half or more of the round’s participants share the same best (lowest) error, the strict “below the median” rule would leave nobody eligible to win, since the shared error is itself the median and none of them beat it. The best-coalition exception covers this: that tied group all win together, even though none of them is strictly below the median. Everyone else loses and funds the prize pool as usual. This only kicks in when the tie is at the lowest error and covers at least half the field, it’s rare in practice.

Payout cap: very few close predictions, or unusually spread-out errors

A few very close predictions among the winners, or unusually spread-out errors, could otherwise concentrate almost the entire prize pool on one or two predictions. A winner’s total payout (stake back plus prize pool share) is capped at a multiple of their stake, currently up to 100× profit in extreme cases. If the uncapped math would pay a winner more than that, the extra is rerouted to the other winners by accuracy weight × stake, not lost.Formulas (for verification):

Prize pool = losing stakes, minus the round’s 20% take.
Payout weight (winners only): $w_i = a_i \times s_i$ (accuracy weight × stake).
Profit cap for winner $i$ : $\text{cap}_i = s_i \times (\text{max ROI} - 1)$ .
Gain: $\text{gain}_i = \min(\alpha \times w_i,\, \text{cap}_i)$ , where $\alpha$ is a single scale factor chosen so the sum of all gains equals the prize pool while no winner exceeds their cap (water-filling).
Total payout: $\text{payout}_i = s_i + \text{gain}_i$ for winners, $0$ for losers.

Show How the extra gets redistributed (water-filling)

A single scale factor α is chosen so that (1) the sum of every winner’s gains equals the prize pool, and (2) no winner exceeds their cap. Each winner first gets a share proportional to

w_i = a_i \times s_i

. If that would push someone over their cap, they get exactly their cap instead, and the rest is shared among the remaining winners by weight, repeating until the pool is fully distributed.

Getting started

Payouts

Scoring & streaks

Programs

Reference