Quote:GreenZeroI was also looking at this progression with roulette betting 1 unit each on the 1st&2nd dozens and then betting on the 2nd&3rd dozens. Since your chances of hitting are about 66% it could work beautifully with this progression since your looking for 2 wins in a row to reset back to the beginning. I haven't done a small simulation on this yet though as a progression system involving 2 dozens get expensive really fast.

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It wouldn't be a carsch.

The idea is to win two in a row. Betting two of three dozens means you lose twice as much per loss and win half as much per win.

Let's say you lose, 1,2,4,7. That's 14×2 so you just lost 28.

And now you win the next two. That's win 22-11 and again the second spin so you made back only 22 units after losing 28.

You would have to win three in a row. And as the progression of losses goes on then the number of winning spins needed goes up.

What you should analyze is the opposite.

Betting just one dozen using the carsch because you only need one win instead of two to profit.

Lose 1,2,4,7 for a fourteen unit loss. Then win 22 on the fifth spin.

Or perhaps a combo carsch. For example if you win the fifth you make a second spin be two dozens by splitting the win.

Example: Lose first four. 1,2,4,7. Then win 22 and divide that for to get the second win so you cover 11 on two of three dozens.

If you win the second spin you add 11 more for a profit of 19. (33-14).

"Standard" method

Units are $20, since winning bets on bank only pay $19

Starting bankroll for each session is 930 units, or $18,600

A session continues until either the bankroll reaches zero or below (and I do allow for "negative bankroll" to make full bets), or 18,695 or above (a profit of 5 units, allowing for the possibility that all of the "winning" hands were bets on bank).

The results:

4883.5 hands per session

92% of the sessions were winners

However, the average result per session was a loss of just over 70 units, which makes sense - 92% of the time, you ended up around +5, and 8% of the time, you ended up around -930; (0.92 x 5) + (0.08 x (-930)) = -69.8.

If you use the "black box" method and treat each session as having bet 930 units, the "house edge" would be about 7.53%.

("How is that possible, since the house edge on straight player bets in baccarat is only 1.25%?" That's the edge per bet - the more you go back and forth between losing everything and reaching your target, the more you expose to that one bet at a time edge.)

Quote:ThatDonGuyI just ran 750,000 sessions under the following conditions:

"Standard" method

Units are $20, since winning bets on bank only pay $19

Starting bankroll for each session is 930 units, or $18,600

A session continues until either the bankroll reaches zero or below (and I do allow for "negative bankroll" to make full bets), or 18,695 or above (a profit of 5 units, allowing for the possibility that all of the "winning" hands were bets on bank).

The results:

4883.5 hands per session

92% of the sessions were winners

However, the average result per session was a loss of just over 70 units, which makes sense - 92% of the time, you ended up around +5, and 8% of the time, you ended up around -930; (0.92 x 5) + (0.08 x (-930)) = -69.8.

If you use the "black box" method and treat each session as having bet 930 units, the "house edge" would be about 7.53%.

("How is that possible, since the house edge on straight player bets in baccarat is only 1.25%?" That's the edge per bet - the more you go back and forth between losing everything and reaching your target, the more you expose to that one bet at a time edge.)

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I hope ThomasK reads the last bit.

Quote:ThomasKBefore I answer your replies individually, please see here the Excel definitions I use to simulate the progression.

While preparing the Excel to share with you, I realized a mistake which seems to be in the definition of the system. The amounts of the progression add up to 93, but you need two losses in a row on the same progression level to be eligible to move to the next level. Therefore the amount lost in the worst case is 186 units.

This means that the equivalent slot machine requires 186 coins in.

