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_posts/2024-07-09-combination-permutation.md

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+---
+title: "Combinations and Probabilities."
+categories:
+  - Basic Probability
+tags:
+  - Probability
+---
+# Permutations and Combinations
+
+An ordered set of elements in $[n]={1,2,...,n}$ is called a *permutation*.
+
+If we consider an ordered subset with r elements in a set of n elements we call this an *r permutation*.
+
+Example(lottery): An urn contains 25 balls. We draw 4 balls from the urn and record their order and number.
+We see that the state space is simple the set {4 permutations}.
+
+```python
+from itertools import permutations
+import random 
+
+sample_space = set(permutations(range(1,26), 4))
+random.sample(sorted(sample_space), 3)
+```
+```
+[(6, 23, 8, 10), (1, 9, 25, 12), (16, 1, 21, 20)]
+```
+How many elements are in the above sample space?
+For the first element we have 25 possibilities, for the second 24, for the third 23 and for the last ball 22 possibilities.
+
+```python
+len(sample_space) == 25 * 24 * 23 * 22
+```
+```
+True
+```
+Let's consider the event that the first two balls are 1 and 2.
+
+```python
+event = {s for s in sample_space if s[0]==1 and s[1]==2}
+random.sample(sorted(event), 3)
+```
+```
+[(1, 2, 9, 13), (1, 2, 5, 13), (1, 2, 7, 3)]
+```
+How many elements are in the above event? For the first and second element we have 1 possibility. For the remaining two elements we have 23 and 22 possibilities.
+
+```python
+len(event) == 1 * 1 * 23 * 22
+```
+```
+True
+```
+In general there are n! permutations of $n$. We will call this "n factorial".
+
+There are $(n)_r = n \times (n-1) \times ... \times (n-r+1) = n!/(n-r)!$ r permutations of n. We call this "n fall r".
+
+```python
+from math import factorial
+
+def get_number_permutations(n): return factorial(n)
+
+def get_number_fall_permutations(n, r): return factorial(n)/factorial(n-r)
+```
+
+```python
+len(sample_space) == get_number_fall_permutations(25, 4)
+```
+```
+True
+```
+Let us consider the above experiment but this time we will ignore the order.
+That means something like $(1,2,3,4,5)$ will be equivalent to $(2,1,3,4,5)$.
+
+```python
+from itertools import combinations
+
+sample_space_wo = set(combinations(range(1,26), 4))
+random.sample(sorted(sample_space_wo), 3)
+```
+```
+[(6, 7, 8, 23), (1, 16, 21, 22), (5, 15, 16, 22)]
+```
+How many elements will this sample space have?
+
+From above we know that the sample space which takes into account order has $(25)_4$ elements.
+
+To get "the order out" we need to simply divide this quantity by the number of 4-permutations.
+
+${25\choose4} = (25)_4/4!$
+
+The generalized formula is
+${n\choose k} = (n)_k/k!$
+```python
+def get_number_k_out_of_n(n,k):
+    return get_number_fall_permutations(n,k)/get_number_permutations(k)
+```
+
+```python
+len(sample_space_wo) == get_number_k_out_of_n(25,4)
+```
+```
+True
+```
+We can apply these concepts to the game of poker.
+Let's consider the starting hand of Holdem Poker.
+
+We have 13 ranks and 4 suits (Club, Heart, Spades & Diamonds). The deck consists of the unique $(rank, suit)$ combinations. This gives the $13 \times 4 = 52$ cards mentioned above.
+
+```python
+from collections import namedtuple
+from itertools import product
+
+ranks = list(range(2,15))
+suits = ["C", "H", "S", "D"]
+Card = namedtuple('Card', ['rank', 'suit'])
+
+cards = [Card(rank, suit) for rank, suit in product(ranks, suits)]
+hands = set(combinations(cards, 2))
+```
+
+```python
+len(hands) == get_number_k_out_of_n(52,2)
+```
+```
+True
+```
+Let's consider the event that we get a pair as a starting hand.
+For that we need the rank of the two cards to be  equal.
+
+```python
+event = {h for h in hands if h[0].rank == h[1].rank}
+```
+
+How many elements to be have in the event?
+
+We choose one length and than can choose two suits so we have
+$\#E = {13 \choose 1}{4 \choose 2}$
+
+```python
+len(event) == get_number_k_out_of_n(13,1)*get_number_k_out_of_n(4,2)
+```
+```
+True
+```
+What's the probability of getting a pair?
+
+```python
+from fractions import Fraction 
+
+Fraction(len(event), len(hands)), f"{len(event)/len(hands) * 100:.2f} %"
+```
+```
+(Fraction(1, 17), '5.88 %')
+```
+Another possible hand is called "suited connector".
+
+That means that the ranks are only seperated by one, for example $((9, H), (10, H))$ is a suited connector.
+
+We want to consider the "best" suited connectors where the minimum rank is larger than 9.
+
+```python
+event = {h for h in hands if abs(h[0].rank-h[1].rank)==1 and min(h[0].rank,h[1].rank)>=9 and h[0].suit == h[1].suit}
+```
+
+We have 
+9, 10, 11, 12 and 13 as the possible lowest rank.
+We can choose one of the suits.
+So in total we have
+${5 \choose 1}{4 \choose 1}$ possibilties.
+
+```python
+len(event) == get_number_k_out_of_n(5,1)*get_number_k_out_of_n(4,1)
+```
+```
+True
+```
+The corresponding probability is easily calculated as:
+
+```python
+Fraction(len(event), len(hands)), f"{len(event)/len(hands) * 100:.2f} %"
+```
+```
+(Fraction(10, 663), '1.51 %')
+```

