Roulette Probability Formula

Posted on 4/9/202222.08.2017by admin

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Roulette Probability Statistics
Roulette Probability Formula
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Roulette Probability Formula Cheat

The odds for a random event, like a die toss or a roulette spin, denote the likelihood of this event taking place. To calculate the odds for winning with a given roulette bet, you need to figure out what its probability is. Then you can use the following formula: Odds for Winning = Probability of Winning / (1 – Probability of Winning). The calculation of roulette wheel probabilities is very simple. Let us take the classic example of a dice with the numbers one to six. How large is the probability that the next throw is the number five? To calculate the odds for winning with a given roulette bet, you need to figure out what its probability is. Then you can use the following formula: Odds for Winning = Probability of Winning / (1 – Probability of Winning).

MichaelBluejay

My software is giving me unexpected results for binomial probabilities.
I'm trying to solve the question, what is the probability of x hits of a certain number on a 37-slot roulette wheel over a certain number of spins?
e.g.:
p = 1/37 (probability of success of one trial)
q = 1-p (probability of failure of one trial)
x = 1 (number of successes)
n = (number of spins)
Binomial formula: n! ÷ ( x! * (n-x)! ) * p^x * q^(n-x)
With n=170, I get 0.0448 in both LibreOffice Calc (spreadsheet) and my Javascript code.
With n=171, LibreOffice gives a '#VALUE!' error and JS says the answer is Infinity.
Surely I haven't gone from 4.5% to Infinity from just one additional spin!
With n=200, JS returns 'NaN' (not a number).
I'm confused.

OnceDear
Administrator

Thanks for this post from:

170! is pretty big at 7.257416e+306
The biggest/largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double. That is, DBL_MAX or approximately 1.8 × 10^308 (if your double is an IEEE 754 64-bit double).
You are close to that limit.

Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

kubikulann

Thanks for this post from:

Maybe use the Sterling approximate for a factorial? Maybe it would allow simplification of the fraction before actually computing 171!
The problem is you ask the software to compute a huge number (171!) to get a much smaller one.
Note: Excel computes COMBIN(171,x) without problem.

Ace2

Thanks for this post from:

The factorials cancel.
No need to use excel for something that can be done by any calculator or even by hand or slide rule.
(36/37)^169 * 170/37 = 0.0448 = a
a * 36/37 * 171/170 = 0.0438
Probability

In 171 spins you expect to hit your 1/37 number 4.62 times. You have a 1 - (36/37)^171 = 99.08 % chance of hitting it at least once, a 4.38% chance of hitting it exactly once and a 94.7 % chance of hitting it twice or more

It’s all about making that GTA

CrystalMath

Thanks for this post from:

Try this:

It avoids getting huge numbers because most terms cancel.

I heart Crystal Math.

MichaelBluejay

Thanks for this post from:

Thanks everyone for the help!
@OnceDear, yes, I should have figured out that one of my variables besides the actual result was going out of bounds. (Duh.)
@CrystalMath, your solution is perfect and exceptionally helpful. Thanks so much!

MichaelBluejay

Okay, I'm stuck again.
So, my understanding of canceling the factorials doesn't jibe with CrystalMath's Java/C code above. (CrystalMath, I'm referring to you in the third person not out of disrespect, but because other people might be helping besides you.)
So I must be missing something. Here's my understanding:
Binomial coefficient:

n = 10
x = 7

So, to do the most cancelation, I should cancel n! with x!, not with (n-x)!
The binomial coefficient formula can be rewritten as:

So, I'll calculate the factorials after canceling, then divide by the factorial of n-x, then multiply by the rest of the binomial probability formula.
After cancelation, I'll be multiplying 8 x 9 x 10. So, my code starts with:

CrystalMath's code is:

So, that's the first thing that doesn't make sense to me. I'm guessing he's taking a shortcut.
However, if I use his code, looking for the probability of at least one hit of a single number over 37 spins, I get 38%. That does jibe with what I get in my spreadsheet. But it's not intuitive (seems low, seems like it should be >50%), and it doesn't jibe with the formula:
1 - (ways to win / total ways) ^ trials = 1 - (36/37)^37 = 64%.
So, that's the second thing I'm confused about.
Finally, when I use my own JS code, my results are even farther off. My understanding for 'at least' is that I find the probability for exactly x matches, then multiply it by the number of trials.

That gives me probability of <0.1%, which is clearly wrong. That's the third thing I'm lost on.

kubikulann

1st thing.
There is a symmetry : C_n^x = C_n^n-x
So you could put some code like x:= max(x,n-x)?
2d thing and 3d thing.
P(at least) is not P(exactly), but it is also not [1-P(exact)]^n or *n. You need to compute P(exact) for all the x’s from x to n, then sum it.

CrystalMath

I see, I wasn't dividing by the x! term.
Looking at this example:

Roulette Probability Statistics

We can simplify this to 10 x 9 x 8/(3 x 2 x 1). So, the loop should just do three steps.
I've modified my code below.

If you want 'at least,' you will need to loop through all possibilities and sum them.