# How to continuously count Random Number generated successes

#### Pfeif0

##### New Member
HI,
I am trying to calculate probabilities of a frequency of occurance; what I'm looking for is probably best described by roulette: Say you are playing based on the first 3rd that has a 12/38 chance every time you play. If you use a specific betting strategy 5,10,15,20,30,45,70 etc what is the overall % chances for each; obviously it is 12/38 for the first try, but the next try you win if you hit the first or second time b/c of the betting etc...
Where my problem lies is every time I try to replicate (F9) and count failures or successes I can't figure out how to keep continually adding successes every time a diffferent random number generates. I'm sure there's a simple way to do that but I am hoping ya'll can help me out.
Thanks,
Keith

### Excel Facts

Why are there 1,048,576 rows in Excel?
The Excel team increased the size of the grid in 2007. There are 2^20 rows and 2^14 columns for a total of 17 billion cells.
Perhaps a review of the principals of probability is in order. At the lowest level, if any one-time try has a favorable probabliltiy of pf, then for two tries, if either one or both establishes a win, then the combined result will be:
For one try, pf.
For only one or the other successful tries, pf + pf
For both successful, pf*pf
So, for two tries, either one or both successful, the probability is
2*pf + pf^2

Does the above show you the way?

I understand your logic...but not sure it is correct after the 2nd try - it seems like for 4 tries it would be 4*pf + pf^4 = which is over 100%...I'm not sure if sheer probability works in this situation where your bet changes based on the result of the previous...and thats why I was hoping for the counting solution in excel

does anyone know how to assign a specific cell to a variable i excel?

I apologize for my misinterpretation of your problem, as I am not into roullete or betting strategies. Perhaps if you would describe the roulette (How many numbers does it have, 38?), how is the 12/38 probability of hitting the first third in one try arrived at? (Are there 12 numbers in one third, or 13?). What is "a betting strategy of 5, 10, 15, 20, 30, 45, 70, etc." mean?

Ummm...no, not quite.

Let the probability of success be p. The corr. prob. of failure is q=(1-p)

Then, the probability of only 1 (but not both) of 2 tries being successful is 2*p*(1-p)
Edit: Strictly speaking, the above is the result of Combin(2,1)*p*q. The general formula for m successes out of n tries would be Combin(n,m)*p^m*q^(n-m). So, with 3 tries, the probability of only 1 success would be Combin(3,1)*p*q^2.

The probability of either 1 or 2 successes out of 2 trials is 1-{prob. of both fail}
=1-q^2 = 1-(1-p)^2 = 1 - 1 + 2p - p^2 = 2p - p^2

Perhaps a review of the principals of probability is in order. At the lowest level, if any one-time try has a favorable probabliltiy of pf, then for two tries, if either one or both establishes a win, then the combined result will be:
For one try, pf.
For only one or the other successful tries, pf + pf
For both successful, pf*pf
So, for two tries, either one or both successful, the probability is
2*pf + pf^2

Does the above show you the way?

Re: How to continuously count Random Number generated succes

I am not quite sure I know what you are doing. Care to clarify the two "etc."s in your post?

Also, what do the numbers in the "betting strategy" represent? What's the logical basis for the sequence of numbers?

I am also having problems with the last statement as to how the probability of success in a subsequent try is dependent on the of an earlier result.

Finally, if this is about something other than roulette, you may want to discuss the real issue at hand and not an analogy.

HI,
I am trying to calculate probabilities of a frequency of occurance; what I'm looking for is probably best described by roulette: Say you are playing based on the first 3rd that has a 12/38 chance every time you play. If you use a specific betting strategy 5,10,15,20,30,45,70 etc what is the overall % chances for each; obviously it is 12/38 for the first try, but the next try you win if you hit the first or second time b/c of the betting etc...
Where my problem lies is every time I try to replicate (F9) and count failures or successes I can't figure out how to keep continually adding successes every time a diffferent random number generates. I'm sure there's a simple way to do that but I am hoping ya'll can help me out.
Thanks,
Keith

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