insane07
• Posts: 6
Joined: Jan 3, 2015
March 25th, 2015 at 7:58:01 PM permalink
Hello - looking for some logic on the following:

I understand there is a house edge in the game betting banker is 1.06%. With the following assumptions:
Banker hits: 0.458597
Player hits: 0.446247
Tie hits: 0.095156

I hope it's reasonable to exclude the tie as that outcome doesn't affect the win/loss proposition of betting banker every time.
Thus I assume the probability of Banker and Player hitting are:
Banker hits = 0.458597 / (0.458597 + 0.446247) = 0.506824

Now the commission obviously plays a role in the house edge, but if you are able to win at a clip rate above 50%, how is it that the Labouchere system ends up failing with a large bankroll. Say, you are playing with a base line of 12.5-12.5 ie. betting \$25 as the base bet and having only two numbers in the initial line to win 1 unit. Table min is \$25 and max is \$25,000. Let's say you have a bankroll of \$400,000 or whatever is required to you playing the table max on all squares of the table.

I'm struggling with understanding how the long term ruin approaches the house edge when you are having a winning outcome more often than a losing outcome...

rdw4potus
• Posts: 7237
Joined: Mar 11, 2010
March 25th, 2015 at 8:26:57 PM permalink
If the game paid 1:1 and you had a hit rate over 50%, you wouldn't need to mess with a betting system - every bet would have a player advantage. But, the game doesn't pay 1:1. Betting banker and ignoring ties, 49.4% of the time you lose all your money. 50.6% of the time you win 95% of your money.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Dalex64
• Posts: 1067
Joined: Feb 10, 2013
March 25th, 2015 at 8:37:32 PM permalink
You can't just use the outcome in your evaluation, you have to use the return.

You can't discount the effect of the commission, it is what gives the banker bet a negative return even though you might win more than half the time.

To exaggerate to make a point, let's invent a dice game where you bet a dollar, win on 1-5, lose on 6. When you lose, you lose your whole dollar. When you win, they take a 99% commission and only pay you one cent.

You are going to win 5/6 of the time, but you should be able to see that in 6 rolls you can expect to win 5 cents and lose a dollar, for a net of -95 cents.

The numbers in baccarat are much closer, but the short pay because of the commision more than compensates for your greater winning percentage.
DeMango
• Posts: 2958
Joined: Feb 2, 2010
March 26th, 2015 at 6:13:32 AM permalink
The reason is negative variance. When you lose say 5 in a row more than once, you will have bets that look like martingale bets. When you lose a few more you are toast.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
ThatDonGuy
• Posts: 6406
Joined: Jun 22, 2011
March 26th, 2015 at 6:19:06 AM permalink
Here's an equivalent in roulette:

Suppose you bet \$100 on "first 12" and 100 on "second 12"
If the number is anything from 1-24, you win \$100 profit (you win \$200 on the bet that won, and lose \$100 on the bet that lost).
If the number is anything else, you lose \$200.

In effect, you are betting \$200 at 1-2 odds that the number will be 1-24, which has a 63% chance of happening (on a double-zero wheel). So why does this system fail? Because you have to take the fact that the bet pays less than even money into account.

It is the same with betting on the banker in baccarat. Because of the commission, the bet isn't even money, but 19-20.
insane07
• Posts: 6
Joined: Jan 3, 2015
March 26th, 2015 at 6:51:00 AM permalink
Makes complete sense, and I completely understand Expected value. For this it would be
-1*.494 + 0.95 * .506 = -0.0133
So I get there is a negative return.

But doesn't this assume that you are betting the same amount each time? And that you are accounting for commission when you bet to win. So if you are looking to win \$100, you actually bet \$105. If you lose, and are looking to then win \$155, you bet \$155*1.05 = \$162.75.....

Even as I'm writing this my mathematical/statistical sense is telling me that it's a negative proposition and I believe the Wizards posts about all betting systems being equally useless. But my degenerate mind can't wrap itself around the theory of you only need to win greater than 33% of the time to clear the lines in the Labouchere betting system.

I guess the flaw comes in the assumption of an unlimited bankroll because that is not true.

I wonder if I could be honoured by the wizard running a billion hand test of how big your bankroll is required to be to not go bust with this assumptions I listed above....
Romes
• Posts: 5605
Joined: Jul 22, 2014
March 26th, 2015 at 6:55:23 AM permalink
Quote: insane07

Makes complete sense, and I completely understand Expected value. For this it would be
-1*.494 + 0.95 * .506 = -0.0133
So I get there is a negative return...

All you need to know right there. Each bet (regardless of size) has a negative return. If you know 'when' you're going to win and can thus bet bigger because you know you're going to win, you're either cheating, or you need to come to the casino with me...
Playing it correctly means you've already won.
ThatDonGuy
• Posts: 6406
Joined: Jun 22, 2011
March 26th, 2015 at 7:31:57 AM permalink
Quote: insane07

I guess the flaw comes in the assumption of an unlimited bankroll because that is not true.

That, and you have just discovered one of the reasons the table has a "maximum bet" - it stops Martingale and other "increase the bet until you eventually win" systems dead in their tracks.

Quote: insane07

I wonder if I could be honoured by the wizard running a billion hand test of how big your bankroll is required to be to not go bust with this assumptions I listed above....

I ran 3 billion hands using an "infinite deck", and the longest run of player wins between two bank wins (ignoring ties) I got was 28. Even with a straight Martingale system, where you only bet just enough so you cover your losses plus make your initial bet amount since your last win, your bet after your 28th loss would be over 268 million times what your first bet was; if it was \$105, you would be betting over 28 billion dollars, and remember, you have also lost over 28 billion dollars up to that point.
insane07
• Posts: 6
Joined: Jan 3, 2015
March 26th, 2015 at 9:49:48 AM permalink
That's great info, thank you. However a martingale has a different system as you're looking to win back all losses in one bet. With the Labouchere system that is not the case. You are looking to lose less than twice the time you win and win it back over a series of bets. Thus that 268 million or 28 billion dollar will be much smaller I believe. Any chance of coding that test?
Dalex64
• Posts: 1067
Joined: Feb 10, 2013
ThatDonGuy
• Posts: 6406
Joined: Jun 22, 2011
March 26th, 2015 at 12:04:28 PM permalink
Quote: insane07

That's great info, thank you. However a martingale has a different system as you're looking to win back all losses in one bet. With the Labouchere system that is not the case. You are looking to lose less than twice the time you win and win it back over a series of bets. Thus that 268 million or 28 billion dollar will be much smaller I believe. Any chance of coding that test?

The number of consecutive losses would be the same regardless of which system you are using.

Assuming a Labouchere where the left and right numbers start at 50, after the Nth consecutive loss, the left number would be 50 and the right number 50 x (N + 1), so after 28 consecutive losses, the sum of the left numbers in the losing bets would be 28 x 50 = 1400, and the right numbers would be (50 + 100 + 150 + ... + 1400) = 20,300, for a bankroll loss of 21,700.

If this were to be followed by a string of wins, the first bet would be 1500 (50 + 1450), and the subsequent bets would all be 1450; 15 consecutive wins would give you 21,755, so you would be ahead by 55.