The chances that the numbers that did not come up will hit in the next x spins.

[FreeRoulette]

1 - (numbers up/ 38)^numbers to spin

Example:
1-(10/38)^2  copy formula into google search

Suppose 10 numbers came up and you want to bet on the other numbers that did not come up for the next 2 spins.

The probability that you will win is (1-(10/38)^2)*100 = 93.07%

[Patrik]

I want to show the weaknesses and strengths behind sleepers.
And I do that using the binomial probability.

A dozen has 50% to hit within two attempts.
That means that a dozen that has slept 16 times is equal to 8 EC.
And a 12-step dozen progression is equal to 6 EC.
That is poor, just for illustration.

One single number has 50% to hit within 25 attempts, just so you know.
I attach the binomial probability chart for you to explore.

Cheers

[FreeRoulette]

Hello Patrik,

Thank you for the chart, those are nice trigger points to know.

[Albalaha]

Very interesting discussion. Recently, I was working on such things and observed that a dozen/column mostly hits within 17 spins. Rather, if we see it break even point wise, 5x of its break even, I.e. 15 spins of no hit, it is most likely to hit within next three spins than not. That works on most bets like in an EC, after 10 spins of no hit, it is most likely win soon in the next break even than not. Remember, it is not a real limit but probability resists any bet going past that, in most cases. Now, this should be taken as probability and not to be used as a fallacy, in any manner.

[Smith]

1 - (numbers up/ 38)^numbers to spin

Example:
1-(10/38)^2  copy formula into google search

Suppose 10 numbers came up and you want to bet on the other numbers that did not come up for the next 2 spins.

The probability that you will win is (1-(10/38)^2)*100 = 93.07%

I'm finding this very confusing. What are "the other numbers that did not come up"?

I don't know what your calculation 1-(10/38)^2 means, but the probability that one of 10 particular numbers comes up in the next 2 spins is far less than 93%. Are you sure you didn't mean 1-(1-10/38)^2? This is the probability that you will get a hit in 2 spins betting 10 numbers, and it's 45.7%.

[FreeRoulette]

1 - (numbers up/ 38)^numbers to spin

Example:
1-(10/38)^2  copy formula into google search

Suppose 10 numbers came up and you want to bet on the other numbers that did not come up for the next 2 spins.

The probability that you will win is (1-(10/38)^2)*100 = 93.07%

I'm finding this very confusing. What are "the other numbers that did not come up"?

I don't know what your calculation 1-(10/38)^2 means, but the probability that one of 10 particular numbers comes up in the next 2 spins is far less than 93%. Are you sure you didn't mean 1-(1-10/38)^2? This is the probability that you will get a hit in 2 spins betting 10 numbers, and it's 45.7%.

You are right, I did mean this 1-(1-10/38)^2. After I posted and saw the 93%, I knew that didn't sound right, but was pulled way and didn't get back to edit it. Thank you for the correction

[Toprengo]

Thank you for this calculation. The number "38" means that it is a double zero wheel or ... I think better if I ask you about it. Thanks for answer.