1. Arrangements and Selections (and how they may be counted)

2. Running our 'megawin.py' demo
Last time we described the idea for a Python program that would compute the probability of winning the MegaMillions lottery jackpot
It was based on a 'recursive' implementation for the function that computes binomial coefficients
0, if n<0 or k<0 or n<k
b(n;k) = , if n==0 or k==0 or n==k
b(n-1;k-1) + b(n-1;k), otherwise {

3. Let's take time now to execute this demo
$ python megawin.py
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NOTE: We can use the 'time' command to find out how long it takes to execute this program:
$ time python megawin.py

4. Output from 'megawin.py'

5. has the potential for faster execution
Can we get 'speedup'?
A formula for evaluating binomial coefficients without resort to recursive function-calls
has the potential for faster execution

6. Special case: binomial(56;5)
Let's first consider counting all arrangements
of 5 integers selected from { 1, 2, 3, ... , 56 }
By an 'arrangement' we mean a way of listing the five selected numbers in some definite order, without repetitions or omissions
Example-selection: { 1, 2, 3, 4, 5 }
Then two possible arrangements are:
( 5, 4, 3, 2, 1 ) and ( 2, 4, 1, 3, 5 )

7. The number of arrangements
There is more than one way to organize the counting of all the arrangements of 5 integers selected from the set { 1, 2, 3, ... , 56 }
Method #1: we make the ordered list directly
Method #2: we first select the 5-member subset, then we arrange its members in an ordered list

8. Method #1 ( __, __, __, __, __ ) 56 ways to choose the 1st integer
55 ways to choose the 2nd integer
54 ways to choose the 3rd number
53 ways to choose the 4th number
52 ways to choose the 5th number
By FCP the total number of arrangements will be
56 x 55 x 54 x 53 x 52

9. Method #2 ( __, __, __, __, __ ) First select the 5-member subset
b(56;5) represents the number of ways to do it
Then arrange these selected integers
5 x 4 x 3 x 2 x 1 = number of ways to do that
There are b(56;5) x 5! ways to do both tasks

10. The two answers must be equal
b(56;5) x 5x4x3x2x1 = 56x55x54x53x52
(method #2 answer) (method #1 answer)
From this equality we can infer that
b(56;5) = 56x55x54x53x52 / 5x4x3x2x1
Or more simply: b(56;5) = 56! / (5! x 51!)

11. General formula for b(n;k)
The same ideas we just employed lead us to a general formula for binomial coefficients:
b(n;k) = n! / k! (n-k)!
So this is the number of ways of choosing k integers from a set containing n integers
(i.e., the number of k-element subsets of {1,...,n}

12. Applying formula in Python program
def binomial( int n, int k ):
num = 1
den = 1
for j in range( n ):
num *= (n – j)
den *= (k – j)
return num / den
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(Trouble if n or k is negative or zero -- or equal)

13. Our Python code

14. Output from 'megawin2.py'
So this revised version of our 'megawin' program ran more than 250 times faster!

15. Poker hands
In the game of 'poker' a player is dealt 5 cards selected from a 'shuffled' deck of 52 cards
How many different poker hands are possible?
The answer is another binomial coefficient:
b(52;5) = 52x51x50x49x48 / 5x4x3x2x1

16. Evaluating without a computer
Look for 'common factors' that can be cancelled:
(52)(51)(50)(49)(48) ((13)(4))((17)(3))((10)(5))(49)((24)(2)) =
(5)(4)(3)(2)(1) (5)(4)(3)(2)(1)
= (13)(17)(10)(49)(24) = (221)(10)(1176) =
So more than two-and-a-half million different poker hands are possible

17. Royal Flush The highest ranking poker hand is one with
'ace', 'king', 'queen', 'jack', 'ten' in a single suit
(i.e., 'hearts', 'spades', 'diamonds' or 'clubs')
What are your chances of being dealt a
'Royal Flush'?
Answer: 4 chances in
(or 1 chance in )

18. One Pair
This is a poker hand in which exactly two cards have the same face-value (i.e., a 'pair') while the other three cards have three other face-values
How likely is this type of poker hand?
Answer: This probability is given by
b(13;1) x b(4;2) x b(12;3) x b(4;1)3 / b(52;5)

19. Personal Identification Numbers
Suppose you are asked to select a PIN that consists of 4 different uppercase letters
How many choices do you have?
Answer: b(26;4) x 4! (= 26! / 22!)
If you choose a PIN at random, what are your chances of including just one vowel (A,E,I,O,U)?

20. Solution #1
( __, __, __, __ )
First choose your vowel: b(5;1) ways to do it
Next choose 3 consonants: b(21;3) ways to do it
Then decide their ordering: 4! ways to do it
Apply the Fundamental Counting Principle
probability = b(5;1)*b(21;3)*4! / b(26;4)*4!

21. Solution #2
( __, __, __, __ )
First decide where your vowel goes: b(4;1) ways
Then decide which vowel to use: b(5;1) ways
Now pick your consonants, in left-to-right order:
21 choices for 1st consonant
20 choices for 2nd consonant
19 choices for 3rd consonant
probability = 4x5x21x20x19 / 26x25x24x23

22. Are these answers equal?
Answer #1: b(5;1)xb(21;3)x4! / b(26;4)x4!
Answer #2: 4x5x21x20x19 / 26x25x24x23

23. Successive repetition
If you toss a coin n times, what are the chances of getting the same side twice in succession?
Answer: 2 chances in 2n
why?
HINT: It's often easier to count outcomes in the complementary event