An Exploitable Anomaly?

I read somewhere that a typical collection of 38 roulette spins will only include 24 distinct numbers. If this is true, it seems that this fact could present an exploitable anomaly.

What is an anomaly?

Because there are 38 slots on a roulette wheel, math and probability theory would tell us that if you played one number straight up, you could expect to see the number hit once every 38 spins. There would be no reason for you to expect to encounter the number 17, for example, any more or less often than once every 38 spins.

If you were to set out to play one number for 38 spins, you would very often find yourself winning (once during the 38 spins). A single number on the inside of the roulette layout pays 35 to 1, this means that you will receive 36 chips back after placing one winning bet of one single chip. If you commit to play the number 17 for 38 spins in a row (stopping after a win), you will have a positive result if your number hits during the first 35 spins (with your expected gain diminishing with each spin). You would break even if your number hit on the 36th spin and you would suffer a small loss if your number came up on the 37th or 38th spin. (Of course, you would lose big if your number never hit during the 38 spins.)

IF the number did exactly as it is expected to do (appear every 38 spins), you would lose in the long run an amount equal to the house edge. The casinos rely on this.

But what about the "fact" that only 24 numbers will appear during the next 38 spins? Shouldn't this give us some potential edge?

Is it exploitable?

If you were to play 1 unit on each of the numbers 1 through 24 at the same time, a win would net you +12 units. You should win this bet 24 times for every 38 spins and if so, again you could expect to lose an amount equal to the house edge. You should win this bet 24 times out of 38 spins because math and probability theory tell us that your 24 numbers should each appear once during 38 spins.

BUT what if, instead of playing the same 24 numbers for each spin, you played the last 24 numbers?

It would make sense that IF all sets of 38 consecutive decisions contain an average of 24 spins, THEN the very next spin is likely to be one of the last 24 numbers. It SEEMS that it is more likely that the next number would be be one of the last 24 numbers than it is that the number would be of any other group of 24 numbers. IF SO, you should win this bet more often than the mathematical expectation and therefore you could eliminate the house edge and perhaps even come out ahead.

In order to fully exploit this fact, we should develop a progression. By using a progression, we are not simply betting that the next decision will be one of the last 24, instead we would be betting that we are not going to see 3 (for example) decisions in a row that are all three (each) not of the last 24 decisions. With a three-step progression, we would only lose when the next decision was not one of the last 24 decisions THREE times in a row.

Here's how a progression would work:

First bet (1u) = 24u, a win on this bet would give us 36 minus 24 = 12units

Next bet (3u) = 72u, a win on this bet would give us 108 minus 96 = 12units
Next bet (9u) = 216u, a win on this bet would give us 324 minus 312 = 12 units

60 decisions per hour
20 wins on first spin
15 wins on 2nd spin
5 wins on 3rd spin

40 wins x 12units = 480 units per hour
less one full series loss = 168 units per hour

The real question is: How often would you see a full-series loss? Is once an hour realistic or would you see more or less?

I wonder what a study of the Zumma roulette book would show?

Would it be difficult to program an Excel spreadsheet to identify if the next number is one of the last 24?

Is 24 the magic number? Or is there a better number to use?

Perhaps a better way of looking at the "units" required for this sort of play is to recognize that the smallest bet you play is 24 units. It is probably best to consider this 1 unit. Looking at this system from this angle, you will be winning .5 (one-half) unit with every spin (not counting losses).

1 unit = 24 chips

With the minimal chip size of 1 dollar, your unit would be $24. Usually, table minimums require a bet of at least $10 on the inside. Table maximums for inside numbers are often limits on the amount you can bet on a single number.

Here's a chart to show requirements and anticipated results:

Chip = $1

Unit = $24

Largest Bet = $216

Net/hr = $168 (with 1 full-series loss per hour)

Bankroll = $312 x 3 = $936 ($1,000)

******************************************

Ratio Between Wins and Full-Series Losses = 12/312 = 1 to 26

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Chip = $5

Unit = $120

Largest Bet = $1,080

Net/hr = $840

Bankroll = $1560 x 3 = $4680 ($5,000)

*******************************************

Double your bankroll in 6 hours.

But, does it work??

ANy thoughts?? Please comment.

Simple Grinds

I thought I'd start a thread of simple systems that should keep you in the game for a long time. These would be systems that allow you to get in on the action often and will win with a little luck. These would also allow you play for a long time without much risk of ruin. The criteria here also includes simplicty, systems that can be easily learned and employed.

Barstow's New Shooter System

(Bet with Him)

page 126

Barstow claims this is like getting 7 to 1 odds on a 15 to 1 proposition. The basic idea is that you bet that the new shooter will make his point and if he doesn't, you double that the next shooter will make his, and if not, you go one more time with a third new shooter (three stage martingale). You lose your series only when there are 4 consecutive losses (because there must have been a loss just prior to your first bet). A full series loss will cost you 7 units so you need to win this bet more often than 7 times for each lost series. Barstow would like us to believe that you sould win this 15 times for each loss. I think the reality should be considerably less, but this should be a reliable grind on any hot table.