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Binomial and Geometric Probability Distributions

Date: 10/20/2005 at 00:02:08
From: Josť
Subject: the probability of rolling _exactly_ one 6 in n rolls

What is the probability of rolling exactly one 6 in n rolls?  I know
how to find the expected value, and I know how to compute it if it's
not n, but I don't know how to compute it for n.  I think it would
have something to do with a probability distribution like 1-(5/6)^n.



Date: 10/20/2005 at 08:36:55
From: Doctor Wilko
Subject: Re: the probability of rolling _exactly_ one 6 in n rolls

Hi Jose,

Thanks for writing to Dr. Math!

It looks like to answer your question, you'd use the binomial 
probability distribution.  

For a dice roll, the probability of getting exactly 1 success (in this 
case a six) out of n trials is calculated by,

  nC1 * (1/6)^1 * (5/6)^(n-1)

Let's think about a concrete example.

If your question was, "What's the probability of rolling exactly one 
six in four rolls?", you could have a six in any of the four spots 
and any other non-six's in any of the remaining spots, i.e., 

  _6_  _2_  _1_  _5_, or

  _2_  _6_  _3_  _3_, or

  _3_  _3_  _6_  _5_, or

  _3_  _2_  _2_  _6_

The probability would then be,

  4C1 * (1/6)^1 * (5/6)^3 ~ 0.3858
   |       |         |
  6 can be |         |
 in any of |         |
 the four  |     Prob. of three
  spots    |     non-six's
           |
           |
        Prob. of 
      getting one 6

About a 38.5% probability of getting exactly one six in four rolls.

This link from our archives will explain more about the binomial 
distribution:

  Binomial Probability Formula
    http://mathforum.org/library/drmath/view/66627.html 


Just for fun, what if the question had been, "What's the probability 
of getting a six for the _first time_ on the nth trial?"

Notice how this wording is different from the first question.  It's 
still exactly one six in four rolls, but this time the six is 
restricted to the last roll.

This time the probability would be calculated as,

  (5/6)^(n-1) * (1/6)      

Looking at a concrete example again, "What's the probability of 
getting a six for the _first time_ on the fourth roll of the die?"

You could have,

 _4_  _3_  _3_  _6_
                 | 
   \        /    |  
     \     /   Last spot reserved for 'success',  
        |       in this case a 6
        |
   (n-1) non-six's
   (4-1) = 3 non-six's

So the probability would be calculated as,

  (5/6)^(3) * (1/6) ~ 0.0965

About 9.5% of getting a six for the first time on the fourth roll of 
a die.

This is referred to as the geometric probability distribution.

Note the difference in probabilities.  It looks like since the six 
can be in any of the four spots in the first example (binomial), the 
probability is higher (4 times in fact!) than when the six is 
restricted to be in the last spot (geometric).

Does this help?  Please write back if you have further questions.

- Doctor Wilko, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Probability

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