Are all bets the same?

It is an accepted fact of Roulette that all possible bet combinations are equally bad. Except for the 5-number combination bet which has a larger house edge.

BUT, what about these two possible bets: (1) betting 35 units, one each on 35 different numbers; or (2) betting 34 units, one each on 34 different numbers?

Think about this, if you lose under the 1st option (35 units), it will take you 35 wins to offset one loss, BUT a loss under the 2nd option can be offset with only 17 wins. Doesn't this seem like an advantage??

More thoughts (on a later day) - Looking at the first scenario: If the wheel hit 38 different numbers, you would have won one unit 35 times and lost 35 units three times for a net loss of 70 units. Looking at the second scenario: If the wheel hit 38 different numbers, you would have won two units 34 times (+68) and loss 34 units 4 times (-126) for a net loss of -68. Is this the advantage?? a mere 2 units? or is my math simply wrong again??

My Latest Conundrum

MY last post got me thinking about the difference between statistics and probability or expected outcomes vs. actual outcomes.

According to the law of big numbers, a large sample of random events will tend to move toward expected outcomes. Therefore, if you were to look at 38 million roulette decisions, you should see that each number appeared about 1 million times and that about 47% of the decisions would be red for example.

Now, IF this proposition is true then it is fair to say that the average gap between repeats would have to be about 38 spins. This means that if you were to take any particular number, like the number 3 for example and see how often it appears, the answer would be once every 38 spins.

IF it is true that numbers tend to appear every 38 spins in the long run, it would seem to me that we have an exploitable event.

Suppose you place one chip valued at $18 on the "high" numbers (outside bet on numbers 19-36), probability and statistics would show that you would probably win this bet almost half the time (because you have almost half the layout covered). Now suppose instead, you placed 18 individual bets on the inside placing one dollar chips on each of the numbers between 19 and 36. Statistics and probability would tell you that you should win almost half of these bets and in fact the results should be identical to the "other" way mentioned above.

Now lets look at a moving target. Suppose you were to place 18 individual one-dollar bets on the 18 numbers which are NOT the last 18 numbers to appear (not considering the zeros just for the sake of this example). Let me re-phrase that. Make a list of the last 18 numbers to appear and place 18 one-dollar bets on the "other" numbers. Would this provide you an advantage? Should you win this bet MORE OFTEN than half of the time?

Now looking back at our first proposition. IF IT IS TRUE that numbers tend to appear every 38 spins then it would seem that while many numbers appear less often than 18 spins many more numbers are appearing at intervals greater than 18 spins. (If the average gap between repeats is 38, then necessarily some number gaps would be greater than 38.)

NOW, IF the average gap is greater than 18, then wouldn't it make sense that by playing 18 numbers which are not the last 18, would give you an advantage? Wouldn't it seem that the next number to come up is more likely to NOT be of the last 18 than TO be of the last 18.

And if so, doesn't this fly in the face of the fact in my most recent thread that the last 38 decisions is likely to only be comprised of 24 numbers??

Something to think about . . .

PRELIMINARY TESTING

I took a look at some spins created by a random number generator. The sample is entirely too small to draw any conclusions, but the results are encouraging.

The First sample i looked at was 1000 spins. I recorded data for 982 spins (skipping the first 18). I looked at each spin and measured the distance since the last appearance of that number. If the number was 18 or less, I counted this as a loss of 20 units. If the number was greater than 18, I counted that as a win of 16 units. The explanation is as follows: if you bet 1 unit on each of the 20 numbers that do not represent the last 18 numbers to show, then you will win a net of 16 units if the next number is not one of the last 18 and you will lose 20 units if the number is among the last 18.

This produced +1960 units for the 982 spin sample. The gain is 2 units per spin (or about than 1/10th of the minimum bet).

If you play with $1 chips, you would need to bet $20 a spin. Each win will net you $16. You would have won $2 per spin or $120 per hour.

If you play with $5 chips, you would need to bet $100 a spin. You would have won $600 per hour.

Not bad at all.

What about drawdowns? I broke my preliminary batch of numbers into hour long sessions. Only 2 sessions ended up in the negative and they were -72 units. I would feel comfortable with a bankroll of 300 units or 400 units.

