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Lea Tutorial
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Introduction
WARNING: Lea 2 is no longer maintained. If you are new to Lea, you are invited to use Lea 3, which has a comprehensive tutorial.
The present tutorial explains how to use the Lea 2 package, through a sequence of simple examples executed in the Python console.
The present tutorial uses the Python language. Before starting, note that if you are not used to Python language or if you want to save typing with a fresh syntax, then you could jump to the Leapp version of the present tutorial. Leapp is an easytolearn probabilistic programming language on top of Python/Lea.
For the impatient reader … : just read the examples (light grey background)! These should be mostly selfexplanatory if you are acquainted with probability distributions.
Prerequisites
In order to execute the statements given in the present tutorial, you have to install the Lea package. Instructions are provided in Installation page. Then, start a Python console and type (>>>
is the Python prompt):
>>> from lea import * >>>
From here, all the examples provided in the present tutorial can be executed. What follow the >>>
prompt are the Python statements or expressions that you type; then, below, the results (if any) of the execution are displayed.
Defining probability distributions
The first thing to do when working with Lea is to define a probability distribution that models a given phenomenon. There are essentially two ways to do that:
 providing a sequence of equiprobable values
 providing a sequence of pairs (values,probability)
Equiprobable values
The simplest way to define a Lea distribution is by providing a sequence of values, each considered as equiprobable. The Lea.fromVals(…)
method does that job, the values passed as arguments being taken as equiprobable. As an example, the following statement defines the result of coin flipping, assuming fair chances to get head or tail :
>>> flip = Lea.fromVals('Head','Tail')
If each value occurs no more than once in the sequence, then we get a uniform probability distribution: each value has the same occurrence probability, which is 1/N where N is the number of given values (N = 2, in the example). The distribution can be displayed simply by entering its name. This displays the different values, in sorted order, with their respective probabilities expressed as rational numbers:
>>> flip Head : 1/2 Tail : 1/2
Note that Lea fully complies with the axioms of probability: each displayed probability is between 0 and 1, the total of probabilities displayed for a given distribution is equal to 1.
If the same value occurs several times in the definition sequence, then its probability is proportional to its number of occurrences. For instance, an unfair coin can be defined by the following statement:
>>> biasedFlip = Lea.fromVals('Head','Tail','Tail')
Since the value 'Tail' is provided twice, it has two more chances to occur than 'Head', which occurs only once:
>>> biasedFlip Head : 1/3 Tail : 2/3
We will show later in this tutorial that this feature of repeated values shall prove handy for determining frequencies of values in a given sample sequence.
You can associate probabilities to Python objects of any type. Consider for example a fair die with six faces. It is natural to model this die with integers from 1 to 6 :
>>> die = Lea.fromVals(1,2,3,4,5,6) >>> die 1 : 1/6 2 : 1/6 3 : 1/6 4 : 1/6 5 : 1/6 6 : 1/6
Note that, as from Lea 2.2, the convenience function V
is provided as an alias to Lea.fromVals
(see here)
Also, by using the method Lea.fromSeq(…)
, any sequence of values (Python's tuple, list, set, iterator, …) can be used to define a probability distribution. All the following statements are equivalent and create the same die distribution as above:
>>> die = Lea.fromSeq(range(1,7)) >>> myList = [1,2,3,4,5,6] >>> die = Lea.fromSeq(myList) >>> myTuple = (1,2,3,4,5,6) >>> die = Lea.fromSeq(myTuple)
or, using Python's list comprehension / generator expressions,
>>> die = Lea.fromSeq([v+1 for v in range(6)]) >>> die = Lea.fromSeq(v+1 for v in range(6)) >>> die = Lea.fromSeq(6v for v in range(6))
and we could continue examples with many other iterable objects, ad nauseam.
For the specific case of sequences of equiprobable integers, which is typical of dice, Lea provides a convenience method, Lea.interval(a,b)
. It generates integers from a
to b
inclusive:
>>> die = Lea.interval(1,6)
In the examples above, we have seen probability distributions with Python's strings and integers being the values having some chances to occur. Actually, there are no limitation on the types of object that can be handled: floats, complexes, vectors, date/times, userdefined objects, … can be set. However, depending of the object defined in the distributions, some operations can become impossible to achieve; if you try to execute such operation then an exception shall be raised. For example, we shall see later that you shall be able to get the mean value on the die distribution; it is impossible however to get a mean on the flip distribution, since there isn't such concept among character strings!
