TheArchitect
TheArchitect
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October 20th, 2010 at 10:30:11 AM permalink
I'm not touting this betting progression as a winner, just simply asking what is the probability of it losing? I know in the long run, it will lose more or less the house edge of whatever game it's applied to, however I just want to know.

This progression is called the "Carsch Progression" and is named from the user who created it, and first posted it on Gamblers Glen. It's a Star System/Fibo variant.

For every loss, you move down the progression, for every win, you repeat the last bet. Win two in a row and start over. WLWL doesn't hurt you.. So how do you calculate the probability of losing the progression when ostensibly, you could play 1000s of decisions and neither win nor lose?

Progression: 1,1,2,3,4,6,9,13,21,32

-TheArchitect
Wizard
Administrator
Wizard
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October 20th, 2010 at 10:40:00 AM permalink
It would depend on how "losing" is defined. Whatever the answer, the answer is probably best arrived at by random simulation.
It's not whether you win or lose; it's whether or not you had a good bet.
TheArchitect
TheArchitect
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October 20th, 2010 at 10:44:17 AM permalink
Thank you for responding Wizard.

Losing would be defined by losing at the 32unit bet.

I can calculate the probability of losing all 10 bets in a row, however it gets tricky when you add in non progression clearing wins, and double wins to clear. I agree, a simulation would likely solve it rather easily.. Of course I'm one of the simulation challenged members of your great forum.

-TheArchitect
TheArchitect
TheArchitect
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October 20th, 2010 at 11:27:02 AM permalink
Does anyone out there have an easy simulation for this? -TheArchitect
weaselman
weaselman
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October 20th, 2010 at 12:08:31 PM permalink
Quote: TheArchitect

I'm not touting this betting progression as a winner, just simply asking what is the probability of it losing? I know in the long run, it will lose more or less the house edge of whatever game it's applied to, however I just want to know.

This progression is called the "Carsch Progression" and is named from the user who created it, and first posted it on Gamblers Glen. It's a Star System/Fibo variant.

For every loss, you move down the progression, for every win, you repeat the last bet. Win two in a row and start over. WLWL doesn't hurt you.. So how do you calculate the probability of losing the progression when ostensibly, you could play 1000s of decisions and neither win nor lose?

Progression: 1,1,2,3,4,6,9,13,21,32

-TheArchitect



Well, I am sure I have made a bunch of technical mistakes below, so, the final answer is probably off, but here is the idea.

Let p be the probability of a single loss, and f(k) the probability of a loss given that you are k steps from the end, and N be the total number of steps (10 in this case).

Then:
f(1) = p + (1-p)^2*f(N)
f(2) = p*f(1) + (1-p)^2*f(N)
...

f(n) = p*f(n-1) + (1-p)^2*f(N)
...

Let F=(1-p)^2*f(N) just for brevity:

f(n) = p*f(n-1) + F = p^2*f(n-2)+pF + F = p^3*f(n-3) + p^2*F + p*F + F = ... = p^(n-1)*f(1) + F*sum([k=0;n-2] p^k)

f(n) = p^n + F*(1-p^(n-2))/(1-p) = p^n + f(N)*(1-p)*(1-p^(n-2))

So F(N) = p^N/(1 - (1-p)*(1-p^(N-2)))

In particular F(10) = p^10/(1-(1-p)*(1-p^8))

For example, if p=0.51, then F(10) = 0.002324 (take this with a grain of salt, because, like I said, I dunno how many errors I made along the way, but the basic idea is, I think, correct).
"When two people always agree one of them is unnecessary"
mkl654321
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October 20th, 2010 at 12:10:21 PM permalink
Quote: TheArchitect

I'm not touting this betting progression as a winner, just simply asking what is the probability of it losing?



Certainty.
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
Ayecarumba
Ayecarumba
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October 20th, 2010 at 2:32:44 PM permalink
Quote: TheArchitect

For every loss, you move down the progression, for every win, you repeat the last bet.



I'm not sure I understand how you move up the progression, if you repeat the last bet with every win.
Simplicity is the ultimate sophistication - Leonardo da Vinci
MathExtremist
MathExtremist
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October 20th, 2010 at 3:57:52 PM permalink
Quote: Ayecarumba

I'm not sure I understand how you move up the progression, if you repeat the last bet with every win.



That was going to be my question. On the other hand, the sequence itself seems very similar to the Fibonacci sequence, and I've had some very, very big wins playing a straight positive progression on the Fibonacci.

One of the interesting things about the Fibonacci is that, when applied to a streak of winning bets, the rate of wager increase and bankroll increase are equal. Contrast that with a 100% parlay where your wager increase is maximal but you always lose the unit you started with unless you hit the table max.
"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
TheArchitect
TheArchitect
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October 20th, 2010 at 7:44:13 PM permalink
I should be more clear with the rules.

When you play this progression and lose a wager, you move to the next level. When you win at any level (except the first 1unit bet) you will have to repeat that beat and win to clear the progression (similar to the fibo where you have to win two consecutive bets to clear). If you lose the "repeat" bet, you do not progress.. So say you make it to the 3 unit bet and win, your next bet is 3 units and lose, you do not progress down, you repeat the 3 unit bet again.. This is where I'm having a difficult time simulating this progression..

In theory, you could find yourself in a WLWL situation for a large number of bets without winning or losing..

Any suggestions?
weaselman
weaselman
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October 21st, 2010 at 5:22:56 AM permalink
Quote: TheArchitect

I should be more clear with the rules.

When you play this progression and lose a wager, you move to the next level. When you win at any level (except the first 1unit bet) you will have to repeat that beat and win to clear the progression (similar to the fibo where you have to win two consecutive bets to clear). If you lose the "repeat" bet, you do not progress.. So say you make it to the 3 unit bet and win, your next bet is 3 units and lose, you do not progress down, you repeat the 3 unit bet again.. This is where I'm having a difficult time simulating this progression..

In theory, you could find yourself in a WLWL situation for a large number of bets without winning or losing..

Any suggestions?


So, what is a "win" then? You got it that you "lose" when you get to the end of the sequence, but if you don't get there, when do you stop, and say that you won? After any two wins in a row?

If so, the probability of losing one bet is:
f1 = p + (1-p)*p*f1
(either you lose straight away (p) or you win (1-p), then lose (p), and then start over (f1)).
Solve it for f1:

f1 = p/(1-p(1-p))

Your sequence is 10 steps long, so the probability of getting to the end of it before returning to the start f1^10, or (p/(1-p(1-p)))^10 (about 2.1% if p =0.51)
"When two people always agree one of them is unnecessary"

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