I thought there were 38 possibilities?
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: roulette probability calculation attempt
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.
Thanks in advance.
I thought there were 38 possibilities?
I don't have emotions and sometimes that makes me very sad.
You second formula looks alright to me. You're just looking at the binomial distribution. If we assume a probability of 1/37 for a success on a single spin then the probability of getting exactly 1 success in 37 spins is 0.3729305
I don't have emotions and sometimes that makes me very sad.
Kav (03-25-2015)
Hi and thanks for the replies.
Yes the second formula is correct I believe. I have corrected my article.
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