Values with explicit probabilities
Another way to define Lea distributions is to provide explicit pairs (value,probability), using Lea.fromValFreqs(…)
method. For instance, the following statement is another way to define the unfair coin shown above.
>>> biasedFlip = Lea.fromValFreqs(('Head',1),('Tail',2))
This way of doing is especially handy for nonuniform definitions. By using Lea.fromValFreqsDict(…)
method, a Python dictionary can be passed also:
>>> myPythonDict = {'Head': 1, 'Tail': 2} >>> biasedFlip = Lea.fromValFreqsDict(myPythonDict)
Note that the displayed probabilities are always simplified to have the lowest common denominator. We shall see later how to display the probabilities as percentages. Note also that values given with probability 0, if any, are removed automatically:
>>> Lea.fromValFreqs(('Head',1),('Tail',99),('SHAZAM!',0)) Head : 1/100 Tail : 99/100
Note that, as from Lea 2.2, the convenience function VP
is provided as an alias to Lea.fromValFreqs
(see here).
Lea provides a couple of convenience methods for a couple of standard probability distributions, viz. boolean, Bernoulli, binomial and Poisson. These are described hereafter.
Random boolean variables
Based on what has been seen above, you can model random boolean variable by defining the probabilities of True and False values, such as the total is 1. For example, you could model a probability of 27 % that it shall rain tomorrow as
>>> rain = Lea.fromValFreqs((True,27),(False,73)) >>> rain False : 73/100 True : 27/100
Note that there is of course a redundancy in such display since the probability of False can be calculated from the probability of True. We shall see later a more handy way to display probabilities of random variables.
Since boolean variables are very common, Lea provides a convenience method, Lea.boolProb(n,d)
that models a statement that has probability n/d to be true. For instance, the example above can be rewritten as
>>> rain = Lea.boolProb(27,100)
Bernoulli distribution
The method Lea.bernoulli(n,d)
defines the value 1 with probability n/d and 0 with probability 1  n/d
>>> Lea.bernoulli(3,10) 0 : 7/10 1 : 3/10
Binomial distribution
The method Lea.binomial(x,n,d)
defines the number of successes among a number x of independent experiments, each having probability n/d of success (see binomial distributions). For example, suppose a biased coin comes up heads with probability 0.3 when tossed. What is the probability of achieving 0, 1,…, 6 heads after six tosses?
>>> Lea.binom(6,3,10) 0 : 117649/1000000 1 : 302526/1000000 2 : 324135/1000000 3 : 185220/1000000 4 : 59535/1000000 5 : 10206/1000000 6 : 729/1000000
Note that Lea.binomial(1,n,d)
is equivalent to Lea.bernoulli(n,d)
Poisson distribution
The method Lea.poisson(m)
defines a Poisson probability distribution having mean m.
>>> Lea.poisson(2) 0 : 6766764161830635520/50000000000000006339 1 : 13533528323661271040/50000000000000006339 2 : 13533528323661271040/50000000000000006339 3 : 9022352215774180352/50000000000000006339 4 : 4511176107887090176/50000000000000006339 5 : 1804470443154836224/50000000000000006339 6 : 601490147718278656/50000000000000006339 7 : 171854327919508192/50000000000000006339 8 : 42963581979877048/50000000000000006339 9 : 9547462662194900/50000000000000006339 10 : 1909492532438980/50000000000000006339 11 : 347180460443451/50000000000000006339 12 : 57863410073908/50000000000000006339 13 : 8902063088294/50000000000000006339 14 : 1271723298328/50000000000000006339 15 : 169563106444/50000000000000006339 16 : 21195388305/50000000000000006339 17 : 2493575095/50000000000000006339 18 : 277063899/50000000000000006339 19 : 29164621/50000000000000006339 20 : 2916462/50000000000000006339 21 : 277758/50000000000000006339 22 : 25251/50000000000000006339 23 : 2196/50000000000000006339 24 : 183/50000000000000006339 25 : 15/50000000000000006339 26 : 1/50000000000000006339
Because the Poisson distribution has, in theory, a infinite number of possible values, Lea provides an approximation where values having a probability below 1e20 are dropped. To keep a probability total equal to 1, the remaining probabilities are all slightly overestimated. The threshold can be changed by issuing the needed precision as a second argument, e.g. Lea.poisson(2,1e30)
.
