Wizard
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Wizard
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September 9th, 2010 at 3:32:28 PM permalink
In Europe, as you may know, if the player makes an even money bet, and the ball lands in zero (on a single-zero wheel), the player may sometimes choose to "imprison" his bet. If the player chooses to imprison the bet, and it would otherwise win on the next spin, then the bet is returned as a push. If the imprisoned bet loses, the bet is lost. Let's call a bet in this state, after a zero, "single imprisoned."

However, if there is a second consecutive zero, then it becomes "double imprisoned." Then a win will move up the bet up a level, back to being single imprisoned. A loss always causes the bet to lose. Another zero would cause the bet to move down a level to "triple imprisoned."

Consider a bet on red, as an example. Let R=red outcome, Z=zero outcome.

Spin 1: Z, bet is single-imprisoned.
Spin 2: Z, bet is double-imprisoned.
Spin 3: R, bet is single-imprisoned.
Spin 4: Z, bet is double-imprisoned.
Spin 5: Z, bet is triple-imprisoned.
Spin 6: R, bet is double-imprisoned.
Etc...

The question is what is the probability of getting a push with infinite levels of imprisonment possible? I've heard casinos in Spain allow for infinite imprisonment, by the way, so this isn't just a theoretical question.
It's not whether you win or lose; it's whether or not you had a good bet.
BigTip
BigTip
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September 9th, 2010 at 4:29:40 PM permalink
Mirage use to have a single zero European ruled table that had the en prison rule.
mkl654321
mkl654321
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September 9th, 2010 at 5:15:05 PM permalink
How could you ever reach a state where the imprisonment was "infinite"? No matter what, there would always be a POSSIBLE sequence of spins that would get the bet out of prison.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
BigTip
BigTip
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September 9th, 2010 at 5:21:10 PM permalink
I don't think the meaning of infinite here means "forever". I think the meaning here means "unlimited". Blackjack tables usually let you split pairs four times. They could change the rule to an "infinite" number of times.
dwheatley
dwheatley
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September 9th, 2010 at 6:26:07 PM permalink
I think the question being asked is: what is the probability of a push, if there is no limit to the 'level' of imprisonment.

While there may be an easier way, I know a Markov analysis would answer this one. We need two absorption states, call them Success & Failure. let states 1 to n represent the imprisonment level, then consider the following transition matrix:

s F 1 2 3 4
-------------------------
S [ 1 0 0 0 0 ...

F [ 0 1 0 0 0 ...

1 [ 18/37 18/37 0 1/37 0 ...

2 [ 0 18/37 18/37 0 1/37 ...

3 [ 0 18/37 0 18/37 0 1/37

...

I think we see a pattern develop. If you find the limiting probabilities of this sucker, you can get the probability of success, which is the probability of a push. I have a computer program on another computer I'll visit next week, if no one finishes the work I'll try to post the answer.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
weaselman
weaselman
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September 9th, 2010 at 6:34:00 PM permalink
If the first roll is zero (1/37), we get a push if either the next roll is red (18/37) or if the next sequence of rolls pushes (P), and then there is another red (18/37):
P = 1/37(18/37 + 18/37*P) = 18/37^2 * (P+1)


Solve this for P:
P=18/(37^2-18)=0.0133234641

Is this wrong?
"When two people always agree one of them is unnecessary"
Wizard
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Wizard
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September 9th, 2010 at 7:08:06 PM permalink
Quote: BigTip

I don't think the meaning of infinite here means "forever". I think the meaning here means "unlimited". Blackjack tables usually let you split pairs four times. They could change the rule to an "infinite" number of times.



Right, I meant what is the probability of a push from the initial state.
It's not whether you win or lose; it's whether or not you had a good bet.
ChesterDog
ChesterDog
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September 9th, 2010 at 7:23:28 PM permalink
Quote: dwheatley

...what is the probability of a push, if there is no limit to the 'level' of imprisonment.



I get 0.01332586 as the probability of a push from the initial state of a bet on red.
Wizard
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Wizard
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September 9th, 2010 at 8:23:57 PM permalink
With a lot of hand-waiving math I'm getting 18/595=1.3899614%. I'm not very confident at this point. A telescoping sums solution seems to be evading me.

No, never mind, that can't be right.
It's not whether you win or lose; it's whether or not you had a good bet.
MathExtremist
MathExtremist
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September 9th, 2010 at 9:20:37 PM permalink
My initial instinct is to model this as a recursive grammar:
Terms:
z = zero, or down one level
r = red, or up one level
b = black, or immediate loss.
And unless I'm mistaken, the generator for all push/success scenarios
A -> zA*r

I haven't determined whether there's a straightforward probability model for that yet, however.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563

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