Name: trudy
Question:
A boy holds 6 pieces of strings in his hands with the ends protruding
above and below. The top ends are tied together in pairs and then the
lower ends are tied together in pairs. What is
the probabilty
that the pieces of string are all joined in one loop?
What is the probability
of obtaining two loops?
Can you generalise this to solve the problem for 2n blades of grass?
Hi Trudy,
Look at the six string ends above his hand. Choose any one of them.
There are 5 choices for the string to tie it to. Once you have done that
there are 4 untied strings. Choose any one of them. There are 3 choices
of the string to tie it to. Once that is done you must tie the remaining
two strings. Thus there are 5
31
ways to tie the string ends above his hand in pairs. In a similar way
there are 531
ways to tie the string ends below his hand in pairs. Hence there are
(531)2 ways
to tie the ends together.
How many of these result in the strings forming one loop? Take the top
end on any string (call it string 1) and choose one of the available
five ends to tie to it. There are 5 choices. Call the string you choose,
string 2. Now look at the bottom end of string 2. It can't be tied to
itself and it can't be tied to the bottom end of string 1 or you will
have a loop. Thus there are 4 choices of the string to tie to it. Call
the string you choose, string 3. Now look at the top end of string 3.
There are three availabe top ends of strings to tie to it, hence 3 choices.
Choose one of them and call it string 4. Again, look at the bottom end
of string 4. You can't tie it to string 1 or you will complete a loop.
Strings 2 and 3 are already tied togethet so there are only two available
string ends to tie to it. Choose one of them and call it string 5. Now
you are out of choices. The top end of string 5 can only be tied to the
one remaining end and the bottom end of that string must be tied to string
1, forming one loop. Hence there are 54321
ways to form one loop.
Hence the probability that the pieces of string are all joined in one loop is
5
54321
is called 5 factorial and written 5!. Similarily n!=n(n-1)(n-2)...21.
531
ic called the double factorial of 5 and written 5!!. If n is odd then
n!!=n(n-2)(n-4)...1
and if n is even then n!!=n(n-2)(n-4)...2.
Using this notation the argument above for 6 strings can be extended to 2n strings to show that the probability that the pieces of string are all joined in one loop is
(2n-1)!/[(2n-1)!!]2
Andrei and Penny