Displaying probability distributions
As seen in the previous examples, a probability distribution displays itself as rows giving the possible values with their respective probabilities expressed as fractions. This has the advantage to give exact probability values, without rounding errors (as found in many probability packages) and with a probability total guaranteed to be 1. However, in several cases, such value may be difficult to interpret ; also, many examples found in literature use floatingpoint notation, which are painful to compare with fractions.
Lea provides two display alternatives for probabilities : floatingpoint and percentages :
>>> print (biasedFlip.asFloat()) Head : 0.333333 Tail : 0.666667 >>> print (biasedFlip.asPct()) Head : 33.3 % Tail : 66.7 %
A standard way to display a probability distribution is to draw its histogram. Lea does not have real plotting features. However, a modest textoriented histogram is provided through the histo()
method:
lea> print (die.histo()) 1 :  2 :  3 :  4 :  5 :  6 :  lea> print (biasedFlip.histo()) Head :  Tail : 
The histogram can be augmented with numerical probability values by using the asString()
method, which provides many customisation means :
lea> print (biasedFlip.asString('/')) Head : 1/3  Tail : 2/3  lea> print (biasedFlip.asString('.')) Head : 0.333333  Tail : 0.666667 
New: As from Lea 2.2, you can display a true bar chart based on the matplotlib package using the plot
method.
Getting individual probabilities
Given a probability distribution, the probability of a given value can be extracted by using the p(…)
method.
>>> biasedFlip.p('Head') 1/3 >>> die.p(5) 1/6
The returned value is a ProbFraction
instance, an object inheriting Python's Fraction
class and which displays itself as a fraction. To display this probability in other formats, use the methods seen before:
>>> print (die.p(5).asFloat()) 0.16666666666666666 >>> print (die.p(5).asPct()) 16.666667 %
To get a genuine floatingpoint number (instead of just display it), you can use Python's float
function or Lea's pmf
method:
>>> pHead = float(biasedFlip.p('Head')) #or pHead = biasedFlip.pmf('Head') >>> pHead 0.3333333333333333
For boolean random variables, starting from version 2.2, Lea provides the special functions P
and Pf
:
>>> P(rain) 27/100 >>> Pf(rain) 0.27
>>> P(biasedFlip == 'Head') 1/3 >>> P(die == 5) 1/6
For Lea version anterior to 2.2 (and in 2.2 also), the probability of 'True' can be displayed as follows:
>>> rain.p(True) 27/100 >>> float(rain.p(True)) 0.27 >>> print ((rain.p(True).asPct()) 27.000000 %
Getting probability distribution data
Consider the following height distribution of a sample of smurfs:
>>> heights = Lea.fromValFreqs((0.5,1),(1.0,2),(1.5,4),(2.0,5),(2.5,5),(3.0,2),(3.5,1)) >>> print ((heights).asPct()) 0.5 : 5.0 % 1.0 : 10.0 % 1.5 : 20.0 % 2.0 : 25.0 % 2.5 : 25.0 % 3.0 : 10.0 % 3.5 : 5.0 %
Lea provides several methods to get the distribution values and probabilities at once. The most general method is vps
, which returns (value,probability weight) pairs in an iterator:
>>> tuple(heights.vps()) ((0.5, 1), (1.0, 2), (1.5, 4), (2.0, 5), (2.5, 5), (3.0, 2), (3.5, 1)) >>> dict(heights.vps()) {0.5: 1, 1.0: 2, 2.0: 5, 3.0: 2, 3.5: 1, 1.5: 4, 2.5: 5}
You can notice that this returns probabilities as integer weights. To get usual probabilities as float numbers, there exist other methods giving the values, p.m.f. (probability mass function) and c.d.f (cumulative distribution function) of a given distribution:

vals
(orsupport
): getting values only>>> heights.vals() (0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5) >>> heights.support() (0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5)

pmf
: getting probabilities only, as floatingpoint values>>> heights.pmf() (0.05, 0.1, 0.2, 0.25, 0.25, 0.1, 0.05)

cdf
: getting cumulative distributive functionNote that the last value returned by>>> heights.cdf() (0.05, 0.15, 0.35, 0.6, 0.85, 0.95, 1.0)
cdf
is always 1.0 (or something close to 1.0, in case of rounding error!).
