Ask The Wizard #259

To clarify for those who might be reading this, in many European casinos, if you place an even money bet on roulette and it lands on zero, that bet gets imprisoned. Should the following spin result in a win for that bet, you will receive your original stake back, but no winnings. If the next spin doesn't favor your bet, your wager is lost.

What occurs to bets that are single-imprisoned if the ball lands on zero? In establishments that only allow for a single imprisonment, the bet would be lost. However, certain casinos permit double imprisonment—here, if a double-imprisoned bet wins, it reverts to being a single-imprisoned bet. If a casino has a rule limiting imprisonment to double, encountering another zero would result in a loss for the double-imprisoned bet. By extension, if a casino accepts triple imprisonment, then a bet once double imprisoned would transition to triple imprisonment upon hitting another zero.

The house edge for scenarios where triple imprisonment is allowed calculates to 1.370120%. Before diving into how I reached this figure, let’s first examine the concepts of single and double imprisonment. Imagine beginning with a wager on red.

Let Z = probability of a zero = 1/37.
Let R = probability of a red number = 18/37.

Single imprisonment

pr(push) = ZR = 0.0131482834.
pr(win) = R = 0.4864864865.
pr(loss) = 1-pr(push)-pr(win) = 0.5003652301.
Expected value = pr(win)-pr(loss) = -0.0138787436.

Double imprisonment

In the case where the initial spin results in a zero, the player doesn't lose but instead may continue to push. This means they can repeat the zero result an unlimited number of times before finally landing on red. To put it differently, the ways to achieve a push are as follows:

ZR, Z(ZR)R, Z(ZR)(ZR)R, Z(ZR)(ZR)(ZR)R, ..

pr(push) = 0.013323464
pr(win) = 18/37 = 0.4864864865.
pr(loss) = 1-pr(push)-pr(win) = 0.5001900494.
Expected value = pr(win)-pr(loss) = -0.0137035629.

Triple imprisonment

First, let’s find the probability p₁Essentially, a single-imprisoned bet can experience a transition to triple imprisonment by facing another two zeros, eventually returning to its original single-imprisoned state. This sequence can unfold as follows:

ZZRR, ZZ(RZ)RR, ZZ(RZ)(RZ)RR, ZZ(RZ)(RZ)(RZ)RR, ..

Put simply, the bet can oscillate between triple and double imprisonment countless times.

Second, let p₂= This is the likelihood that a single-imprisoned wager reaches the first or second imprisonment level and subsequently returns to being single imprisoned.

let p₂= ZR + p₁= 0.013323464.

The player has the ability to loop back from zero to the first imprisonment level infinitely. Thus, the probability of achieving a push is as follows:

ZR + Z p₂R + Z p₂ p₂R + Z p₂ p₂ p₂R + .. =

Z × (1/(1- p₂)) × R = 0.013325830.

pr(push) = 0.013325830.
pr(win) = 18/37 = 0.4864864865.
pr(loss) = 1-pr(push)-pr(win) = 0.5001876839.
Expected value = pr(win)-pr(loss) = -0.0137011974.

Infinite imprisonment

Although you didn’t inquire, Maff, I've heard that in Spain, they allow for infinite imprisonment. Let's denote p as the probability of a push, which also describes the likelihood of commencing from a certain level of imprisonment, dropping deeper into subsequent levels, and then eventually ascending back to that starting point.

p = ZR + ZpR + ZppR + ZpppR + ..

p²- p + ZR = 0

By the Quadratic Formula p = (1-(1-4*RZ)^1/2)/2 = 0.0133258620.

pr(push) = 0.0133258620.
pr(win) = 18/37 = 0.4864864865.
pr(loss) = 1-pr(push)-pr(win) = 0.5001876515.
Expected value = pr(win)-pr(loss) = -0.0137011650.

Zeros ignored imprisonment

Lastly, in some casinos, zeros are ignored after the first zero that causes imprisonment. In such cases, the probability of reaching a push is simply calculated as (1/37)×(1/2) = 0.0135135135.

The table below outlines the four distinct types of rules we are discussing.

A heartfelt thank you to ChesterDog and weaselman for their assistance with the math involved. Additionally, my gratitude extends to professor G. Artico and polarprof.it for their contribution to the HTML representation of the summation formula provided above.

This inquiry was brought up and deliberated within the discussion forum on my affiliated site. Wizard of Vegas .

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This inquiry was brought up and deliberated within the discussion forum on my affiliated site. Wizard of Vegas .

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This inquiry was brought up and deliberated within the discussion forum on my affiliated site. Wizard of Vegas .