Russian Roulette

How it works

You start with a $100 bankroll, choose a wheel of 5, 6, 8, 10, or 20 numbers, stake $5, $10, $20, or $50, and mark one number as yours. Each spin lands uniformly on one active number. Land on a marked number and you are out - the whole stake is gone. Land on anything else and that number is removed from the wheel for good, your multiplier compounds, and you may spin again or cash out (cashing out is available any time after your first safe spin).

Before every spin you answer a probability question about the current wheel - survival odds over one or several spins, symmetry arguments over your marked numbers, and the like, all computed from the live wheel state. Answer wrong and you must mark an ADDITIONAL number before you are allowed to spin: a wrong answer genuinely worsens your odds, not just a score.

The payout mechanism is printed on the wheel: with s safe numbers out of n active, the fair multiplier for a safe spin is n/s (the reciprocal of your survival probability), and the game offers exactly 97% of fair, rounded to 2 decimals. That missing 3% is the house edge, taken on every single spin.

How scoring works

Cash out pays wager x the product of every offered multiplier you have locked in. Each safe spin multiplies in offeredMultiplier = round(0.97 x fairMultiplier, 2 decimals), where fairMultiplier = activeCount / (activeCount - markedCount).

Landing on any marked number pays $0 - the compounded multiplier and the stake vanish together.

Example multipliers from the formula: 8 active, 1 marked: fair 8/7 = 1.1429, offered 1.11. 6 active, 1 marked: fair 6/5 = 1.2, offered 1.16. 6 active, 2 marked: fair 6/4 = 1.5, offered 1.46.

Every spin is a -3% EV trade

Suppose your current cash-out value is V. Taking one more spin returns V x offered with probability s/n and 0 otherwise. Since offered = 0.97 x (n/s), the EV is (s/n) x 0.97 x (n/s) x V = 0.97V. Every spin, on every wheel size, at every stage, converts your position into 97% of itself in expectation.

That makes the whole game a chain: after k spins your expected value is wager x 0.97^k. Ten spins grind expectation down to 0.97^10, about 0.737 of the stake. The compounding multiplier makes the position LOOK like it is growing; the expectation says it is shrinking by 3% per pull. The only +EV button on the screen is Cash Out.

Wrong answers are the real house edge

A wrong answer forces an extra marked number. On a 6-number wheel with 1 marked, your per-spin death probability is 1/6, about 16.7%. Mark a second and it jumps to 2/6, 33.3% - double. The offered multiplier does rise to compensate (6 active, 2 marked offers 1.46 versus 1.16), and the EV identity still holds at 0.97 per spin, but your ruin probability - the chance you lose everything before cashing out - climbs sharply.

The questions themselves are no-replacement drills. The three-spin survival probability on a wheel with s safe of n active is (s/n) x ((s-1)/(n-1)) x ((s-2)/(n-2)) - both numerator and denominator shrink each spin because safe numbers leave the wheel while your marked numbers stay. Reusing the same denominator is the exact error the question is built to catch.

Wheel size trades variance, not edge

A 20-number wheel with 1 marked risks only 1/20 = 5% per spin and offers 0.97 x 20/19, about 1.02x - slow, low-variance decay. A 5-number wheel risks 1/5 = 20% per spin for 0.97 x 5/4 = 1.21x - fast and violent. Both are exactly 0.97 EV per spin. Pick size by how many questions you want to practice per stake, because that is the only thing it changes.

A worked example

Wheel of 8, stake $10, one number marked.

Spin 1: survival 7/8 = 0.875. Fair multiplier 8/7 = 1.1429; offered 0.97 x 1.1429 = 1.1086, rounded to 1.11. You survive; position $10 x 1.11 = $11.10, and the number you landed on is removed.

Spin 2: 7 active, 1 marked, survival 6/7 = 0.857. Fair 7/6 = 1.1667; offered 1.13. EV check before spinning: (6/7) x 1.13 x $11.10 = $10.75, versus $11.10 for cashing out - the spin costs about 3% of your position in expectation.

You spin anyway and survive: $11.10 x 1.13 = $12.54. Chance of having survived both spins from the start: (7/8) x (6/7) = 6/8 = 75%.

You cash out $12.54, up $2.54 - having beaten the odds, not the edge. Expectation for the two-spin plan was $10 x 0.97^2 = $9.41.

Common mistakes

• Reusing the starting denominator on multi-spin survival questions. The wheel shrinks: it is (7/8) x (6/7), not (7/8) x (7/8).

• Reading the compounding multiplier as growing EV. Offered is 97% of fair, so each spin multiplies your expected value by exactly 0.97 no matter what the on-screen number does.

• Guessing on questions. A wrong answer marks an extra number - on a 6-wheel that doubles your per-spin ruin probability from 1/6 to 2/6.

• Choosing a big wheel because it feels safer per spin. 5% per spin at 1.02x and 20% per spin at 1.21x are the same 0.97 EV; only the variance differs.

• Never cashing out. EV after k spins is wager x 0.97^k; ride long enough and either the wheel takes you or the compounding edge does.

Why interviews test this

This is the quant-interview double-or-nothing structure in pure form: a sequence of bets, each individually near-fair but slightly negative, with an option to stop. Interviewers pose it to see whether candidates evaluate the per-step EV (0.97 here) rather than the seductive path of the running multiplier, and whether they recognize optimal stopping - when every continuation is -EV, the optimal stop is now.

The shrinking wheel is also a clean without-replacement conditional-probability generator, the same math as urn problems and depleting order books: survival probabilities chain as products of changing fractions, and the most common error - frozen denominators - is precisely what phone screens are calibrated to catch.