SECOND BATCH

The second batch of randomly generated numbers represented 282 spins. The total gain was 660 units (2.34 per spin) or slightly higher than the first batch average (2 units per spin). Perhaps the best way to look at this is a gain of about 1/10th unit per spin or 6 units per 60 spins or 6u/hour.

(I remind anyone who reads this that the preliminary test samples are very small at this point.)

THIRD BATCH

The 3rd batch was 982 spins which produced 1622 units or 1.65 units per spin.

Totals through 3 batches:

2,246 spins = 4,242 units or 1.89 units per spin (113.4 units per hour).

FOURTH BATCH

982 spins with a total gain of 2212 or 2.25 units per spin.

Totals through 4 batches:

3,228 spins = 6,454 units or 1.999 units per spin

Average = 120 units per hour (based on 60 spins per hour)

FIFTH BATCH

982 spins with a total gain of 2052 or 2.09 units per spin.

Totals through 5 batches:

4,210 spins = 8,506 units or 2.02 units per spin

Average = 121.2 units per hour

SIXTH BATCH (abnormally high results)

456 spins with a total gain of 1860 units or 4.078 units per spin.

Totals through 6 batches:

4,666 spins = 10,366 units or 2.22 units per spin

Average = 133.2 units per hour.

SEVENTH BATCH

979 spins with a total gain of 1947 units or 1.988 units per spin.

Totals through 7 batches:

5,645 spins = 12,313 units or 2.18 units per spin

(very impressive through 5,645 spins*)

*but see "THE ERROR" below

PRACTICAL CONSIDERATIONS

I have to admit that covering 20 numbers within a few seconds may be challenging. The task is not only to cover 20 numbers but to select those numbers as well. The key is to cover all except the last 18 to show. With a partner, one could be placing all the numbers as the other person reads the board and removes the most recent 18. If this works, I'm sure I can find a way to get the coverage I need.

Another idea might be to have a small laminated copy of the layout and to "x" out the last eighteen numbers with a dry-erase pen. Then use this to show you where to place your bets. Another thought would be that you do not need to place very bet, it would make sense that your chances of winning are the same if you placed a bet every other spin. So, if it took you a while to get your info together, you could just play a table minimum outside bet on one of the even chances and then place your 20 inside bets on the next spin.

The above numbers are actually quite staggering. Using $1 chips, you could expect to win 130 units per hour. Using $5 chips, you could expect to win $654 per hour. All with a bankroll of only 400 chips ($400 or $2,000).

The worst-case scenario (in the first 7 batches tested) is an hour with losses of 288 units ($288 or $1,440 using $5 chips).

The best hour (in the first 7 batches tested) is +396 units ($396 or $1,980).

ONE POTENTIAL FLAW

One potential flaw in my test numbers is that I am counting as wins any number that has not appeared in the last 18 spins. I am covering 20 numbers and leaving the last 18 uncovered. However, if the last 18 spins include any repeats, then 20 numbers would not be enough coverage. For example, 18 spins might only represent 14 numbers, if so, then I would have to cover 24 numbers to cover all numbers not included in the last 18 spins. This would mean that the only ideal time to bet is when the last 18 spins have no repeats.

THE ERROR (OF MY WAYS)

I've come to accept that the "one potential flaw" mentioned above is in fact the critical flaw of my testing. The very impressive numbers above are based on a scoring system where I assigned a +16 to each win (repeat gap larger than 18) and a -20 to each loss (repeat gap 18 or less). The problem is that this would only be accurate if the last 18 spins represented exactly 18 distinct numbers (which should rarely be the case). In order to truly test this tehory, I would have to look at the last 18 numbers to appear (not the last 18 spins). The question becomse, how many spins does it take to produce 18 numbers, my guess is something like 21 or 22. If I were to go back and re-score the test numbers using 21 or 22 spins, the result would be dramatically different (obviously worse). A different way to look at the test nukbers is that 18 spins probably only represents about 16 numbers on average (maybe even less). So, to cover all the other numbers, I would have to place 22 bets. This means each win would be only a +14 and each loss would be a -22. No doubt this would also drastically reduce the results above.

SO

So, does the law of big numbers help us to find an exploitable anomoly? Is the average gap between hits really 38 or is it more like around 24?

Back to the drawing board . . .

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

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

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.