_posts/2024-07-09-introduction-probability.md

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+---
+title: "Introduction to basic probability."
+categories:
+  - Basic Probability
+tags:
+  - Probability
+---
+# Introduction
+
+In this notebook I want to implement some of the ideas from [this lecture](https://www.youtube.com/watch?v=2Xwk6yNq9og) in Python.
+The approach of implementation is inspired by [the great notebook of Peter Norvig](https://github.com/norvig/pytudes/blob/main/ipynb/Probability.ipynb).
+
+Let us start with the notion of a *sample space*. A sample space describes the set of all possible outcomes when performing an experiment.
+
+Let us consider a simple coin throw. The sample space is simply $\{H, T\}$. We can get either Heads or Tails as a result.
+
+```python
+sample_space_coin = {"H", "T"}
+```
+
+Let's consider the experiment that we throw a coin three times.
+The state space contains all possible combinations of Heads and Tails at each turn.
+
+```python
+from itertools import product
+
+sample_space_coins = set(product(sample_space_coin,repeat=3))
+sample_space_coins
+```
+```
+{('H', 'H', 'H'),
+ ('H', 'H', 'T'),
+ ('H', 'T', 'H'),
+ ('H', 'T', 'T'),
+ ('T', 'H', 'H'),
+ ('T', 'H', 'T'),
+ ('T', 'T', 'H'),
+ ('T', 'T', 'T')}
+```
+
+An *event* is a possible outcome of our experiment.
+Let's consider the event that Heads showed at least two times up.
+
+```python
+event = {s for s in sample_space_coins if s.count("H") >= 2}
+event
+```
+```
+{('H', 'H', 'H'), ('H', 'H', 'T'), ('H', 'T', 'H'), ('T', 'H', 'H')}
+```
+The *frequentist* approach states the following:
+If all possible outcomes are equally likely an event E occurs with frequency $\#E/\#S$ where $\#E$ is the number of elements in the Event and #S is the number of elements in the State Space.
+For our particular example we can do it very simply in Python.
+
+```python
+from fractions import Fraction
+
+Fraction(len(event)/len(sample_space_coins))
+```
+
+Similar we can consider a repeated dice throw.
+We identify each side of the dice with a number from 1-6.
+That means our sample space for a single dice throw is $\{1,...,6\}$
+
+```python
+sample_space_dice = {1,2,3,4,5,6}
+```
+
+```python
+sample_space_dices = set(product(sample_space_dice, repeat=2))
+sample_space_dices
+```
+```
+{(1, 1),
+ (1, 2),
+ (1, 3),
+ (1, 4),
+ (1, 5),
+ (1, 6),
+ (2, 1),
+ (2, 2),
+ (2, 3),
+ (2, 4),
+ (2, 5),
+ (2, 6),
+ (3, 1),
+ (3, 2),
+ (3, 3),
+ (3, 4),
+ (3, 5),
+ (3, 6),
+ (4, 1),
+ (4, 2),
+ (4, 3),
+ (4, 4),
+ (4, 5),
+ (4, 6),
+ (5, 1),
+...
+ (6, 2),
+ (6, 3),
+ (6, 4),
+ (6, 5),
+ (6, 6)}
+```
+Let's consider the event that the sum of two dice throws is four or less.
+
+```python
+event = {s for s in sample_space_dices if sum(s) <= 4}
+event
+```
+```
+{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}
+```
+
+```python
+Fraction(len(event), len(sample_space_dices))
+```
+```
+Fraction(1, 6)
+```