You can use these methods, for example, to plot histograms using your preferred Python chart package. To get value probability pairs, use Python's zip function:
>>> tuple(zip(heights.vals(),heights.pmf())) ((0.5, 0.05), (1.0, 0.1), (1.5, 0.2), (2.0, 0.25), (2.5, 0.25), (3.0, 0.1), (3.5, 0.05)) >>> dict(zip(heights.vals(),heights.pmf())) {0.5: 0.05, 1.0: 0.1, 2.0: 0.25, 3.0: 0.1, 3.5: 0.05, 1.5: 0.2, 2.5: 0.25}
Important: the methods vps
, pmf
and cdf
without argument return tuples with an order consistent with the order of values returned by vals
. For instance, the first values of pmf
corresponds to the first value of vals
, etc. What is this order? This depends on the values:
 if the values are sortable, then the values are sorted in increasing order (e.g. integers, floats, strings, …)
 if the values are not sortable, then an arbitrary order, consistent from call to call, is used (e.g. complexes, matrices, …)
Note that the order you use to pass your sequence of values to build a Lea distribution is irrelevant. If these values can't be sorted, then they will most probably be mixed up, similarly to what happens in Python's dictionaries and sets.
pmf
and cdf
can be called with a value given as argument: then, the method returns the probability or c.d.f. for the given value.
>>> heights.pmf(2.0) 0.25 >>> heights.cdf(2.0) 0.6
Note that these values can be found also by using comparison operators:
>>> Pf(heights == 2.0) 0.25 >>> Pf(heights <= 2.0) 0.6
Probability distribution indicators
The following standard indicators are available on Lea probability distributions : mean, variance, standard deviation, mode and entropy. The three first indicators are demonstrated in the following examples:
>>> die = Lea.fromVals(1,2,3,4,5,6) >>> die.mean 3.5 >>> die.var 2.9166666666666665 >>> die.std 1.707825127659933
Note: in Lea 1.x versions, you have to use the method call notation, e.g. die.mean()
.
Note that these three indicators require that 1° the values can be subtracted together, 2° the differences can be multiplied by floatingpoint numbers and 3° the weighted differences can be added to the values. These conditions are verified, among others, for distributions having values with numeric, matrix and datetime types. These are NOT verified, among others, for strings, functions, classes, objects without overloaded operations.
The mode attribute returns a tuple giving the value(s) having the highest probability:
>>> die.mode (1, 2, 3, 4, 5, 6) >>> heights = Lea.fromValFreqs((0.5,1),(1.0,2),(1.5,4),(2.0,5),(2.5,5),(3.0,2),(3.5,1)) >>> heights.mode (2.0, 2.5)
This indicator requires of course that the values can be compared together.
The entropy provides a measure of the degree of randomness of the distribution. For a given set of values, it is maximum for a uniform probability distribution (i.e. all values have equal probabilities); in this case, it is equal to the logarithm in base 2 of the number of values, the result is expressed in bits. The entropy
method is illustrated in the following statements.
>>> Lea.boolProb(1,2).entropy 1.0 >>> die.entropy 2.584962500721156 >>> dieU = Lea.fromVals(*(1,2,3,4,5,6,6)) >>> dieU.entropy 2.521640636343318
You can verify that the entropy value of the loaded die (dieU
) is lower than the fair die's (die
) : the chances that the loaded die gives a 6 are twice than the chances of other numbers; so it is "less random" than the fair die.
The entropy is null for distributions having a certain, unique, value (i.e. with probability = 1).
>>> Lea.boolProb(1,1).entropy # certain True 0.0 >>> Lea.boolProb(0,1).entropy # certain False 0.0 >>> Lea.fromVals(*(1,)).entropy # certain 1 0.0
Note that the entropy calculation does not involve the values of the distribution; only the probabilities play a role. So, contrarily, to other indicators seen above, the entropy can always be calculated, whatever the type of values.
