Here is a Python function to calculate the solution to the interrupted game problem: ```python import math def interrupted_game_solution(p, r): n = p + r - 1 numerator = math.factorial(n) denominator = math.factorial(p-1) * math.factorial(r-1) * math.factorial(n-p-r+1) probability = numerator / denominator return probability ``` To use it: ```python p = 3 # player 1 needs 3 more wins r = 5 # player 2 needs 5 more wins probability = interrupted_game_solution(p, r) print(probability) ```

1. Dr. Carlos Rodríguez Contreras
UNAM
Probability - The Science of
Uncertainty and Data

2. The Science of Probability
 Probability is that arm of science which deals with the
understanding of uncertainty from a mathematical
perspective.
 The foundations of probability are about three centuries
old and can be traced back to the works of Laplace,
Bernoulli, et al.
 However, the formal acceptance of probability as a
legitimate science stream is just a century old.
 Kolmogorov (1933) firmly laid the foundations of
probability in a pure mathematical framework.

3. History of Probability
 The mathematical idea of randomness and probability are
relatively new concepts, in ancient times people didn't
believe in chance, everything was the work of the Gods.
 They also believed future could be divined by using lucky
charms or rolling dice, the first dice were made of bones.
 The astragalus or talus bone, the
ankle bone, usually taken from
goats or sheep. Were used in
games of chance and skill since at
least 3500 BC.
 When tossed in the air, they will
land with one of four sides face up,
sometimes referred to as Camel,
Goat, Horse and Sheep.

4. History of Probability in 6 minutes
A video of Catetham Maths

5. Nature of Probability
The world is full of uncertainty: accidents, storms, unruly financial
markets, noisy communications. The world is also full of data.
Probabilistic modeling and the related field of statistical inference
are the keys to analyzing data and making scientifically sound
predictions.
“one mathematical and the other philosophical,
reveal the double root of the mathematical
theory of probability”
Blaise Pascal

6. de Mere’s Paradox
When throwing one dice [sic], there is a 1-in-6 chance of a
‘one’ Appearing.
In four rolls, therefore, there is a 4-in-6 chance to get at least
one ‘one’ value.
If a pair of dice is rolled, there is a 1-in-36 chance of two
‘ones’ resulting.
In 24 rolls, the chance of at least one pair of ‘ones’ is thus 24-
in-36.
Both equal 2/3.
When throwing one dice [sic], there is a 1-in-6 chance of a
‘one’ Appearing.
In four rolls, therefore, there is a 4-in-6 chance to get at least
one ‘one’ value.
If a pair of dice is rolled, there is a 1-in-36 chance of two
‘ones’ resulting.
In 24 rolls, the chance of at least one pair of ‘ones’ is thus 24-
in-36.
Both equal 2/3.
The apparent difference between probability theory and reality is referred to as
the Paradox of Chevalier de Mere.

7. de Mere’s Paradox

8. The Origins
In July of 1654 Blaise Pascal wrote to Pierre Fermat about a gambling
problem which came to be known as the Problem of Points: Two
players are interrupted in the midst of a game of chance, with the
score uneven at that point. How should the stake be divided? The
ensuing correspondence between the two French mathematicians
counts as the founding document in mathematical probability, even
though it was not the first attempt to treat games of chance
mathematically.
Photographic reproduction of Philippe de
Champaigne’s painting of Blaise Pascal
(1623–1662)
Photographic reproduction of painting
of Pierre de Fermat
(1601–1665)
Antoine Gombaud a.k.a.
Chevalier de Mere
(1607–1684)

9. The Unfinished Game (The Problem of
Points) and the need of Counting

10. The Unfinished Game (The Problem of
Points) and the need of Counting

11. Counting

12. Counting Rule for Combinations
A combination is an outcome of an experiment where x
objects are selected from a group of n objects and where
order does not matter
Where:
nCk = number of combinations of x objects selected from n objects
n! =n(n - 1)(n - 2) . . . (2)(1)
k! = k(k - 1)(k - 2) . . . (2)(1)
0! = 1 (by definition)
!
!( )!n k
n n
C
k n kk
 
  
 
 


13. Counting Rule for Permutations
• A permutation is an outcome of an experiment where x
objects are selected from a group of n objects and
where order matter
where:
nPk = number of permutations of x objects selected from n
objects
n! =n(n - 1)(n - 2) . . . (2)(1)
k! = k(k - 1)(k - 2) . . . (2)(1)
0! = 1 (by definition)
!
( )!n kP
n
n k



14. The Unfinished Game Solution
n=p+r-1
p = number of trials the advantaged player needs to win
r = number of trials the disadvantaged player needs to win
1
0 2
r
n k
n
k
C


1
0 2
r
n k
n
k
C



15. Briagoberto needs to win 3 more rounds to win the game and
Catarrín needs to win 5 more rounds. Briagoberto’s probability of
winning is:
99/128
If pot is $25.00
Briagoberto’s share of the pot is then
99/128 ($25) =$19.34
1
0 2
r
n k
n
k
C


n=p+r-1
The case of
Briagoberto
vs
Catarrín

16. Important Terms
 Probability – the chance that an uncertain event will
occur (always between 0 and 1)
 Experiment – a process of obtaining outcomes for
uncertain events
 Experimental Outcome – the most basic outcome
possible from a simple experiment
 Sample Space – the collection of all possible
experimental outcomes

