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.
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.
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.
P = 1/37(18/37 + 18/37*P) = 18/37^2 * (P+1)
Solve this for P:
Is this wrong?
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.
...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.
No, never mind, that can't be right.
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.