Generation of random samples
Since Lea copes with randomness, you can legitimately expect means to generate random samples! The random
method serves this purpose.
Here is how to draw a random value from the fair die defined above:
>>> die.random() 3
You could interpret this as: "Append .random()
after a probability distribution and you get the associated random variable!"
Note: since we cope with randomness, the actual values you get here (as well as in the other examples of the present section) may be different! To generate a sample with a given number of random values, you provide the required sample's size:
>>> die.random(20) (5, 6, 3, 3, 3, 4, 4, 5, 1, 2, 1, 1, 3, 4, 4, 2, 2, 2, 1, 5)
The distribution of random values shall of course respect the given probability distribution. Let us make a random sample from the unfair die defined above.
>>> dieU.random(20) (6, 1, 6, 4, 5, 1, 2, 4, 6, 6, 3, 6, 5, 4, 6, 1, 6, 6, 4, 3)
We see that the value 6 seems to occur more often, as we could have expected it. Generally speaking, as the random sample size grows, the frequency of occurrence of each value is expected to match closer and closer the probability distribution. In order to measure this trend  and to verify the correctness of Lea's random generator , the fromSeq
method can be used as a frequency counter. To illustrate this, let us generate a sample of 42,000 random values with the fair die and let us count the frequency of each value.
>>> sample1 = die.random(42000) >>> Lea.fromSeq(sample1) 1 : 6933/42000 2 : 6942/42000 3 : 7059/42000 4 : 6956/42000 5 : 7039/42000 6 : 7071/42000
We notice that the frequencies are all close to 7000/42000, which is 1/6. So far, so good!
Now, let us proceed similarly with the unfair die.
>>> sample2 = dieU.random(42000) >>> Lea.fromSeq(sample2) 1 : 5869/42000 2 : 5958/42000 3 : 6050/42000 4 : 6100/42000 5 : 6112/42000 6 : 11911/42000
You can verify here that the frequencies of 1, 2, 3, 4 and 5 are close to 1/7, while the frequency of 6 is close to 2/7.
The random sampling works with any distribution, whatever the type of values. Here is an example with tossing of an unfair coin.
>>> biasedFlip = Lea.fromValFreqs(('Head',1),('Tail',2)) >>> print (biasedFlip.asFloat()) Head : 0.333333 Tail : 0.666667 >>> biasedFlip.random(10) ('Tail', 'Tail', 'Tail', 'Tail', 'Tail', 'Head', 'Tail', 'Head', 'Tail', 'Tail') >>> print (Lea.fromSeq(biasedFlip.random(100000)).asFloat()) Head : 0.333200 Tail : 0.666800
You can verify that the frequency of Head in the 100,0000 tosses sample is close to the defined probability.
Certain stochastic processes require that a given value does not appear twice in a random sample. These are named "draws without replacement", which include many lottery games. The randomDraw
mthod is dedicated to such processes. As an example, imagine a fair lottery game with 45 balls, from which 6 balls shall be drawn.
>>> lottery = Lea.fromSeq(range(1,46)) >>> lottery.randomDraw(6) (13, 24, 25, 10, 1, 26)
This method guarantees that all drawn values are different. As a special case, if the argument of this method is the number of values, we get a random permutation of all the values. Note that this can be done also by omitting the argument:
>>> lottery.randomDraw(45) (44, 42, 36, 4, 10, 2, 14, 19, 22, 39, 12, 41, 18, 7, 31, 5, 38, 45, 21, 8, 13, 1, 3, 23, 32, 35, 20, 40, 11, 29, 27, 15, 30, 37, 16, 28, 43, 17, 6, 34, 25, 24, 9, 33, 26) >>> lottery.randomDraw() (40, 41, 36, 27, 15, 26, 31, 24, 9, 7, 21, 16, 13, 2, 10, 34, 35, 30, 29, 33, 3, 23, 42, 32, 37, 45, 22, 20, 5, 25, 17, 43, 28, 4, 18, 39, 14, 1, 11, 8, 19, 12, 6, 38, 44)
Should the argument exceed the number of values, then an exception is raised.