Here now the components of the Excel for Carsch with Parlay on intitial bet and its winning and corrected coin in for the slot machine:

Prepare section A.probability(hit) =18/38

#attempts =MAX(L:L)

amount wagered =SUM(H:H)

payed =SUM(I:I)

return =B6/B5

house edge =1-B7

coin in =SUM(M:M)

cash out =SUM(N:N)

return =B11/B10

house edge =1-B12

Prepare section B.random won? key state wager payed wager per attempt payed per attempt attempt # coin in cash out fin

=RAND() =IF(D2<=$B$1;"won";"lost") =G2&E2 1 =VLOOKUP(F2;Q:W;5;0) =IF(E2="won";2*H2;0) =IF(G2=1;H2;H2+J1) =IF(G2=1;I2;I2+K1) 1 =IF(G2=1;186;0) =IF(G3=1;MIN(1;I2);0)*(186+K2-J2)

=RAND() =IF(D3<=$B$1;"won";"lost") =G3&E3 =VLOOKUP(F2;Q:W;4;0) =VLOOKUP(F3;Q:W;5;0) =IF(E3="won";2*H3;0) =IF(G3=1;H3;H3+J2) =IF(G3=1;I3;I3+K2) =IF(G3=1;L2+1;L2) =IF(G3=1;186;0) =IF(G4=1;MIN(1;I3);0)*(186+K3-J3)

Because of the localization of my Excel you might need to try this version. Semicolons ";" are replaced by commas ",". I'm not able to test whether it works properly:random won? key state wager payed wager per attempt payed per attempt attempt # coin in cash out fin

=RAND() =IF(D2<=$B$1,"won","lost") =G2&E2 1 =VLOOKUP(F2,Q:W,5,0) =IF(E2="won",2*H2,0) =IF(G2=1,H2,H2+J1) =IF(G2=1,I2,I2+K1) 1 =IF(G2=1,186,0) =IF(G3=1,MIN(1,I2),0)*(186+K2-J2)

=RAND() =IF(D3<=$B$1,"won","lost") =G3&E3 =VLOOKUP(F2,Q:W,4,0) =VLOOKUP(F3,Q:W,5,0) =IF(E3="won",2*H3,0) =IF(G3=1,H3,H3+J2) =IF(G3=1,I3,I3+K2) =IF(G3=1,L2+1,L2) =IF(G3=1,186,0) =IF(G4=1,MIN(1,I3),0)*(186+K3-J3)

