Can Excel's BINOMDIST function accurately calculate roulette probabilities?

[Roulette]

Hello I have a basic but quite difficult question about roulette probabilities (assuming roulette has 37 numbers 0-36). I need a general formula for calculating the probability of an event, given specific parameters:
We need to calculate the probability P(e) of the event E = [Bet B appearing X times in N trials(ie. spins) ]
We know:
N= the numbers of spins
X=the number of times a specific bet wins/appears in those N trials (spins)
P(b) = the probability of bet B for a single spin/trial.
How do we calculate the probability of [Bet B appearing X times in N trials(spins) ]?

For example what is the probability of a specific number appearing exactly 1 time in 37 spins, given it’s probability is 1/37?
What’s the formula?
Can we in the same way calculate, let’s say the probability of 12 specific numbers (probability 12/37) coming 2 times in 5 spins etc.?

I have tried to devise a formula which you can see here: http://www.roulette30.com/2014/01/calculating-probability-roulette.html
P(e) = (n!/(x!(n-x)!)) P(b)^x
But I think it is incorrect since it gives an extremely low probability of a specific number appearing exactly 1 time in 37 spins.

Maybe this is the correct formula?
P(e) = (n!/(x!(n-x)!)) P(b)^x (1-P(b))^n-x
But this again gives an extremely low probability of a specific number appearing exactly 1 time in 37 spins, if my calculations are correct.

Thanks in advance.

 

[PeroK]

This is covered by the Binomial Distribution, perhaps using the Poisson approximation:

http://en.wikipedia.org/wiki/Binomial_distribution

 

[PeroK]

Maybe this is the correct formula?
P(e) = (n!/(x!(n-x)!)) P(b)^x (1-P(b))^n-x
But this again gives an extremely low probability of a specific number appearing exactly 1 time in 37 spins, if my calculations are correct.

Thanks in advance.

What do you get for this? I get 0.373 for a number appearing exactly once in 37 spins. And 0.362 that it doesn't appear. And 0.186 that it appears twice.

That formula is correct. If you do x = 0, 1, 2 ... and add them all up it will come to 1.

 

[Roulette]

Hi PeroK,

Thanks for taking the time to reply and actually do the calculations.
Maybe I did an error in the calculation of P(e) = (n!/(x!(n-x)!)) P(b)^x (1-P(b))^n-x . I found it much lower, but you are probably correct.
And yes I did consult wikipedia about it.

And one more question, what kind of calculator do you use for such complicated functions and often very long numbers? (I use google's scientific calculator)

Thanks again

 

[PeroK]

Hi PeroK,

Thanks for taking the time to reply and actually do the calculations.
Maybe I did an error in the calculation of P(e) = (n!/(x!(n-x)!)) P(b)^x (1-P(b))^n-x . I found it much lower, but you are probably correct. So this is the correct equation to calculate the described probability?
And yes I did consult wikipedia about it.

And one more question, what kind of calculator do you use for such complicated functions and often very long numbers? (I use google's scientific calculator)

I found a BINOMDIST function in Excel!