Note that Lea's random
and randomDraw
methods internally use the standard Python random
module. This may be useful to know: you may wish to be able to produce the very same pseudorandom sequence for a demo, a course, a test suite, etc.; the Python random.seed
function can be helpful for this purpose:
>>> import random
>>> random.seed(0) >>> die.random(10) (4, 4, 1, 3, 5, 4, 4, 3, 4, 3) >>> random.seed(1) >>> die.random(10) (2, 5, 1, 3, 1, 4, 4, 4, 6, 4) >>> random.seed(0) >>> die.random(10) (4, 4, 1, 3, 5, 4, 4, 3, 4, 3) >>> random.seed(1) >>> die.random(10) (2, 5, 1, 3, 1, 4, 4, 4, 6, 4)
Calculus with probability distributions
Context
The examples seen so far work with explicitly defined probability distributions. In many situations however, a stochastic phenomenon results from the combination of other stochastic phenomenon's, which are easier to model and which have known probability distributions. Examples are:
 adding the results of N thrown dice,
 counting the number of successes of N trials,
 evaluating the truth of a given condition made on stochastic events,
 evaluating joint occurrences of N stochastic events,
 applying a given function on a random variable,
 calculating a given function with random variables as arguments.
The traditional approach requires detailed calculations, made manually or by customised programs; this is tedious and errorprone. Fortunately, Lea has several features to make these derivations of probability distribution easy and generic.
The following sections shall present the most basic techniques, assuming that
all random events are independent.
This is an important assumption to take in mind. It allows for using simple methods to derive probabilities; in particular, it is usually well suited to model gambling situations or stateless systems.
For more complex situations, the random events at hand cannot be assumed independent. The occurrence of one random event may influence the occurrence of another random event. In general terms, the knowledge of some information can impact the probability distribution of future (or unknown) events. Lea provides some tools for that situations. These are presented in an "advanced tutorial", which covers topics like joint probability, conditional probability and Bayesian inference.
Arithmetic operations
We shall see here how Lea instances can be put in arithmetic expressions to produce new Lea instances. The most basic cases involve expressions mixing one Lea instance with constant values. In such cases, the independence assumption is obviously true since a constant is something certain.
Here are some examples:

Subtracting 3 from the result of a die:
>>> die1 = Lea.fromVals(1,2,3,4,5,6) >>> die1  3 2 : 1/6 1 : 1/6 0 : 1/6 1 : 1/6 2 : 1/6 3 : 1/6

Doubling the result of a die
>>> 2 * die1 2 : 1/6 4 : 1/6 6 : 1/6 8 : 1/6 10 : 1/6 12 : 1/6

Dividing the result of a die by 3, using floatingpoint arithmetic
>>> die1 / 3 0.333333333333 : 1/6 0.666666666667 : 1/6 1.0 : 1/6 1.33333333333 : 1/6 1.66666666667 : 1/6 2.0 : 1/6
For Python 2.x users, note that Lea (from ver 1.3) uses the "true division", as Python 3. For integer division, use double slash as in the example below.

Dividing the result of a die by 3, truncated to integers
>>> die1 // 3 0 : 2/6 1 : 3/6 2 : 1/6

Taking the residue of a die result modulo 2 (i.e. the parity bit)
>>> die1 % 2 0 : 1/2 1 : 1/2

Adding the result of a die with itself
>>> die1 + die1 2 : 1/6 4 : 1/6 6 : 1/6 8 : 1/6 10 : 1/6 12 : 1/6
Note that the last expression returns the same distribution as 2 * die1
.
Now, to get the distribution resulting from the sum of two dice, you have to create a new die instance, by issuing a new Lea.fromVals(…)
expression or, simpler, by "cloning" the first die:
>>> die2 = die1.clone()
From now, die2
represents a second die having the same probability distribution as die1
but its throw is assumed to be an event independent of the throw of the first die. Then, the following statement produces the distribution of the sum of the two dice.
>>> die1 + die2 2 : 1/36 3 : 2/36 4 : 3/36 5 : 4/36 6 : 5/36 7 : 6/36 8 : 5/36 9 : 4/36 10 : 3/36 11 : 2/36 12 : 1/36
lea> print ((die1+die2).histo()) 2 :  3 :  4 :  5 :  6 :  7 :  8 :  9 :  10 :  11 :  12 : 
Actually, as you could expect it, the expression has been used to browse all the combination pairs from two dice values (an operation known as the cartesian product); these pairs have been reduced by addition and the occurrences of each sum value has been counted.