Prepare section C.key state input next wager sub-state fin

=R2&S2 1 won =R4 1 1

=R3&S3 1 lost =R6 1 1

=R4&S4 2 won 1 =2*U2 2

=R5&S5 2 lost =R6 =2*U3 2

=R6&S6 3 won =R4 1 3

=R7&S7 3 lost =R8 1 3

=R8&S8 =U8*10+V8 =S2 =R10 1 =V2

=R9&S9 =U9*10+V9 =S3 =R12 1 =V3

=R10&S10 =U10*10+V10 =S4 1 =2*U8 =V4

=R11&S11 =U11*10+V11 =S5 =R12 =2*U9 =V5

=R12&S12 =U12*10+V12 =S6 =R10 1 =V6

=R13&S13 =U13*10+V13 =S7 =R14 1 =V7

=R14&S14 =U14*10+V14 =S8 =R16 2 =V8

=R15&S15 =U15*10+V15 =S9 =R18 2 =V9

=R16&S16 =U16*10+V16 =S10 1 =2*U14 =V10

=R17&S17 =U17*10+V17 =S11 =R18 =2*U15 =V11

=R18&S18 =U18*10+V18 =S12 =R16 2 =V12

=R19&S19 =U19*10+V19 =S13 =R20 2 =V13

=R20&S20 =U20*10+V20 =S14 =R22 3 =V14

=R21&S21 =U21*10+V21 =S15 =R24 3 =V15

=R22&S22 =U22*10+V22 =S16 1 =2*U20 =V16

=R23&S23 =U23*10+V23 =S17 =R24 =2*U21 =V17

=R24&S24 =U24*10+V24 =S18 =R22 3 =V18

=R25&S25 =U25*10+V25 =S19 =R26 3 =V19

=R26&S26 =U26*10+V26 =S20 =R28 4 =V20

=R27&S27 =U27*10+V27 =S21 =R30 4 =V21

=R28&S28 =U28*10+V28 =S22 1 =2*U26 =V22

=R29&S29 =U29*10+V29 =S23 =R30 =2*U27 =V23

=R30&S30 =U30*10+V30 =S24 =R28 4 =V24

=R31&S31 =U31*10+V31 =S25 =R32 4 =V25

=R32&S32 =U32*10+V32 =S26 =R34 6 =V26

=R33&S33 =U33*10+V33 =S27 =R36 6 =V27

=R34&S34 =U34*10+V34 =S28 1 =2*U32 =V28

=R35&S35 =U35*10+V35 =S29 =R36 =2*U33 =V29

=R36&S36 =U36*10+V36 =S30 =R34 6 =V30

=R37&S37 =U37*10+V37 =S31 =R38 6 =V31

=R38&S38 =U38*10+V38 =S32 =R40 9 =V32

=R39&S39 =U39*10+V39 =S33 =R42 9 =V33

=R40&S40 =U40*10+V40 =S34 1 =2*U38 =V34

=R41&S41 =U41*10+V41 =S35 =R42 =2*U39 =V35

=R42&S42 =U42*10+V42 =S36 =R40 9 =V36

=R43&S43 =U43*10+V43 =S37 =R44 9 =V37

=R44&S44 =U44*10+V44 =S38 =R46 14 =V38

=R45&S45 =U45*10+V45 =S39 =R48 14 =V39

=R46&S46 =U46*10+V46 =S40 1 =2*U44 =V40

=R47&S47 =U47*10+V47 =S41 =R48 =2*U45 =V41

=R48&S48 =U48*10+V48 =S42 =R46 14 =V42

=R49&S49 =U49*10+V49 =S43 =R50 14 =V43

=R50&S50 =U50*10+V50 =S44 =R52 21 =V44

=R51&S51 =U51*10+V51 =S45 =R54 21 =V45

=R52&S52 =U52*10+V52 =S46 1 =2*U50 =V46

=R53&S53 =U53*10+V53 =S47 =R54 =2*U51 =V47

=R54&S54 =U54*10+V54 =S48 =R52 21 =V48

=R55&S55 =U55*10+V55 =S49 =R56 21 =V49

=R56&S56 =U56*10+V56 =S50 =R58 32 =V50

=R57&S57 =U57*10+V57 =S51 =R60 32 =V51

=R58&S58 =U58*10+V58 =S52 1 =2*U56 =V52

=R59&S59 =U59*10+V59 =S53 =R60 =2*U57 =V53

=R60&S60 =U60*10+V60 =S54 =R58 32 =V54

=R61&S61 =U61*10+V61 =S55 1 32 =V55

Finalize section B.

BEWARE: Excel might take quite some time to calculate the simaulation. Start with a few thousand rows before copying down the whole worksheet.

Mark the cells D3 through N3.

Copy these cells.

Paste these cells down the worksheet.

Simulate.

By refreshing the sheet, the random rolls in column D get updated.

The overall results are displayed in section A, i.e. columns A and B.

