The house edge with En Prison is commonly 1.35% on qualifying even-money bets, but that number is not automatic. It depends on what the table does when zero appears again while the bet is imprisoned. If repeat zero keeps the wager in prison, the edge is $1/74$, or about 1.351%. If the next zero loses the whole imprisoned wager, the edge rises to $19/1369$, or about 1.388%.
That difference is small, but it is real. “En Prison” describes a settlement process, not one globally identical rule. Read the posted rule before assigning a percentage.
This page focuses on the mathematics. For the physical dealer procedure and which bets qualify, see the separate En Prison rule guide.
Start with ordinary single-zero roulette
A single-zero wheel has 37 pockets:
- 18 red;
- 18 black;
- one green zero.
The same count applies to the other even-money pairs: odd/even and 1–18/19–36. Zero belongs to neither side.
On a standard even-money bet without a favorable zero rule:
$$ EV=\frac{18}{37}(+1)+\frac{18}{37}(-1)+\frac{1}{37}(-1) $$
The red and black terms cancel:
$$ EV=-\frac{1}{37}=-0.027027 $$
So the standard house edge is:
$$ \text{House edge}=\frac{1}{37}=2.7027% $$
A $100 bet therefore has an expected loss of about $2.70, although any individual spin still ends with a $100 win or $100 loss.
En Prison changes only the zero branch. The 18 winning numbers, 18 losing numbers, and even-money payoff remain the same.
The common recursive version: 1.351%
Under a favorable recursive version:
- You place $1 on red.
- Zero lands, so the $1 is marked En Prison.
- If red lands on a later spin, the $1 stake is returned with no profit.
- If black lands, the $1 is lost.
- If zero lands again, the bet remains imprisoned and waits for another spin.
Once the first zero has occurred, later zeros only delay the decision. Ignore those delays and look at the first future non-zero result. Among the 36 non-zero pockets, 18 release the stake and 18 lose it.
The conditional probability of losing the imprisoned bet is therefore:
$$ P(\text{loss after prison})=\frac{18}{18+18}=\frac{1}{2} $$
The conditional expected loss after the original zero is half a unit:
$$ E(\text{loss}\mid 0)=\frac{1}{2} $$
Zero occurs on the original spin with probability $1/37$, so:
$$ \text{House edge}=\frac{1}{37}\times\frac{1}{2} $$
$$ \text{House edge}=\frac{1}{74}=1.35135% $$
That is the familiar En Prison figure. It is mathematically the same expected loss as La Partage, where the table immediately takes half of an even-money wager when zero lands.
The timing differs. La Partage settles immediately. En Prison carries the original stake into one or more later spins.
A second-zero loss changes the answer
Some authorized rules are less favorable. Colorado’s roulette regulation, reproduced in 1 CCR 207-1-22, allows the player to take back half after zero or leave the whole bet in prison for another spin. Under that rule, if the following spin is another zero, the entire bet is lost.
After the first zero, the next spin has:
- 18 outcomes that release the stake: net result 0;
- 18 opposite-side outcomes that lose the stake: net result $-1$;
- one zero that also loses the stake: net result $-1$.
The conditional expected value of choosing prison is:
$$ EV(\text{prison}\mid 0)=\frac{18}{37}(0)+\frac{19}{37}(-1) $$
$$ EV(\text{prison}\mid 0)=-\frac{19}{37} $$
The original spin reaches that branch only once in 37 spins:
$$ EV=\frac{1}{37}\times\left(-\frac{19}{37}\right) $$
$$ EV=-\frac{19}{1369}=-1.38787% $$
The house edge is therefore about 1.39%, not exactly 1.35%.
Take half or leave it in prison?
Under the Colorado-style second-zero-loss rule, the player may choose between:
- taking half the stake back immediately, for a certain loss of 0.5 units; or
- imprisoning the full stake, with a conditional expected loss of $19/37\approx0.5135$ units.
Compare the two conditional values after zero:
$$ EV(\text{take half})=-0.5 $$
$$ EV(\text{leave in prison})=-\frac{19}{37}\approx-0.5135 $$
Because $-0.5$ is greater than $-0.5135$, taking half back has the better expectation under that exact rule.
The difference is:
$$ \frac{19}{37}-\frac{1}{2}=\frac{1}{74}\approx1.351% $$
That is 1.351% of the original wager conditional on already having hit zero. On a $100 imprisoned decision, taking half back is worth about $1.35 more in expectation than choosing prison under the second-zero-loss rule.