Other similar expressions may be used to calculate the distribution of the difference of two dice, their product, etc. A more sophisticated example below calculates the square of the difference of the two thrown dice.
>>> (die1die2)**2 0 : 3/18 1 : 5/18 4 : 4/18 9 : 3/18 16 : 2/18 25 : 1/18
Note that the same Lea instance may appear at several places in the same expression. The values are processed consistently, as shown in the following:
>>> die1 + 2*die2  die1  (die2 + die2) 0 : 1
Repeating operations
In some situations, you will want to repeat an operation a given number of times, like throwing N dice, and adding the results.
For instance, what is the probability of getting a total of 13 when throwing 4 dice?
From the techniques seen so far, you could write:
>>> (die1 + die1.clone() + die1.clone() + die1.clone()).p(13) 35/324
This is of course tedious to type. Here is an alternative, shown essentially to demonstrate that Lea integrates smoothly with Python's building blocks:
>>> diexN = sum(die1.clone() for i in range(4)) >>> diexN.p(13) 35/324
Now, to alleviate the job, Lea provides the times
method:
>>> die1.times(4) 4 : 1/1296 5 : 4/1296 6 : 10/1296 7 : 20/1296 8 : 35/1296 9 : 56/1296 10 : 80/1296 11 : 104/1296 12 : 125/1296 13 : 140/1296 14 : 146/1296 15 : 140/1296 16 : 125/1296 17 : 104/1296 18 : 80/1296 19 : 56/1296 20 : 35/1296 21 : 20/1296 22 : 10/1296 23 : 4/1296 24 : 1/1296 >>> die1.times(4).p(13) 35/324
This emulates the distribution of the sum of 4 thrown dice; then it returns the same result as the previous expression. Note that this is not only shorter to write but also faster to execute because the method uses a dynamic programming technique, that is a dichotomic algorithm with far less calculations than the former naive expression. To illustrate that point, you may notice that Lea has no particular problem to calculate the distribution resulting of 50 thrown dice:
>>> die1.times(50) 50 : 1/808281277464764060643139600456536293376 51 : 50/808281277464764060643139600456536293376 52 : 1275/808281277464764060643139600456536293376 53 : 22100/808281277464764060643139600456536293376 54 : 292825/808281277464764060643139600456536293376 …
Note that the method for binomial distribution (see above) is no more than a shortcut to avoid to use times(n)
method explicitly. So,
>>> Lea.binom(6,3,10)
is equivalent to
>>> Lea.bernoulli(3,10).times(6)
By default, times(n)
method performs additions. Any other twoary operator or two arguments function can be specified as a second argument. For instance, here is how you can get the probability that the product of 4 thrown dice is 5:
>>> from operator import mul >>> die1.times(4,mul).p(5) 1/324
Note : This probability of 1/324 can be checked by noticing that 5 is a prime number : this means that it shall be reached only by one of 4 permutations of the form 1x1x1x5; hence the probability is 4/6^4 = 1/324.
Comparison operations
Usual comparison operators may be used on any Lea instances, resulting in Boolean distributions giving the probabilities of each truth value.
>>> die1 == 4 False : 5/6 True : 1/6 >>> die1 <= 4 False : 1/3 True : 2/3
To get only the probability value that the expression is true, you can use the P
and Pf
functions seen earlier, available as of Lea 2.2:
>>> P(die1 == 4) 1/6 >>> P(die1 <= 4) 2/3 >>> Pf(die1 <= 4) 0.6666666666666666
For version anterior to 2.2, you can use the p
and pmf
methods seen earlier:
>>> (die1 == 4).p(True) 1/6 >>> (die1 <= 4).p(True) 2/3 >>> (die1 <= 4).pmf(True) 0.6666666666666666
Comparison operators may be used in expressions, as shown in the following examples involving the two independent dice.