The other finite state machines.key state input next wager sub-state fin

=R2&S2 1 won =R4 1 1

=R3&S3 1 lost =R6 1 1

=R4&S4 2 won 1 1 2

=R5&S5 2 lost =R6 1 2

=R6&S6 3 won =R4 1 3

=R7&S7 3 lost =R8 1 3

=R8&S8 =U8*10+V8 =S2 =R10 1 =V2

=R9&S9 =U9*10+V9 =S3 =R12 1 =V3

=R10&S10 =U10*10+V10 =S4 1 1 =V4

=R11&S11 =U11*10+V11 =S5 =R12 1 =V5

=R12&S12 =U12*10+V12 =S6 =R10 1 =V6

=R13&S13 =U13*10+V13 =S7 =R14 1 =V7

=R14&S14 =U14*10+V14 =S8 =R16 2 =V8

=R15&S15 =U15*10+V15 =S9 =R18 2 =V9

=R16&S16 =U16*10+V16 =S10 1 2 =V10

=R17&S17 =U17*10+V17 =S11 =R18 2 =V11

=R18&S18 =U18*10+V18 =S12 =R16 2 =V12

=R19&S19 =U19*10+V19 =S13 =R20 2 =V13

=R20&S20 =U20*10+V20 =S14 =R22 3 =V14

=R21&S21 =U21*10+V21 =S15 =R24 3 =V15

=R22&S22 =U22*10+V22 =S16 1 3 =V16

=R23&S23 =U23*10+V23 =S17 =R24 3 =V17

=R24&S24 =U24*10+V24 =S18 =R22 3 =V18

=R25&S25 =U25*10+V25 =S19 =R26 3 =V19

=R26&S26 =U26*10+V26 =S20 =R28 4 =V20

=R27&S27 =U27*10+V27 =S21 =R30 4 =V21

=R28&S28 =U28*10+V28 =S22 1 4 =V22

=R29&S29 =U29*10+V29 =S23 =R30 4 =V23

=R30&S30 =U30*10+V30 =S24 =R28 4 =V24

=R31&S31 =U31*10+V31 =S25 =R32 4 =V25

=R32&S32 =U32*10+V32 =S26 =R34 6 =V26

=R33&S33 =U33*10+V33 =S27 =R36 6 =V27

=R34&S34 =U34*10+V34 =S28 1 6 =V28

=R35&S35 =U35*10+V35 =S29 =R36 6 =V29

=R36&S36 =U36*10+V36 =S30 =R34 6 =V30

=R37&S37 =U37*10+V37 =S31 =R38 6 =V31

=R38&S38 =U38*10+V38 =S32 =R40 9 =V32

=R39&S39 =U39*10+V39 =S33 =R42 9 =V33

=R40&S40 =U40*10+V40 =S34 1 9 =V34

=R41&S41 =U41*10+V41 =S35 =R42 9 =V35

=R42&S42 =U42*10+V42 =S36 =R40 9 =V36

=R43&S43 =U43*10+V43 =S37 =R44 9 =V37

=R44&S44 =U44*10+V44 =S38 =R46 14 =V38

=R45&S45 =U45*10+V45 =S39 =R48 14 =V39

=R46&S46 =U46*10+V46 =S40 1 14 =V40

=R47&S47 =U47*10+V47 =S41 =R48 14 =V41

=R48&S48 =U48*10+V48 =S42 =R46 14 =V42

=R49&S49 =U49*10+V49 =S43 =R50 14 =V43

=R50&S50 =U50*10+V50 =S44 =R52 21 =V44

=R51&S51 =U51*10+V51 =S45 =R54 21 =V45

=R52&S52 =U52*10+V52 =S46 1 21 =V46

=R53&S53 =U53*10+V53 =S47 =R54 21 =V47

=R54&S54 =U54*10+V54 =S48 =R52 21 =V48

=R55&S55 =U55*10+V55 =S49 =R56 21 =V49

=R56&S56 =U56*10+V56 =S50 =R58 32 =V50

=R57&S57 =U57*10+V57 =S51 =R60 32 =V51

=R58&S58 =U58*10+V58 =S52 1 32 =V52

=R59&S59 =U59*10+V59 =S53 =R60 32 =V53

=R60&S60 =U60*10+V60 =S54 =R58 32 =V54

=R61&S61 =U61*10+V61 =S55 1 32 =V55key state input next wager sub-state fin

=R2&S2 1 won 1 1 1

=R3&S3 1 lost =R6 1 1

=R4&S4 2 won 1 =2*U2 2

=R5&S5 2 lost =R6 =2*U3 2

=R6&S6 3 won 1 1 3

=R7&S7 3 lost =R8 1 3

=R8&S8 =U8*10+V8 =S2 1 1 =V2

=R9&S9 =U9*10+V9 =S3 =R12 1 =V3

=R10&S10 =U10*10+V10 =S4 1 =2*U8 =V4

=R11&S11 =U11*10+V11 =S5 =R12 =2*U9 =V5

=R12&S12 =U12*10+V12 =S6 1 1 =V6

=R13&S13 =U13*10+V13 =S7 =R14 1 =V7

=R14&S14 =U14*10+V14 =S8 1 2 =V8

=R15&S15 =U15*10+V15 =S9 =R18 2 =V9

=R16&S16 =U16*10+V16 =S10 1 =2*U14 =V10

=R17&S17 =U17*10+V17 =S11 =R18 =2*U15 =V11

=R18&S18 =U18*10+V18 =S12 1 2 =V12

=R19&S19 =U19*10+V19 =S13 =R20 2 =V13

=R20&S20 =U20*10+V20 =S14 1 3 =V14

=R21&S21 =U21*10+V21 =S15 =R24 3 =V15

=R22&S22 =U22*10+V22 =S16 1 =2*U20 =V16