This is a good example of why the rule text matters. The word “En Prison” alone is not enough to identify the best choice.
Three zero treatments, three values
| Treatment after initial zero | What happens on a repeat zero? | Edge on qualifying even-money bets |
|---|---|---|
| La Partage | No second spin; half is lost immediately | 1/74 = 1.351% |
| Recursive En Prison | Bet stays imprisoned again | 1/74 = 1.351% |
| One-spin prison, second zero loses | Whole imprisoned stake is lost | 19/1369 = 1.388% |
| Ordinary single-zero roulette | Initial zero loses the whole bet | 1/37 = 2.703% |
Other procedures may exist. A table could impose a maximum prison duration, treat a repeat zero as another half loss, or use a different approved settlement. The percentage must be derived from that branch structure rather than copied from a generic roulette chart.
The wider French roulette rules guide explains how En Prison and La Partage fit into French-style play.
Worked session comparison
Suppose a player makes 500 separate $20 even-money wagers. Total original action is:
$$ 500\times$20=$10{,}000 $$
Expected loss under ordinary single-zero settlement:
$$ $10{,}000\times\frac{1}{37}=$270.27 $$
Expected loss with recursive En Prison or La Partage:
$$ $10{,}000\times\frac{1}{74}=$135.14 $$
Expected loss with a second-zero-loss prison rule:
$$ $10{,}000\times\frac{19}{1369}=$138.79 $$
| Rule | Expected loss on $10,000 original action |
|---|---|
| Standard single-zero | $270.27 |
| Recursive En Prison | $135.14 |
| Second-zero-loss En Prison | $138.79 |
The difference between the two prison versions is only $3.65 across $10,000 of original wagers. The larger difference is between either favorable zero rule and ordinary single-zero settlement.
Expected loss is not a forecast of the session result. A player can finish ahead or lose far more than these amounts because even-money roulette has high short-term variance relative to its small average edge.
What counts as “action” when a bet is imprisoned?
The standard house-edge calculation uses the original amount risked as its denominator:
$$ \text{House edge}=\frac{\text{expected loss}}{\text{original wager}} $$
An imprisoned stake is unresolved, but it is not usually treated as a fresh additional wager each time zero delays settlement. Calling every prison spin another $20 of action would overstate the amount newly committed.
Time is still affected. Under recursive En Prison, an original bet occasionally requires extra spins before it is released or lost. The expected number of wheel spins needed to reach a non-zero decision after imprisonment is:
$$ E(S)=\frac{1}{36/37}=\frac{37}{36}\approx1.0278 $$
Here $S$ counts spins after the bet enters prison. Most imprisoned bets settle on the first following spin; repeated zeros create the small extra delay.
That delay matters operationally because the dealer must preserve the bet’s status, prevent unauthorized changes, and settle it separately from new action.
Why En Prison helps only selected bets
The rule generally applies to:
- red or black;
- odd or even;
- low 1–18 or high 19–36.
It does not normally reduce the edge on straight-up numbers, splits, streets, corners, six-lines, dozens, or columns. Those bets still lose on zero under their ordinary rules.
This creates an important distinction:
- single-zero roulette with En Prison on even-money bets: about 1.35% to 1.39% on those qualifying wagers, depending on repeat-zero treatment;
- other bets on the same wheel: normally 2.70% unless another specific rule changes them.
A sign advertising French Roulette does not automatically prove every outside bet receives the most favorable treatment. Check the layout and house rules.
En Prison changes expectation, not prediction
The lower edge does not make red more likely after zero. The next spin still has 18 red pockets, 18 black pockets, and zero on a balanced single-zero wheel. The benefit comes from the contract: the casino does not collect the full original stake immediately on the first zero.
No staking progression changes this calculation. Doubling after a loss, repeating a color, following wheel sectors, or using previous-spin patterns changes wager size or selection but not the expected value of the approved payout rule.
For standard wheel probabilities, read roulette odds. For the cost of other bet families, use the main roulette house edge guide or the roulette odds calculator.
The practical conclusion is precise: quote 1.35% only after confirming that the repeat-zero procedure supports it. If the second zero loses the whole imprisoned bet, the correct edge is about 1.39%. Both are much better than the ordinary 2.70% single-zero edge, but neither gives the player a positive expectation.