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

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

Quote:TheArchitectI'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).

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

Certainty.

Quote:TheArchitectFor 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.

Quote:AyecarumbaI'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.

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?

Quote:TheArchitectI 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)