=R23&S23 =U23*10+V23 =S17 =R24 =2*U21 =V17

=R24&S24 =U24*10+V24 =S18 1 3 =V18

=R25&S25 =U25*10+V25 =S19 =R26 3 =V19

=R26&S26 =U26*10+V26 =S20 1 4 =V20

=R27&S27 =U27*10+V27 =S21 =R30 4 =V21

=R28&S28 =U28*10+V28 =S22 1 =2*U26 =V22

=R29&S29 =U29*10+V29 =S23 =R30 =2*U27 =V23

=R30&S30 =U30*10+V30 =S24 1 4 =V24

=R31&S31 =U31*10+V31 =S25 =R32 4 =V25

=R32&S32 =U32*10+V32 =S26 1 6 =V26

=R33&S33 =U33*10+V33 =S27 =R36 6 =V27

=R34&S34 =U34*10+V34 =S28 1 =2*U32 =V28

=R35&S35 =U35*10+V35 =S29 =R36 =2*U33 =V29

=R36&S36 =U36*10+V36 =S30 1 6 =V30

=R37&S37 =U37*10+V37 =S31 =R38 6 =V31

=R38&S38 =U38*10+V38 =S32 1 9 =V32

=R39&S39 =U39*10+V39 =S33 =R42 9 =V33

=R40&S40 =U40*10+V40 =S34 1 =2*U38 =V34

=R41&S41 =U41*10+V41 =S35 =R42 =2*U39 =V35

=R42&S42 =U42*10+V42 =S36 1 9 =V36

=R43&S43 =U43*10+V43 =S37 =R44 9 =V37

=R44&S44 =U44*10+V44 =S38 1 14 =V38

=R45&S45 =U45*10+V45 =S39 =R48 14 =V39

=R46&S46 =U46*10+V46 =S40 1 =2*U44 =V40

=R47&S47 =U47*10+V47 =S41 =R48 =2*U45 =V41

=R48&S48 =U48*10+V48 =S42 1 14 =V42

=R49&S49 =U49*10+V49 =S43 =R50 14 =V43

=R50&S50 =U50*10+V50 =S44 1 21 =V44

=R51&S51 =U51*10+V51 =S45 =R54 21 =V45

=R52&S52 =U52*10+V52 =S46 1 =2*U50 =V46

=R53&S53 =U53*10+V53 =S47 =R54 =2*U51 =V47

=R54&S54 =U54*10+V54 =S48 1 21 =V48

=R55&S55 =U55*10+V55 =S49 =R56 21 =V49

=R56&S56 =U56*10+V56 =S50 1 32 =V50

=R57&S57 =U57*10+V57 =S51 =R60 32 =V51

=R58&S58 =U58*10+V58 =S52 1 =2*U56 =V52

=R59&S59 =U59*10+V59 =S53 =R60 =2*U57 =V53

=R60&S60 =U60*10+V60 =S54 1 32 =V54

=R61&S61 =U61*10+V61 =S55 1 32 =V55

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Thank you so much for the correction as ill be honest it's not easy for me to understand all this high level stuff for math isn't my strong suit. Thank you again for all the work and info. I can tell this forum is way beyond my math intelligence. What have I done??? LOL

Quote:darkozQuote:GreenZeroI was also looking at this progression with roulette betting 1 unit each on the 1st&2nd dozens and then betting on the 2nd&3rd dozens. Since your chances of hitting are about 66% it could work beautifully with this progression since your looking for 2 wins in a row to reset back to the beginning. I haven't done a small simulation on this yet though as a progression system involving 2 dozens get expensive really fast.

link to original post

It wouldn't be a carsch.

The idea is to win two in a row. Betting two of three dozens means you lose twice as much per loss and win half as much per win.

Let's say you lose, 1,2,4,7. That's 14×2 so you just lost 28.

And now you win the next two. That's win 22-11 and again the second spin so you made back only 22 units after losing 28.

You would have to win three in a row. And as the progression of losses goes on then the number of winning spins needed goes up.

What you should analyze is the opposite.

Betting just one dozen using the carsch because you only need one win instead of two to profit.

Lose 1,2,4,7 for a fourteen unit loss. Then win 22 on the fifth spin.