>>> P(die1 < die2) 5/12 >>> P(die1 <= die2) 7/12 >>> P(die1 == die2) 1/6 >>> P(die1 != die2) 5/6
The first example (die1 < die2
) shows that, after throwing two dice, there are 5/12 chances that one given die has a value strictly less than the other's. This assumes that you are able to distinguish between the two dice, for example using colors (e.g. green die + blue die). If you make no difference between dice, then the probability that one of them is strictly less than the other is 5/6, as shown in the last example.
As more advanced examples, the following expressions involve multiple occurrences of die1
and die2
.
>>> P((die1die2)**2 == die1**2  die2**2) 1/6 >>> P((die1die2)**2 == die1**2 + die2**2) 0 >>> P((die1die2)**2 == die1**2 + die2**2  2*die1*die2) 1
Let us remark that the last expression is evaluated as certainly true, as we can expect from the wellknown algebra rule.
For evaluating the probability of a range of values, you can put two inequalities connected with a logical and (see next section).
For testing the equality of distribution against a set of given values, the isAnyOf(…)
shall be used. As an example, here is how to calculate the probability of a "craps":
>>> P((die1+die2).isAnyOf(2,3,12)) 1/9
The method isNoneOf(…)
calculates the inverse distribution, that is the probability of having none of the values in a given set:
>>> P((die1+die2).isNoneOf(2,3,12)) 8/9
Logical operations
Boolean probability distributions (as obtained for example by using comparison operator, as seen just above) can be combined by usual logical operators.
These are detailed in the following table, where a and b are standing for expressions returning Lea boolean distributions or simple Python booleans:
Lea expression  meaning 

~a  not a 
a & b  a and b 
a ⎮ b  a or b 
a ^ b  a xor b 
Examples:
>>> P(~(die1 == 3)) 5/6 >>> P(~(die1 == 3)  (die1 == 3)) 1 >>> P((die1 == 4) & (die2 == 2)) 1/36 >>> P((die1 <= 3) & (die2 > 3)) 1/4
WARNINGS:

Due to Python precedence rules, it is extremely important to embed expressions connected by logical operators with parentheses. If parentheses are missing, an exception shall be raised.

The Python's augmented comparison between Lea instances (e.g. a < b < c) DOES NOT WORK! Actually, for several technical reasons related to Python design, such expressions do not (and cannot) return any sensible result. These can be replaced by using a logical AND (e.g. (a < b) & (b < c) ).
Applying functions
If you need special processing beyond arithmetic or logical operator, you can apply a given function on the values of a probability distribution by using the map
method; this builds a new probability distribution on the function results. For instance, this is how to transform the integer values of a die into floating point numbers by invoking the standard Python float
function:
>>> die1.map(float) 1.0 : 1/6 2.0 : 1/6 3.0 : 1/6 4.0 : 1/6 5.0 : 1/6 6.0 : 1/6
You could of cause use your own functions:
>>> def parity(x): … return "odd" if x%2==1 else "even" >>> die1.map(parity) even : 1/2 odd : 1/2
and chain calls
>>> die1.map(parity).map(len) 3 : 1/2 4 : 1/2
These last examples show that the number of values of the resulting distributions can be lower than the one in the distribution passed to the function. This is normal since several values can be mapped to the same resulting value.
If you need to call functions with multiple arguments, then a special method, Flea.build
, shall be used. The goal of the following example is to determine the probability distribution that results from keeping the die having the smallest value among two thrown dice.
>>> Flea.build(min,(die1,die2)) 1 : 11/36 2 : 9/36 3 : 7/36 4 : 5/36 5 : 3/36 6 : 1/36
and here is what happens when keeping the die having the greatest value:
>>> Flea.build(max,(die1,die2)) 1 : 1/36 2 : 3/36 3 : 5/36 4 : 7/36 5 : 9/36 6 : 11/36
What's next?
Thank you for reading the present tutorial!
You should be able now to define your own probability distributions and combine them to model many phenomenon's with uncertainties, assuming independence of the random events. You can find more examples in the Examples page of the Wiki.
If you are looking for special topics, like draws without replacement, joint distributions, marginalisation, conditional probabilities, then jump to Lea advanced tutorial: Part 1.
Please send your comments, critics, suggestions, bug reports,… by Email to pie.denis@skynet.be . Thanks!
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