I need clarification

Please help. I know the odds don't change, and the casino has the advantage. But I've read this progression before which seems to think it can make the odds favorable. I know that's not going to happen, but don't know how to find out why it's wrong. I will paste the progression here:

It takes 2 wins (repeat or parlay) to reset the progression cycle to step 1. It also takes a single loss (parlay) or single NET loss (repeat) to advance to the next step.

. . . . Bet . . . . Loss . . . Win
1) 5 repeat . . . .-5 . . . +10
2) 5 parlay . . . -10 . . . +10
3) 10 repeat . . -20 . . . +10

4) 10 parlay . . -30 . . . +10
5) 20 repeat . . -50 . . . +10
6) 20 parlay . . -70 . . . +10

Now, the Wizard Of Odds says, ignoring pushes, you have a 52.5% chance of losing any hand. You’ve also got a 47.5% chance of winning. The odds of winning 2 hands in a row are about 1 in 4.5 tries. Or twice in 9 tries.

Looking at the chart of the progression, you can say there is a 2/3 chance of any progression cycle ending with a win in the first 3 steps. That’s because 3 tries (steps) times 1 in 4.5 tries, = 3/4.5 = 2/3.

Another way to say this is that on average, of every 3 cycles played, 2 will end in steps 1-3 (with a win) and the 3rd will continue on to steps 4-6. So for 9 cycles played, you’ll have 6 wins in steps 1-3 and 3 cycles continue on to steps 4-6.

Same thing goes for the 3 of 9 cycles that reach steps 4-6. You’ll have 2 wins in steps 4-6, and one time you’ll lose step 6 and the progression cycle ends with a loss.

Totaled up, for each 9 cycles you play, you’ll have 8 wins (+$80) and 1 loss (-$70) for a net gain of $10.

What happened to the “house edge”?

The odds of losing 6 hands in a row are about 1 in 48 tries. The odds of winning 2 hands in a row are 1 in about 4.5 tries. 48/4.5 = just over 10. So the odds of losing 6 hands in a row before winning 2 in a row are about 1 in 10. Since each step pays $10 for a win, on average you should win $90 and lose $70 for every 10 cycles you play. That’s a net gain of $20.

Well, you are trying at least to take a mathematical approach. Everyone is assuming knows there is a flaw in your thinking. This will be frustrating for you, as people realize that they have to tediously study what you are trying to lay out in your own fashion. That fashion seems to be unique. It sure is Greek to me.

Apparently you're trying to re-invent the wheel here, this math has already been done - the old math. Your "new math" isn't going to cut it, looks like to me.

You do seem to be an intelligent person. The sagest advice I have seen in this site I will repeat to you, ala "word to the wise". When it comes to looking at betting systems: Yes, various combinations of bets sometimes have you "wondering". But the math you need to do isn't some convoluted homegrown proof, instead you need to demonstrate that it is possible to defeat the Expected Value of a bet. That it is undefeatable is the underlying money machine of every casino in the world, who, although not knowing much about math themselves, will let you make any bet combination you like. Eagerly so. Go right ahead and bet Right-side and the Don't side at the same time right in front of them. Hedge all your bets. Come up with fanasticly complicated parlay combinations. Martingale yourself to the max [they do mean the max tho]. The casino can just sit back and know their Golden Rule. No Combination Beats the EV of Each Bet.

This is a modified Martingale.

The trouble with Martingales is that when you win, aster risking a bunch of money, you're only up one unit. This system has the advantage of continuing so that when it succeeds, you're up a good few bucks.

The Martingale looks attractive because its hard to believe you can have X losses in a row. Ditto for this system.

Note: in both systems, failure comes when the losing streak causes the next bet to exceed you bankroll, or the table limit.

It happens more often that you can imagine.

Quote: tommymarks

It takes 2 wins (repeat or parlay) to reset the progression cycle to step 1. It also takes a single loss (parlay) or single NET loss (repeat) to advance to the next step.

. . . . Bet . . . . Loss . . . Win
1) 5 repeat . . . .-5 . . . +10
2) 5 parlay . . . -10 . . . +10
3) 10 repeat . . -20 . . . +10

4) 10 parlay . . -30 . . . +10
5) 20 repeat . . -50 . . . +10
6) 20 parlay . . -70 . . . +10

The odds of losing 6 hands in a row are about 1 in 48 tries. The odds of winning 2 hands in a row are 1 in about 4.5 tries. 48/4.5 = just over 10. So the odds of losing 6 hands in a row before winning 2 in a row are about 1 in 10. Since each step pays $10 for a win, on average you should win $90 and lose $70 for every 10 cycles you play. That’s a net gain of $20.

I may be reading your progression incorrectly but it seems that you may be overlooking the results of the first hand played. A win on this hand will result in a $5 gain, not $10. Assuming a $5 wager, the total amount lost after a second loss (the first repeat) is now $10. In order to come out $10 ahead, you'd now have to place a $20 wager on the third hand and win. It seems your wager amounts farther along in the progression are much too low to achieve the win amount you'd need to get ahead at that point in the progression.

Again, I may be missing something in your description but that seems to be the error from what I can see - your math is wrong on the amount won and wagered in your progression.

Example of play:

1) bet 5, if win 'repeat' bet of 5 for a total +10 if win; if second (repeat) looses you are even and begin all over again; if first 5 loss then go to:
2) bet 5, if win 'parlay' for a bet of 10, if win +15 -5(from first step) for total +10; if loss on either bet you are down 10 total then go to:
3) bet 10, if win 'repeat' bet of 10, if win +20 -10(from first 2 steps) for total +10; if second (repeat) looses you begin step 3 again; if first 10 loss then go to:
4) CONTINUE THE PROGRESSION

A quick note to highlight in the original post:
"""It takes 2 wins (repeat or parlay) to reset the progression cycle to step 1. It also takes a single loss (parlay) or single NET loss (repeat) to advance to the next step."""

Quote: tommymarks

Totaled up, for each 9 cycles you play, you’ll have 8 wins (+$80) and 1 loss (-$70) for a net gain of $10.

Just curious. How is this system working out for you at the craps tables?

The immediate jump to logical fallacy is the reason there are casinos. People are easily fooled by them and they permeate all the news shows today. It's a Hell of a great way to win an argument though.

Shit, saw a lady parked at the church today. Her car had a "Put the Christ back in Christmas" sign and me and Jesus took a bigger look and both thought put the handicapped back in the handicapped spot you lazy Bitch.

Quote: tommymarks

Please help.
(a...z)
Totaled up, for each 9 cycles you play, you’ll have 8 wins (+$80) and 1 loss (-$70) for a net gain of $10.

I see it closer to 7 wins (+$70),
1 loss (-$70) and 1 push ($0) for a net gain of $0 without doing an exact calculation.

You do have lots of rounding going on here in your math calcs.
Why is that?
Not a proper way to calculate the expected value and house edge of any gambling betting system, IMHO.
Did you just simulate this instead
to see if your sims matched your about values?

Did you forget to account for the WL sequence? (probability)
WW = +$10 (0.225625)
LL = -$10 and now to level 3 bet (0.275625)
LW = $0 and a $10 bet at level 2 (0.249375)
WL = $0 net win (0.249375)
start over
"if second (repeat) looses you are even and begin all over again"
Good Luck