Losing streaks in terms of an EC

[Bayes]

Most people who play roulette and look at the marquee have a rough idea of the streaks that can occur on the even chances; they know, for example, that longer streaks are less common than shorter streaks, and they get familiar with the kinds of patterns which arise.

But if you're playing some other location, such as a double street or corner, you don't get that feel  for the streaks (in particular, the losing runs) because they're not shown on the marquee.

The following table compares the losing streaks of all bets from 1 number up to 12, in terms of an EC losing streak. In other words, the probability of each losing EC streak was calculated, and then this was used to find the corresponding losing streak for the bet of interest. So for example, suppose you wanted to find the losing streak of a single number, which is equivalent to an EC losing streak of 2:

[attachimg=1]

The numbers along the top row of the table represent all bets from 1 number up to a doz/col (12 numbers), and the numbers down the leftmost column represent the EC losing streaks (from 1 to 16). So to find the the losing streak of a single no. equivalent to an EC streak of 2, first locate row 2/ and move to the right 1 column. The answer is 53.

Another example: What is the equivalent losing streak, betting on a double street (6 numbers) of a run of 10 losses on an EC?

Locate the 10th row of the table, then move along 6 columns to the right. The answer is 41 spins.

[Gizmotron]

I'm just guessing here but is there a chance that "EC" stands for even chance?

Last time I checked an even chance bet was 18 numbers. I don't understand the explanation and the chart.

[GARNabby]

Last time I checked, an even chance bet was 18 numbers. I don't understand the explanation and the chart.

After on average 26 tries, there's half a chance of Ken getting his one number?  (Wait-time probability.)

I'm still waiting for someone to explain the labby, etc, in terms of the straight-up neg/pos bet-progressions.  People get "carried away" with that stuff, but won't take the time to see exactly where that's supposed to be getting them.

[Bayes]

Gizmo, further explanation forthcoming, but I just realized I made a mistake in the table; I used the wrong number in the formula, which overestimated the losing streaks. I'm going to delete the original post and re-post with the corrected table.

[Bayes]

Ok, I can't delete the first post, so here is the corrected table:

Please use THIS table for reference, NOT the table in my first post, which had an error.

[attachimg=1]

[Bayes]

Gizmo, yes EC = even chance.

Take the number in the top left-hand corner of the table (26). 26 is the number of spins of any single number (say 17) required in order to have a 50% chance of it hitting. Notice that this equivalent to a single EC 'miss'. Now taking the left-hand column, the numbers go from 1 to 16, which represent the losing streaks of any EC. The 2/ row means I calculated the chance of losing 2-in-a-row for an EC, which is approximately 25%.

Now, to get the entries for that row in each of the remaining columns, I set this number (0.25) equal to the probability of getting n misses in a row for betting 1 number, 2 numbers, 3 numbers etc, then I 'solved' the formula for n, which gives the number of consecutive spins the bet must go missing in order for it to be equal to the probability of 0.25.

See?

example: how long should the losing streak of a single street (3 numbers) be in order for it to have the same probability of an EC losing 2-in-a-row? For a single zero wheel, the chance of say red losing 2-in-a-row is:

(1 - 18/37)²

This has to be set equal to the chance of single street losing 'n' in a row, which is

(1 â€" 3/37)ⁿ

So,

(1 â€" 18/37)² = (1 â€" 3/37)ⁿ

but we want to know what n is, so we have to get it on its own using a little algebra.

2×log(1 â€" 18/37) = n×log(1 â€" 3/37)

Therefore, n = 2×log(1 â€" 18/37) / log(1 â€" 3/37)

Which gives n = 15.76 (16, rounded up)

If you do this for each EC streak (2,3,4,....) and each bet on 1,2,3... etc, then you get the table.

[Bayes]

So what's the point of the table?

It's just a way of seeing how likely a losing streak is of any bet by comparing it to those which are more familiar. For example, you might think that if betting on a double street, then a losing streak of 30 spins is pretty unlikely, but referring to the table, you can see that it's equivalent to an EC losing streak of 8, which isn't that rare at all.

Or knowing that over 80% of the time, EC losing streaks are less than or equal to 3, you can now see that for over 80% of the time, a bet on a single split will have losing streaks of less than or equal to 36.

[Gizmotron]

Is There a formula to calculate the odds for when things change from continuing to not continuing?

Example: 1)  A streak of reds end. 2) the current sequence is not a streak of red or black but changes to a streak of blacks.

[GARNabby]

Here's a pretty-good calculator for all of that sort of possibility: link:://stattrek.com/tables/binomial.aspx .

[GARNabby]

Is There a formula to calculate the odds for when things change from continuing to not continuing?

Example: 1)  A streak of reds end. 2) the current sequence is not a streak of red or black but changes to a streak of blacks.

Sure, but not a single formula.  Much more work involved, even on something only like determining the number of spins required, on average, to have all the numbers show up.  And with such information in hand, its value is dependent on your real strategies to beat the game.  (This is the point where i, myself, stop giving away my work.)

[Gizmotron]

GARNabby, I think it's wise to keep effective strategies to yourself. People will go into gold rush mode. It's  a sure opportunity killer.

[Bayes]

Is There a formula to calculate the odds for when things change from continuing to not continuing?

Example: 1)  A streak of reds end. 2) the current sequence is not a streak of red or black but changes to a streak of blacks.

Gizmo, I don't really understand the question, can you give a short example.

2) the current sequence is not a streak of red or black ...

huh? it has to be one or the other! 

[Gizmotron]

Bayes - " huh? it has to be one or the other!"

When something is not a streak it's generaly understood that the condition is chaotic.

When no dozens sleep and singles are also scarcely seen this is a useful condition as long as it keeps happening. There is no reason why. There is no way to know how long it will continue. If any thing continues, even if it's not a streak, it can be exploited while it continues.

[Skakus]

 
Great chart you posted there, Bayes. Thanks.

Let's bet 3 numbers and stop on 1 win,

if you bet 3 numbers 12 times you profit/break even/lose.

If you bet 3 numbers 16 times you profit/break even/lose/lose.

Question:
If I opt for betting 16 times should I increase the stake for the last four bets making it profit/break even/profit/lose?

Or flat bet and maintain the original scenario of profit/break even/lose/lose?

[Bayes]

When something is not a streak it's generaly understood that the condition is chaotic.

When no dozens sleep and singles are also scarcely seen this is a useful condition as long as it keeps happening. There is no reason why. There is no way to know how long it will continue. If any thing continues, even if it's not a streak, it can be exploited while it continues.

To me, a streak is an unbroken sequence of a particular bet, like BBBBBB...

So how do you define it? when does a sequence become a streak? when does it stop being a streak and become 'chaotic'?