Probability: Marbles, Coins, Cards

Date: 05/24/97 at 00:00:44
From: Jean-Pierre Berard
Subject: Help wanted

Hello,
 
I want to write a little program to calculate permutations, 
combinations, variations with repetitions, and so on.  I thought that 
writing the formulas down would make the reasoning come clear.
 
Right away, I was confronted with my own shortcomings in this area of 
math! In an effort to overcome this, could you please explain the 
following problems to me?
 
1) A bag contains 10 marbles, which are identical except for their 
color. There are 4 blue marbles and 6 red marbles. If marbles are 
drawn at random from the bag, what is the probability of drawing the 
following colored marbles in the order indicated: red, blue, blue, 
red.
    
2) The directors of a company can not decide who should be chairman 
and minute secretary. If there are 20 directors.
    
a) How many different selections are possible?.
b) What is the probability that any two given directors will become
   chairman and minute secretary?.
c) What is the probability that any two given directors will become
   chairman or minute secretary?.

3) A card player draws 5 cards from a full deck of 52 cards.
    
a) What is the probability that their hand of cards contains an ace of
   spades, a 6 of hearts, a 5 of clubs, a king of hearts and a 9 of 
   diamonds?
b) What is the probability that their hand of cards contains an ace of
   spades, a 6 of hearts, a 5 of clubs, a king of hearts and any
   diamond?            

4) In a coin tossing experiment, an unbiased coin is flipped 8 times. 
What is the probability that 5 heads will result?.

Many thanks,
Jean-Pierre Berard

Date: 05/24/97 at 07:55:49
From: Doctor Mitteldorf
Subject: Re: Help wanted

Dear Jean-Pierre,
    
I'm not sure how you want to program these.  One very powerful and 
general method is called "Monte Carlo".  You use the computer's number 
crunching capability, and you just simulate the entire experiment 
perhaps a million times, keeping track of how many times the computer 
achieves a successful outcome by whatever criterion you have.  This 
method is a bit inelegant and slow, but useful in the many 
circumstances when there is no obvious formula or reasoning that will 
make the problem simpler.  The answer is approximate, and gets better 
the more trials you simulate. You can do a Monte Carlo simulation of 
any of the 4 problems you've described, but they are also simple 
enough that they can all be done with analytic methods.   I don't 
think you need a computer.

1) The probability of drawing the first red is 4/10, which leaves 3 
red and 6 blue. The probability of drawing the second blue is 6/9, 
which leaves 3 red and 5 blue. The probability of drawing the third 
blue is 5/8 which leaves 3 red and 4 blue. The probability of drawing 
the fourth red is 3/7 (tt doesn't matter what this leaves.)

The probability of the entire sequence is just the product of the 
probability of each individual drawing:
     
4/10 * 6/9 * 5/8 * 3/7 = 360/5040 = 1/14

2a) There are 20 possibilities for chairman, and having chosen 
chairman there are 19 remaining possibilities for secretary.

b) The probability of any given combination of two people for chair 
and secretary is 1/380.

c) I'm not sure what you mean by this question.  Do you mean the 
probability that either Andrew will be chairman OR Bill will be 
secretary?  Out of the 380 possibilities, there are 19 in which Andrew 
is chair, and 19 in which Bill is secretary, but these two sets 
overlap by 1, in which both conditions are met.  The probability that 
either condition is met is thus 37/380.

3a) The probability of drawing these 5 cards in order is 1 in 
(52*51*50*49*48). But they don't have to be drawn in that particular 
order.  How many different orders are there?  This is the number of 
ways to order 5 cards, which is 5! or 5*4*3*2*1.So the probability of 
having the given hand is:
 
     (5*4*3*2)/(52*51*50*49*48)

This is just the inverse of the number called "52 choose 5" and
sometimes written as a large pair of parentheses enclosing a 52 and a 
5 directly underneath it.  The formula for "n choose r" is

n!/((n-r)!*r!)

3b) This proability is just 13 times as large as (a).

4) If you count as a separate case each sequence (e.g. head-tail-head-
head-head-tail-tail-head) then the total number of possible outcomes 
for 8 tosses is 2^8.  This comes from the fact that each separate toss 
has 2 outcomes, and there are 8 tosses.

You want to know, of these 2^8, how many have 5 heads and 3 tails? 
The answer is "8 choose 5" or (8*7*6) / (3*2*1). This is because there 
are 8! ways of arranging 8 coins if they are all different, but 
they're not all different:  5 of them are heads and indistinguishable, 
so you've overcounted by the orderings of 5 things, a factor of 5!.  
Also, 3 of them are tails and indistinguishable, so you've overcounted 
by a factor 3!

The probability you want is "8 choose 5" divided by 2^8. 

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/