17. Sample Space
 The Sample Space is the collection of all
possible outcomes
e.g., All 6 faces of a die:
e.g., All 52 cards of a bridge deck:

18. Events
Experimental outcome – An outcome from
a sample space with one characteristic
 Example: A red card from a deck of cards
Event – May involve two or more outcomes
simultaneously
 Example: An ace that is also red from a deck of cards

19. Visualizing Events
 Contingency Tables
 Tree Diagrams
Full Deck
of 52 Cards
Red Card
Black Card
Not an Ace
Ace
Ace
Not an Ace
Sample
Space
2
24
2
24
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Sample
Space

20. Experimental Outcomes
 A automobile consultant records fuel type and vehicle type for a sample of
vehicles
2 Fuel types: Gasoline, Diesel
3 Vehicle types: Truck, Car, SUV
6 possible experimental outcomes:
e1 Gasoline, Truck
e2 Gasoline, Car
e3 Gasoline, SUV
e4 Diesel, Truck
e5 Diesel, Car
e6 Diesel, SUV
Car
Truck
Truck
Car
SUV
SUV
e1
e2
e3
e4
e5
e6
Diesel
Gasoline

21. Probability Concepts
• Mutually Exclusive Events
– If E1 occurs, then E2 cannot occur
– E1 and E2 have no common elements
Black
Cards
Red
Cards
A card cannot be
Black and Red at
the same time.
E1 E2
Black
Cards
Red
Cards

22.  Independent Events
E1 = heads on one flip of fair coin
E2 = heads on second flip of same coin
Result of second flip does not depend on the result of the first
flip.
 Dependent Events
E1 = rain forecasted on the news
E2 = take umbrella to work
Probability of the second event is affected by the occurrence of
the first event
Independent vs. Dependent Events

23. Rules of Probability
0 ≤ P(Ei) ≤ 1
For any event Ei
where:
k = Number of individual outcomes
in the sample space
ei = ith individual outcome
Rule 1 Rule 2
1)P(e
k
1i
i 
Rules for
Possible Values
and Sum
Individual Values Sum of All Values

24. • The probability of an event Ei is equal to
the sum of the probabilities of the
individual outcomes forming Ei.
• That is, if:
Ei = {e1, e2, e3}
then:
P(Ei) = P(e1) + P(e2) + P(e3)
Addition Rule for elementary events
Rule 3

25. Complement Rule
• The complement of an event E is the collection of
all possible elementary events not contained in
event E. The complement of event E is
represented by E.
• Complement Rule:
Or,
P(E)1)EP( 
1)EP(P(E) 
E
E

26. E1 E2 E1 E2+ =
2
6
Addition Rule for Two Events
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
 Addition Rule:
Rule 4
P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)
Don’t count common
elements twice!

27. P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)
Addition Rule Example
= 26/52 + 4/52 - 2/52 = 28/52
Don’t count
the two red
aces twice!
Black
Color
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52

28. Addition Rule for Mutually Exclusive Events
• If E1 and E2 are mutually exclusive, then
P(E1 and E2) = 0
P(E1 or E2) = P(E1) + P(E2) - P(E1 and
E2)
P(E1 or E2) = P(E1) + P(E2)
= 0
if mutually
exclusive
Rule 5
E1 E2

29. • Conditional probability for any
two events E1 , E2:
Conditional Probability
Rule 6)P(E
)EandP(E
)E|P(E
2
21
21 
0)P(Ewhere 2 

30. • What is the probability that a car has a CD player,
given that it has AC ?
i.e., we want to find P(CD | AC)
Conditional Probability Example
Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a CD player (CD).
20% of the cars have both.

31. Conditional Probability Example
No CDCD Total
AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
Of the cars on a used car lot, 70% have air conditioning
(AC) and 40% have a CD player (CD).
20% of the cars have both.
.2857
.7
.2
P(AC)
AC)andP(CD
AC)|P(CD 

32. For Independent Events:
• Conditional probability for
independent events E1 , E2:
Rule 7
)P(E)E|P(E 121  0)P(Ewhere 2 
)P(E)E|P(E 212  0)P(Ewhere 1 

33. Multiplication Rules
• Multiplication rule for two events E1 and E2:
If E1 and E2 are independent, then
and the multiplication rule simplifies to
Rule 8
Rule 9
)E|P(E)P(E)EandP(E 12121 
)P(E)E|P(E 212 
)P(E)P(E)EandP(E 2121 

34. Tree Diagram Example
Diesel
P(E2) = 0.2
Gasoline
P(E1) = 0.8
Truck: P(E3|E1) = 0.2
Car: P(E4|E1) = 0.5
SUV: P(E5|E1) = 0.3
P(E1 and E3) = 0.8 x 0.2 = 0.16
P(E1 and E4) = 0.8 x 0.5 = 0.40
P(E1 and E5) = 0.8 x 0.3 = 0.24
P(E2 and E3) = 0.2 x 0.6 = 0.12
P(E2 and E4) = 0.2 x 0.1 = 0.02
P(E3 and E4) = 0.2 x 0.3 = 0.06
Truck: P(E3|E2) = 0.6
Car: P(E4|E2) = 0.1
SUV: P(E5|E2) = 0.3

35. Your turn now!!!
Challenge:
code a function to calculate the
Interrupted Game Solution