Boozer’s Delight Craps System: Math and Risk
Hedge system · Exact outcome table · Expected-loss review
Boozer’s Delight is an old low-volatility craps hedge: $5 Pass Line, $15 Don’t Pass, then $10 in Pass Line Odds after a point. It can produce many zero or small-positive point outcomes, but the come-out roll leaves a net expected loss.
Sam’s verdict
The system is useful as a math lesson in hedging, not as a way to beat craps or obtain “free” casino drinks. It reduces many resolved-point swings by holding opposing contract bets, but you still pay the combined expected cost of the Pass and Don’t Pass wagers. If you want to watch a table, watching without betting is cheaper.
How the Boozer’s Delight system is set up
- Before the come-out roll, bet $5 on Pass Line.
- At the same time, bet $15 on Don’t Pass.
- If 4, 5, 6, 8, 9, or 10 establishes a point, take $10 Odds behind the Pass Line.
- Leave both flat contract bets until the point or 7 resolves the cycle.
Terminology correction: the $10 behind Pass Line is taking Odds, not laying Odds. This analysis assumes the common bar-12 Don’t Pass rule, where a come-out 12 pushes.
Every possible net result
| Outcome | $5 Pass | $15 Don’t Pass | $10 Pass Odds | Net |
|---|---|---|---|---|
| Come-out 7 or 11 | +$5 | −$15 | Not placed | −$10 |
| Come-out 2 or 3 | −$5 | +$15 | Not placed | +$10 |
| Come-out 12 | −$5 | Push | Not placed | −$5 |
| Point 4 or 10 wins | +$5 | −$15 | +$20 | +$10 |
| Point 5 or 9 wins | +$5 | −$15 | +$15 | +$5 |
| Point 6 or 8 wins | +$5 | −$15 | +$12 | +$2 |
| 7 before any point | −$5 | +$15 | −$10 | $0 |
The table explains the system’s appeal: once a point exists, a seven-out breaks even and a made point produces a small profit. The cost is concentrated on the come-out 7, 11, and barred 12.
Expected loss per completed cycle
The easiest verification is to add the expected losses of the two flat bets:
- $5 Pass Line × 1.414% ≈ $0.0707 expected loss.
- $15 Don’t Pass × 1.364% ≈ $0.2045 expected loss.
- $10 true Odds has 0% house edge.
Total expected loss is approximately $0.275 per completed come-out/point cycle. An exact outcome-weighted calculation gives −109/396 dollars, also about −$0.275. Small individual results do not remove that negative expectation.
Why Pass and Don’t Pass do not cancel perfectly
The bets are not exact opposites on the come-out roll. Under bar 12, Pass loses while Don’t Pass pushes on 12. The unequal $5/$15 sizing and the added Pass Odds then create the point-specific profits shown above. Hedging changes the distribution of results; it does not create a player advantage.
If a table bars 2 instead of 12, the Don’t Pass exception moves to 2 and the outcome table must be adjusted. Always read the printed bar rule.
Bankroll, table minimums, and practical problems
- The system requires the casino to accept simultaneous Pass and Don’t Pass wagers at those amounts.
- A higher table minimum forces the entire construction upward.
- The $10 Odds amount produces clean standard payouts for every point.
- Side bets and tips add cost. Larger true Odds change the outcome table and swings but add no expected loss when paid exactly; the flat-bet cost remains.
- More cycles per hour increase expected dollar loss over time.
About the name and “free drinks” story
The legacy name came from the idea that a player could occupy a table with limited swings while receiving casino beverage service. Drinks, comps, service rules, and required play are never guaranteed. Alcohol also weakens bankroll discipline and judgment. Do not use a negative-expectation betting system as a reason to drink more or stay longer.
If your real goal is to learn the flow, use the free CrapsPit practice game or watch a live table without wagering. For the underlying contract-bet math, read the Pass Line guide.
What changes if Pass Odds rise from $10 to $20?
Keep the $5 Pass and $15 Don’t Pass flats. With $20 Pass Odds, a made 4/10 now gives $30 net profit, a made 5/9 gives $20, and a made 6/8 gives $14. A seven-out gives $10 net loss: the $10 net flat gain is offset by losing $20 Odds. Come-out results are unchanged because Odds have not yet been placed.
For point 6 alone, the $10-Odds version averages (5 × $2 + 6 × $0) ÷ 11 = $10/11 after the point is established. The $20-Odds version averages (5 × $14 − 6 × $10) ÷ 11 = the same $10/11. This favorable conditional mean does not include the preceding come-out risk. Across complete cycles, the negative flat-bet expectation remains; the larger Odds merely redistribute point-stage gains and losses.
