Blackjack Bonus Wagering Strategy

Worked Solution

How to Think About It: This is a classic wagering requirement problem that shows up in sports betting promotions and casino bonuses. The key tension: you got $1000 of free money, but you have to churn through $10,000 of bets on a negative-EV game to unlock it. The question is whether the bonus is large enough to overcome the house edge you'll lose while meeting the turnover requirement -- and, more subtly, whether the amount you actually end up wagering is under your control. It is: busting early truncates the churn, and that changes the expected-value answer.

Quick Estimate: The house edge per dollar bet is $0.55 - 0.45 = 0.10$, i.e., you lose 10 cents per dollar wagered in expectation. If you complete the full $10,000 turnover requirement, you expect to lose roughly $10,000 \times 0.10 = \$1,000$. But you started with a $1,000 bonus, so your expected profit is approximately $\$1,000 - \$1,000 = \$0$. This is roughly break-even -- *if you always survive to complete the full turnover*. Bet sizing changes how much you expect to wager at all, and that is where the strategy question gets interesting.

Approach: Use the optional-stopping identity $E[\text{final bankroll}] = \text{starting bankroll} - \text{edge} \times E[\text{total wagered}]$, then compare bet-sizing policies on both expected profit and ruin risk.

Formal Solution:

The key identity. Every dollar wagered has expected P&L $-0.10$, regardless of bet size or sequencing. So if $W$ denotes the total dollars you wager before you stop (either by clearing the turnover or by busting), optional stopping gives:

$$E[\text{final bankroll}] = 2{,}000 - 0.10 \, E[W], \qquad E[\text{profit}] = 1{,}000 - 0.10 \, E[W],$$

since $1,000 of the starting $2,000 was your own deposit. Expected profit is maximized by minimizing $E[W]$ -- and $W$ reaches $10,000 only on paths where you survive; on paths where you bust, $W$ stops early.

Minimum betting: variance-minimizing, EV $\approx 0$. If you bet $b$ per hand, you need $N = 10{,}000/b$ hands. Expected loss is $1,000 regardless of $b$. The per-hand P&L is $\pm b$, with variance $b^2\left[1 - (0.10)^2\right] = 0.99\,b^2$, so the total P&L has variance $\approx 0.99 \times 10{,}000 \times b = 9{,}900\,b$ and standard deviation $\approx 99.5\sqrt{b}$.

With $b = 1$: standard deviation $\approx \$100$ against a $\$2,000$ bankroll and $\$1,000$ expected loss -- you essentially never bust. Then $E[W] = 10{,}000$ exactly, so $E[\text{profit}] \approx \$0$: min-betting locks in break-even with near-certainty. By the law of large numbers your realized loss concentrates tightly around $\$1,000$, and you reliably cash out close to your original $\$1,000$ deposit.

Aggressive betting: EV-maximizing. Busting is not just a risk here -- it is the only way to escape the mandatory negative-EV churn. On paths where you bust early, you stop wagering, so $E[W] < 10{,}000$ and $E[\text{profit}] > 0$.

Concrete example -- an all-in ladder (each bet is the full bankroll, capped at the remaining turnover): bet $\$2,000$; if you win, bet $\$4,000$; if you win again, bet the final $\$4,000$ needed to reach $\$10,000$ of turnover.

• $E[W] = 2{,}000 + 0.45 \times 4{,}000 + 0.45^2 \times 4{,}000 = \$4{,}610$
• $E[\text{final}] = 2{,}000 - 0.10 \times 4{,}610 = \$1{,}539$, so $E[\text{profit}] = +\$539$
• Outcome distribution: bust with probability $0.55 + 0.45 \times 0.55 \approx 0.80$ (forfeit everything, including your $\$1,000$ deposit); finish with $\$4,000$ w.p. $\approx 0.111$; finish with $\$12,000$ w.p. $\approx 0.091$.

To maximize expected profit, bet large. Aggressive sizing cuts expected turnover from $\$10,000$ to under $\$5,000$, converting the promotion from break-even to solidly positive expected value -- at the price of walking away with nothing about $80\%$ of the time.

Part 3 -- how ruin probability factors in. Ruin plays opposite roles for the two objectives:

• For *expected profit*, ruin is helpful: it truncates a wagering obligation whose every dollar costs you 10 cents in expectation. Bigger bets mean more ruin risk, lower $E[W]$, and higher EV.
• For *realized outcomes*, ruin is catastrophic: you forfeit your own $\$1,000$ deposit. Minimum bets drive ruin probability to essentially zero and let the law of large numbers deliver the $\approx \$0$ EV with tiny variance.

Intermediate bet sizes interpolate: as bets grow, EV rises and survival probability falls. The right choice depends on risk preference -- a risk-neutral player bets as big as the house allows; a risk-averse player grinds minimum bets for a near-certain wash.

Answer: (1) The promotion is worth taking only if you are willing to gamble: played with minimum bets it is roughly break-even ($EV \approx \$0$); played aggressively it has genuinely positive expected value. (2) To maximize expected profit, bet large: by optional stopping, $E[\text{profit}] = 1{,}000 - 0.10 \times E[\text{total wagered}]$, and busting early on unlucky paths truncates the turnover, so $E[W] < 10{,}000$ -- e.g., the all-in ladder yields $E[\text{profit}] \approx +\$539$, versus $\approx \$0$ for minimum bets, which force $E[W] = 10{,}000$. (3) Ruin risk is the price of that EV: the ladder busts $\approx 80\%$ of the time, while minimum betting makes ruin negligible and reliably returns roughly your $\$1,000$ deposit. Bet sizing is a dial between EV-maximization (big bets) and variance/ruin-minimization (minimum bets).