A 1% house edge is the headline number on every casino game, and it is a number no player ever sees. Not because it is hidden, but because it is small next to the thing it sits inside. Here is the arithmetic, with the numbers for a coin-flip bet.

One bet, two numbers

Take a bet that wins with probability p and pays m times the stake (profit m − 1), otherwise loses the stake. Per unit staked:

  • expected return: p · m, so the edge is e = 1 − p·m
  • standard deviation: σ = sqrt(p · m² − (p·m)²)

For a roll-under-50 dice bet on a 9,999-result engine, p = 4,999 ÷ 9,999 and m = 1.98:

p, m = 4999/9999, 1.98 mean = p*m # 0.98990 e = 1 - mean # 0.01010 (1.01 cents per dollar) sigma = (p*m*m - mean**2) ** 0.5 # 0.990 (99 cents per dollar) print(sigma / e) # 98.0

The swing is 98 times the cost. That ratio is the whole story.

The square-root rule

Over N independent bets the expected loss is N·e (linear) and the standard deviation of the total is σ·√N. Their ratio grows as √N, so the expected loss clears k standard deviations at

N = (k · σ / e)²

With σ/e = 98 and k = 2 (the usual "this is not noise" line), N ≈ 38,400. For a clean ×2 bet at exactly 1% it is 39,996. Call it forty thousand bets.

What the player sees before that

The chance of being ahead after N bets is a binomial tail: you are ahead when wins · m > N.

from math import comb def p_ahead(N, p=4999/9999, m=1.98): k = int(N // m) + 1 # wins needed to be above break-even  return sum(comb(N, w) * p**w * (1-p)**(N-w) for w in range(k, N+1)) for N in (100, 1000, 10000, 40000): print(N, round(p_ahead(N), 3)) # 100 0.460 · 1000 0.363 · 10000 0.154 · 40000 0.020

After a hundred bets, 46% of players are up. After a thousand, more than a third. The edge does not reach one standard deviation until about 9,600 bets and two until about 38,000.

Higher payouts hide it longer

Because σ grows with the payout and e does not, every setting has its own crossing. For a "limbo" style bet that wins with chance 0.99/T and pays T, the variance per unit is 0.99·T − 0.9801, so the two-sigma crossing is 39,600·T − 39,204 rounds: about 40,000 at ×2, 357,000 at ×10, 3.9 million at ×100. At 900 rounds an hour the ×100 figure is about 545 eight-hour days. Same price, hidden for longer.

Why the house never waits

The house adds everyone up. A thousand players making forty bets each are forty thousand bets in one ledger, two standard deviations below zero on a day when none of them has seen anything. The law of large numbers favours whoever makes the most bets, and the house makes all of them.

Two practical readings. A session proves nothing about a game, in either direction. And the swing is the thing to size a bankroll for: σ · √N · stake is what you should expect to be up or down by, two sessions in three.

The full per-setting tables, with the exact binomial figures, are in the Betkyo Journal piece this note summarises: https://betkyo.com/en/blog/how-many-bets-before-the-edge-shows-priced-the-40000-rolls-that-hide-a-1-percent-house-edge/

18+. Written with AI assistance; every figure above is a calculation you can rerun from the snippets.