When a lottery ticket is worth more than it costs: pricing a roll-down in 20 lines of Python

Most lottery arithmetic is depressing. A 6-of-46 game with a $2 ticket, $4,000 for five numbers, $150 for four, $5 for three and a free ticket for two returns about 74 cents on the dollar before you even think about the jackpot. But one game, Massachusetts' Cash WinFall (2004–2012), had a cap: when the jackpot reached $2 million and nobody won it, the money "rolled down" into the smaller prizes. On 8 February 2010 the match-five paid $22,096 and the match-four $807.52. The question is what a ticket was worth that night, and how sure a big buyer could be of collecting it.

The odds are hypergeometric: the chance of matching k of 6 drawn numbers from 46 is C(6,k)·C(40,6−k)/C(46,6).

from math import comb, sqrt
N = comb(46, 6)                      # 9,366,819
p = {k: comb(6, k) * comb(40, 6 - k) / N for k in (5, 4, 3, 2)}
std = {5: 4000, 4: 150, 3: 5}
roll = {5: 22096, 4: 807.52, 3: 26.85}   # 8 Feb 2010, Inspector General's letter
ev = lambda prizes: sum(prizes[k] * p[k] for k in prizes)
print(ev(std), ev(roll))              # 0.395 vs 2.141 cash per $2 ticket

So the same ticket, at the same odds, went from 40 cents of cash to $2.14. Buy 200,000 of them ($400,000) and the expectation is $428,000 in cash plus about 29,000 free tickets. That is the edge the syndicates in the story were collecting.

Two things the expectation hides. First, variance. Each prize tier's count is approximately Poisson, so the standard deviation of the cash is the square root of the sum of (count × prize^2):

n = 200000
mean = n * ev(roll)
sd = sqrt(sum(n * p[k] * roll[k] ** 2 for k in roll))
print(round(mean), round(sd))         # 428247, 51652

A $28,000 expected profit with a $52,000 standard deviation ends behind about 29% of the time, even when the roll-down happens.

Second, whether it happens. The roll-down only fires if nobody matches all six, and the more tickets in the drawing, the likelier someone does:

for sold in (470_000, 1_300_000, 2_500_000):
    print(sold, round(100 * (1 - (1 - 1 / N) ** sold), 1), "%")
# 4.9 %, 13.0 %, 23.4 %

By 2007 the syndicates themselves had pushed sales past 1.3 million tickets a roll-down, so each of them was carrying a 13% chance per drawing that the jackpot would be hit and the prizes would revert to standard. It happened once in 45 roll-downs, on 10 July 2008, and every big bettor lost hundreds of thousands that night.

Where the surplus came from is the real point: every ticket bought while the jackpot was small funded a jackpot that 44 times out of 45 nobody won, and the cap handed that money to whoever held tickets on the night. A casino game has no such pool; a provably fair dice round pays out of its own stake by a fixed formula, so its edge is a constant that never changes sign. That is the whole difference between an edge you can read in a brochure and one that cannot exist.

The full case, with the Inspector General's figures and the syndicates' totals, is on the Betkyo Journal: https://betkyo.com/en/blog/jerry-and-marge-selbee-27-million-in-tickets-the-2-million-roll-down-and-the-lottery-ticket-worth-more-than-it-cost/?utm_source=devto&utm_medium=referral&utm_campaign=journal

18+. Written with AI assistance and checked against the sources cited in the article.