Or perhaps a combo carsch. For example if you win the fifth you make a second spin be two dozens by splitting the win.

Example: Lose first four. 1,2,4,7. Then win 22 and divide that for to get the second win so you cover 11 on two of three dozens.

If you win the second spin you add 11 more for a profit of 19. (33-14).

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Thank you for the correction. You could make the progression more aggressive to make up the loss to keep it at 2 wins instead of 3.

Technically, the house edge itself doesn't change as a result of any progression. It's the same on every bet regardless of size, assuming all outcomes are equally random.

What can change is the player's expected returns on a game if he is using a progression that always results in the highest wager being a win. That's a big "if." But with a large enough bankroll, low enough initial wager, high enough maximum wager, and gentle enough progression to allow for more than 10 steps, a system somewhat like Carsch as described in this thead and as applied to two dozens bets on roulette could conceivably be sustained through any "2 in a row" drought a player would ever likely suffer in his lifetime. It therefore could create an expectation that average wagers on winning bets would be slightly higher than on losing bets.

the Carsch progression got me thinking about parlays

and no, of course I'm not claiming that this changes the HA in any way

but it is kinna interesting to me anyway

if you successfully parlay the 2nd to last number in the progression - $21 - you have a win equalling $63___________I'm not counting previous losses

but if you successfully parlay the last number in the progression - $32 you have a win equalling $96

the risk in the larger parlay attempt has increased by only $11 - but your win has increased by $33

the size of the win has increased by three times the size of the increase in the risk

.

Quote:lilredrooster______________

the Carsch progression got me thinking about parlays

and no, of course I'm not claiming that this changes the HA in any way

but it is kinna interesting to me anyway

if you successfully parlay the 2nd to last number in the progression - $21 - you have a win equalling $63___________I'm not counting previous losses

but if you successfully parlay the last number in the progression - $32 you have a win equalling $96

the risk in the larger parlay attempt has increased by only $11 - but your win has increased by $33

the size of the win has increased by three times the size of the increase in the risk

.

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lilredrooster,

If you still are interested, I have prepared my explanation over here. If you don't find any major flaws in it, I would say that your gut feeling has a mathematical foundation.

Quote:JackSpadeThe only way you're going to dent the house edge, assuming you win no more often than your actual statistical odds of winning, is if your average wager on winning bets is higher than on losing bets.

Technically, the house edge itself doesn't change as a result of any progression. It's the same on every bet regardless of size, assuming all outcomes are equally random.

What can change is the player's expected returns on a game if he is using a progression that always results in the highest wager being a win. That's a big "if." But with a large enough bankroll, low enough initial wager, high enough maximum wager, and gentle enough progression to allow for more than 10 steps, a system somewhat like Carsch as described in this thead and as applied to two dozens bets on roulette could conceivably be sustained through any "2 in a row" drought a player would ever likely suffer in his lifetime. It therefore could create an expectation that average wagers on winning bets would be slightly higher than on losing bets.

link to original post

hello

sorry but I don't understand. English is not my first language

Quote:JackSpadeBut with a large enough bankroll, low enough initial wager, high enough maximum wager, and gentle enough progression to allow for more than 10 steps, a system somewhat like Carsch as described in this thead and as applied to two dozens bets on roulette could conceivably be sustained through any "2 in a row" drought a player would ever likely suffer in his lifetime. It therefore could create an expectation that average wagers on winning bets would be slightly higher than on losing bets.

I have a one month period tested (2 dozens) manually from outcomes of a single zero casino. could you please take a look at it and tell me if the way you described would work on those outcomes? I apologize if I misunderstood you.

10/22

+ + - + - * - + - + + * + + * - - + * + - + + 0 - + + + + - * + + - + * - + - - - + + - + * + - - - + + * + * - + + + + - * + - + * + + + + * + + + + + + + * + + + + + * + + + + + - - * + + + - * + + - + + + * + + + + + - + + - + 0 * - + + - * - + 0 + + + - * 0 + + - 0 * + + + + + * + * - + + 0 + * - + + - * + + + * + + - + + * + - + - + - + + + * - - + + + * 0 + + + + - + - * + + - ***

explainer

the * shows the end of a day and start of new day. the *** is end of the month. the 0 is Zero and a loss

thank you