March 10th, 2015 at 2:55:11 AM
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Can someone please help me with odds of this "system" I came up with.

I would be playing at a no commission table as either player or banker I suppose

I need to figure out the odds of my system working while taking into account the risk of hitting an exact pattern of "win, loss, loss" within the system.

By "working" I mean what would be the odds of a catastrophic failure (which I would define as losing on the 6th "tripled bet" aka the ninth overall bet).

Thank you!

Here is the system:

After 3 consecutive losses, begin tripling bet (5,5,5,15,45,135,405,etc.)

Net profits on triple bets will be the bet ("Y") minus initial bet ("X"), divided by 2 and then you subtract an additional ("X") from that remainder.

Payouts would be as follows:

Lose 5

Lose 5

Lose 5

Bet 15--> Net 0

Bet 45--> Net 15

Bet 135--> Net 60

Bet 405--> Net 195

Bet 1215--> Net 600

Bet 3645--> Net 1815

I would be playing at a no commission table as either player or banker I suppose

I need to figure out the odds of my system working while taking into account the risk of hitting an exact pattern of "win, loss, loss" within the system.

By "working" I mean what would be the odds of a catastrophic failure (which I would define as losing on the 6th "tripled bet" aka the ninth overall bet).

Thank you!

Here is the system:

After 3 consecutive losses, begin tripling bet (5,5,5,15,45,135,405,etc.)

Net profits on triple bets will be the bet ("Y") minus initial bet ("X"), divided by 2 and then you subtract an additional ("X") from that remainder.

Payouts would be as follows:

Lose 5

Lose 5

Lose 5

Bet 15--> Net 0

Bet 45--> Net 15

Bet 135--> Net 60

Bet 405--> Net 195

Bet 1215--> Net 600

Bet 3645--> Net 1815

March 10th, 2015 at 3:19:22 AM
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Let's just say there's a 50% chance for any of the bets to win. [Realistically, it's like 45% player, 45% banker, and 10% ties (roughly).]

To figure out the chance of having N losses in a row where you have a 50% chance of losing, you do 0.5^N.

1 loss: 0.5^1 = 0.5 = 50%

2: 0.5^2 = 0.25 = 25%

3: .. = 12.5%

4: 6.25%

5: 3.125%

6: 1.5625%

etc.

I don't know what happens when there's a tie (does your bet lose?). If it does lose, then you'd do 0.55^N where N is the number of losses in a row.

1 loss = 0.55^1

2 losses in a row = 0.55^2

3: 0.55^3

etc.

If a tie is a push, well, it's something like:

X = banker win % chance (ie: 0.4585)

Y = player win % chance (ie: 0.4462)

Chance of banker win, excluding ties, is X/(X+Y). Chance of player win is Y/(X+Y).

Those numbers are for regular baccarat, not no-commission, according to https://wizardofodds.com/ask-the-wizard/baccarat/

For non-commission, you'd have to look up the win %'s for that game, since I believe, there's a specific hand (or two?) where the banker loses [or pushes?] instead of winning, so both have less than 50% chance of occurring, excluding ties.

To figure out the chance of having N losses in a row where you have a 50% chance of losing, you do 0.5^N.

1 loss: 0.5^1 = 0.5 = 50%

2: 0.5^2 = 0.25 = 25%

3: .. = 12.5%

4: 6.25%

5: 3.125%

6: 1.5625%

etc.

I don't know what happens when there's a tie (does your bet lose?). If it does lose, then you'd do 0.55^N where N is the number of losses in a row.

1 loss = 0.55^1

2 losses in a row = 0.55^2

3: 0.55^3

etc.

If a tie is a push, well, it's something like:

X = banker win % chance (ie: 0.4585)

Y = player win % chance (ie: 0.4462)

Chance of banker win, excluding ties, is X/(X+Y). Chance of player win is Y/(X+Y).

Those numbers are for regular baccarat, not no-commission, according to https://wizardofodds.com/ask-the-wizard/baccarat/

For non-commission, you'd have to look up the win %'s for that game, since I believe, there's a specific hand (or two?) where the banker loses [or pushes?] instead of winning, so both have less than 50% chance of occurring, excluding ties.

March 10th, 2015 at 5:59:41 AM
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This is a duplicate of a post in the Betting Systems folder.

There is no need to have the same post twice, especially as it is going to generate two different sets of replies.

There is no need to have the same post twice, especially as it is going to generate two different sets of replies.

March 10th, 2015 at 10:49:12 AM
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Quote:ThatDonGuyThis is a duplicate of a post in the Betting Systems folder.

There is no need to have the same post twice, especially as it is going to generate two different sets of replies.

Yeah, just noticed the duplication myself. I'm closing this thread; please reply to the url TDG posted above.

If the House lost every hand, they wouldn't deal the game.