An almighty coincidence

Issue 45
December 2007

Imagine picking the four hymn numbers out of a hat. First note that four-hymn combinations with one 1-digit number and three 3-digit numbers come in four types: the one digit number can occur in first, second, third or last place of the selection. So the overall chance of picking such a combination is equal to:

Chance of picking a combination with 1-digit number in 1st place
+ chance of picking a combination with 1-digit number in 2nd place
+ chance of picking a combination with 1-digit number in 3rd place
+ chance of picking a combination with 1-digit number in 4th place.

Each of the terms in this sum is equal to

  \[ 9/999 \times 900/999 \times 900/999 \times 900/999, \]    
so the overall chance of picking four hymn numbers such that one of them has 1 digit and the others have three digits is
  \[ 9/999 \times (900/999)^3 \times 4 = 0.0263. \]    

Now for the chance of picking a combination with two 2-digit numbers and two 3-digit numbers. There are 6 ways in which to choose the positions of the two 2-digit numbers within the string of four numbers, so this time the selection comes in 6 different types:

  • $2233;$

  • $3322;$

  • $2332;$

  • $3223;$

  • $2323;$

  • $3232.$

The chance of picking a combination of each individual type is

  \[ (90/999)^2 \times (900/999)^2, \]    
so the overall chance is
  \[ (90/999)^2 \times (900/999)^2 \times 6 = 0.0395 \]    
document

In general, the number of ways you can choose a set of $k$ positions within a sequence of length $n$ is

  \[ \frac{n!}{k!(n-k)!}, \]    
where $n! = n \times (n-1) \times ... \times 2 \times 1.$ In our examples, we first had $n=4$ with $k=1$, giving
  \[ \frac{4!}{1!3!} = 4, \]    
and then $n=4$ with $k=2$, giving
  \[ \frac{4!}{2!2!} = 6. \]    

Return to main article.