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Probability of Independent Events

Date: 02/07/97 at 18:36:51
From: Carl J. LaGrassa
Subject: Probability question

Given that the probability of an event is known, is there a formula
which can give the probability of an event in a given number of 

To use specific examples, I am interested in the probability of an 
event occurring in 16k, 32k, and 64k when the probability of 
occurrence is 1/32000.

Thanks for your consideration.

Date: 02/08/97 at 05:31:03
From: Doctor Mitteldorf
Subject: Re: Probability question

Dear Carl,

There aren't many formulas for probability questions.  The big 
question here is whether your 16k copies of this problem are all 
independent of one another.  Does the occurrence of the event in one 
place forebode an enhanced probability that it will appear in another 

With that caveat, I can give you a formula: for independent events, 
each with probability p, the probability of n such events is:

                1 - (1-p)^n

For your first example, this is 1-(31999/32000)^16000.  Think of it 
this way: The first one has a (31999/32000) chance of not failing. If 
it doesn't fail, the second one has a (31999/32000) chance, etc. If p 
is small, this turns out to be approximately equal to e to the power 

- Doctor Mitteldorf,  The Math Forum
  Check out our web site!   

Date: 02/08/97 at 22:02:25
From: Carl J. LaGrassa
Subject: Re: Probability question

I'm afraid I don't understand the answer.  I was speaking of 
independent events.  I do not understand the notation "^" or the 
reference to "e".

Specifically, in video poker a royal flush will occur randomly with a
frequency of 1/32000 hands played.  I am looking for the probability 
of getting an occurrence in a given sample size.

Date: 02/09/97 at 06:21:46
From: Doctor Mitteldorf
Subject: Re: Probability question

Dear Carl,

The symbol ^ is used on computers to indicate "to the power of".  
x^2 means x squared.  I use it here because it's not easy to send 
raised exponents over the internet.

The point is simply that if two events are independent, you can 
multiply their separate probabilities to get the probability of both 
happening.  If the probability of a single hand coming out not-a-
royal-flush is 31999/32000, then the probability of two coming out 
not-a-royal-flush is (31999/32000)^2 and the probability of three 
coming out not-a-royal-flush is (31999/32000)^3.

To get the probability that none of the 16000 trials will come out a 
royal flush, simply multiply sixteen thousand times the probability 
that each separately will come out not-a-royal-flush.  This can be 
written (31999/32000)^16000.  Take the number (31999/32000) and 
multiply it by itself 16000 times.

The statement that there will be at least one royal flush in the 16000 
trials is just the opposite of the statement "none of the trials will 
produce a royal flush."  So you can get that probability by 
subtracting that product from 1.


e is the base of natural logarithms and is approximately equal to 
2.71828182845... One definition of e is that if you take larger and 
larger numbers n and calculate the nth power of (1+1/n), the answers 
come out closer and closer to e.  This is something you can try on a 

It also works that the nth power of (1-1/n) comes closer and closer to 
1/e as n gets larger and larger.  Try it!

You can use this last fact to come up with an approximate expression 
for (31999/32000)^(16000).  The thing you're raising to a power, 
(31999/32000), is very close to 1.  Think of it as 1-1/n, where 
n = 32000.  You're not raising it to the nth power, but to half n.  
We use the fact that anything to the power of half n is just the 
square root of the thing raised to the nth power.  (Try proving that 
by multiplying two copies of the thing together, and simplifying what 
you get.)

So (31999/32000)^16000 is very close to e^-0.5, or 1/e raised to the 
power 1/2.

Thanks for writing back.  Please let me know if this is clear now.

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Probability

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