Bankroll Safety Analyzer

The Bankroll Safety Analyzer computes even-money gambler’s-ruin risk from three inputs: Bankroll (default $1000), Unit Size (default $10), and Win Probability (default 52%). Bankroll Units = floor(bankroll / unit) = floor(1000/10) = 100. Risk of Ruin is 100% if p ≤ 0.5, otherwise (q/p)^units × 100% with q = 1 − p. On the defaults, p = 0.52, q = 0.48, ruin = (0.48/0.52)^100 × 100% ≈ 0.0334%. You have a hundred $10 lives against a 52% even-money coin, and the probability of going to zero before infinity is 0.0334%.

That 0.0334% is not a 100-hand risk. Classical ruin is the probability that a gambler who stops at 0 and who may play forever (or until an infinitely rich opponent absorbs them) ever hits 0. A finite session of 200 hands has a smaller chance of touching zero; the formula is the infinite-horizon number, which is the conservative one. If you set Win Probability to 50% or below, the tool returns 100% ruin: with no edge, or a house edge, you are absorbed almost surely if you play long enough.

Raise the bankroll to $2000, unit still $10: units = 200, ruin = (12/13)^200 × 100% ≈ 0.0011%. Double the units, square the (q/p) factor, ruin drops by a factor of about 30 in this region. Cut the unit to $5 on a $1000 roll: 200 units again. Raise p to 55%: q/p = 0.45/0.55 ≈ 0.81818, and 0.81818^100 is about 1.7×10^{-9}, so ruin prints 0.0000% at four decimals. The 52% default is the interesting one because 0.48/0.52 = 12/13 ≈ 0.92308 still leaves a visible 0.0334% at 100 units.

The model is even-money: win +1 unit or lose −1 unit. Decimal odds of 1.91 are not even money; do not type 52% from a 1.91 book and expect this ruin number to match Kelly or a skewed random walk. Use the Bet Size Optimizer for Kelly and this page for the coin-flip ruin skeleton.

If p = 52% but you are the underdog in disguise (vig), you should type the true p after vig, not the 52% you advertised to yourself.

Gambler’s ruin is the ancestor of every bankroll lecture. Two players exchange one unit per trial; the game ends when one is broke. If the opponent is infinitely rich (the house, or the market), the finite player’s ruin probability with win chance p > 1/2 and i units is (q/p)^i. Christiaan Huygens posed ruin problems in 1657; Abraham de Moivre and later classic texts treated the finite two-player case; William Feller’s An Introduction to Probability Theory and Its Applications, Volume I, is the modern classroom source for the infinite-opponent formula SorteCalc uses.

The 52% default is a sports-bettor’s sketch of a small edge at even money, not a casino game. No table game pays even money at 52% after vig — European roulette even money is 18/37 ≈ 48.65%. If you type 48.65%, p ≤ q and ruin prints 100%. That is the correct infinite-horizon answer for a negative-edge even-money game: play long enough, go broke with probability 1. Recreational players hate that sentence; the math does not care. A 52% even-money proposition is closer to a mildly sharp sports bet after you have already beaten the close, or a made-up classroom coin.

Risk of ruin in poker literature (Mason Malmuth, later bankroll chapters in every hold’em book) often uses a different model: normally distributed session results, or a diffusion approximation. Those give a different number than (q/p)^N. Do not mix them. This analyzer is the simple random-walk ruin, discrete, even money, no upper absorbing barrier (you do not stop at a profit target). An upper barrier (stop at 2N) would lower the probability of hitting 0; omitting it is conservative for a player who never cashes out.

Limits: even-money only, independent trials, constant p, integer units via floor, infinite horizon, no simultaneous bets, no correlation. A 100-unit bankroll at 52% is “safe” on this measure (0.0334%) and still able to lose 20 units in a weekend. Safety here means absorption probability, not drawdown comfort. Thorp and later Kelly writers would also ask about the growth rate; ruin can be tiny while volatility is intolerable.

Use the page to see the exponential in the unit count. 50 units at 52%: (12/13)^50 × 100% ≈ 1.83%. 100 units: 0.0334%. That drop is why “100 units” became a slogan. The slogan assumes p > 1/2 and even money. Without both, it is folklore.

A lecture on how to compute even-money gambler’s-ruin risk starts from a simple random walk, not from a weekend of two hundred hands. Bankroll 1000, unit 10, win chance 52 percent. The number of units is the integer part of 1000/10, namely 100. When p exceeds one half, the infinite-horizon ruin probability against an infinitely rich opponent is (q/p)^N with q = 1 − p. Here p = 0.52, q = 0.48, q/p = 12/13 ≈ 0.92308, and (0.48/0.52)^100 × 100 percent ≈ 0.0334 percent. You have a hundred ten-dollar lives against a 52 percent even-money coin, and the chance of absorption at zero is 0.0334 percent.

If the win chance is 50 percent or below, ruin is 100 percent. With no edge, or a house edge, you are absorbed almost surely if you play long enough and never stop at a profit target. European roulette even money is 18/37 ≈ 48.65 percent; type that and the page correctly prints certain ruin. The 52 percent default is a sports-bettor’s sketch of a small edge at even money, or a classroom coin, not a table game. Recreational players hate the certain-ruin sentence; the mathematics does not care. Raise the bankroll to 2000 at the same unit: 200 units, ruin ≈ 0.0011 percent. Cut the unit to 5 on a 1000 roll and you have 200 units again. Raise p to 55 percent and (0.45/0.55)^100 is negligible at four decimals.

Christiaan Huygens posed ruin problems in 1657. Abraham de Moivre treated the finite two-player case. William Feller’s An Introduction to Probability Theory and Its Applications, Volume I, is the modern classroom source for the infinite-opponent formula. Derivation: let r_i be ruin from i units. Then r_0 = 1, r_i tends to 0 as i grows if p > q, and r_i = p r_{i+1} + q r_{i−1}. The characteristic equation has roots 1 and q/p; boundedness picks r_i = (q/p)^i. The two-player formula with opponent capital A recovers (q/p)^i in the limit A to infinity when p > q. This page takes that limit: the house does not go broke. At p = 1/2 the finite-A formula is 1 − i/A, which tends to 1 as A grows, consistent with the 100 percent branch.

Finite sessions are smaller risks. Classical ruin is the probability of ever hitting zero, which is the conservative number. Fifty units at 52 percent give about 1.83 percent; one hundred units give 0.0334 percent. That drop is why “100 units” became a slogan. The slogan assumes p > 1/2 and even-money steps of plus or minus one. Decimal odds of 1.91 with a 52 percent win chance are not this walk: the up-step is +0.91, not +1. An upper barrier — stop at a profit — would lower the chance of hitting zero; omitting it is conservative for a player who never cashes out.

Assumptions: plus or minus one, constant p, independent trials, infinite opponent, integer units, no simultaneous bets. A leftover 9 on a 1009 roll is discarded by the integer part; it is still 100 units. Safety here means absorption probability, not drawdown comfort. Thorp and later growth-rate writers would also ask about volatility; ruin can be tiny while a weekend 20-unit drawdown is intolerable. The 0.0334 percent is exactly (0.48/0.52)^100 times 100 under the even-money model, and nothing else.