Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Probability of an Even Number of 6's

Date: 05/03/2003 at 05:45:36
From: Liya Andres
Subject: A fair die

Dear Dr. Math,

A fair die is thrown n times. Show that the probability that there is 
an even number of sixes is

   1/2 * [1+(2/3)^n]

Note: 0 is considered an even number.

What I find confusing about it is how to get the probability of an 
EVEN number of sixes. I'm also confused about where the 2/3 that 
is found in the equation came from.

I was only able to get the probability of getting at least one six for 
n tosses, which I calculated as 1-(5/6)^n, and the probability of 
getting no sixes at all, which I calculated as (5/6)^n. But I can't 
figure out how to work this into showing the equation is true.


Date: 05/03/2003 at 07:35:59
From: Doctor Mitteldorf
Subject: Re: A fair die

Liya -

Here's a shortcut, where you don't have to calculate the probability 
of 0, 2, 4, etc. 

We'll use induction on n. In a proof by induction, we show that the 
statement is true for n = 0, then use the fact that it's true for n to 
show that it's also true for n + 1. Thus the proof for n = 0 implies 
it is true for n = 1, which implies it is true for n = 2, etc.

For n = 0, the formula works out to p=1, which is the right answer 
since 0 sixes is the only possibility.

Now suppose it is true for n = n. Then the probability of an even 
number of sixes is 1/2 * [1+(2/3)^n] and for an odd number of sixes it 
is 1 - 1/2 * [1+(2/3)^n]. We can calculate the probability of an even
number of sixes for n + 1 tosses as coming about in two ways: either
the result for the previous n tosses was even, and the next toss is
not a six (5/6 * 1/2 * [1+(2/3)^n]), or else the result for the
previous n tosses was odd, and the next toss is a six 
(1/6 * (1 - 1/2 * [1+(2/3)^n]). Add these two expressions and 
simplify, and you get  

   1/2 * [1+(2/3)^(n+1)]

which is the original formula as it applies to (n + 1) tosses.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/04/2003 at 05:24:46
From: Liya Andres
Subject: A fair die

Thank you so much!

Is there a way to show how the given formula was derived, using the 
concepts of probability instead of induction? 

-Liya


Date: 05/04/2003 at 05:40:49
From: Doctor Mitteldorf
Subject: Re: A fair die

Dear Liya,

I can't do it for a general n, but I can show you how it would be done 
using a particular n for an example.  Let's say n = 5.

The probability of rolling r sixes out of 5 rolls comes from two 
parts.

The first part comes from the probability of (six) and the probability 
of (not six). These are 1/6 ad 5/6. And the factor we need is 
(1/6)^r * (5/6)^(5-r). This is because the (1/6) probability came up 
r times, while the (5/6) part came up (5-r) times. 

The second factor comes from the number of different orderings that 
are possible. For example, if r = 5, then you are talking about every 
roll being a 6. There is only one way to do that. But for r = 4, there 
is one roll that is not a 6, and it could be the first, the second, 
etc. There are 5 different positions for that one (not-six). For r = 
3, it's a little more complicated: There are 3 sixes and 2 (not-sixes) 
to mix together in any order. There are 10 ways to do this, such as 
66NNN, 6N6NN, 6NN6N, 6NNN6, N66NN, etc. (Can you list the other 5?)

In general, the number of different orderings is the combinatorial
formula C(5,r), sometimes written as 5Cr, with the 5 and the r as
subscripts in the front and back.  The way to calculate 5Cr is

                  5!
   C(5,r) = -------------
             (5 - r)! r!

So we can put this all together: The probability of getting r heads 
out of 5 is C(5,r)* (1/6)^r * (5/6)^(5-r).

I suggest you calculate this for r = 0, r = 2, and r = 4 and add them 
up to see that the number you get agrees with your formula for the
proability of an even number.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/04/2003 at 09:19:35
From: Liya Andres
Subject: Thank you (A fair die)

Dear Doctor Mitteldorf,

Just want to let you know how much I appreciate your quick and helpful 
replies. Thank you! :)

-Liya
Associated Topics:
High School Probability

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/