Chance and Simulations

Simulation

Simulation is the process of using computer to mimic a physical experiment

The process

The process: Step 1: What to Simulate

Decide which quantity you want to simulate. For example, you might decide that you want to simulate the outcomes of tosses of a coin. Each Simulated value will be Head or a Tail

The process: Step 2: Simulating one Value

Figure out how to simulate one value of the quantity you specified in Step 1. In our example, you have to figure out how to simulate the outcome of one toss of a coin. If your quantity is more complicated, you might need several lines of code to come up with one simulated value. Typically we will define a function that returns the simulated value

The process: Step 3: Number of Repitions

Decide how many times you want to simulate the quantity. You will have to repeat the simulation in Step 2 that many times.

The process: Step 4: Simulating Multiple Values

Finally, put it all together as follows:

1. create and empty array in which to collect the simulated values. We will call this the collection array

2. crete “repetitions sequence,” that is, a sequence whose length is the number of repetitions you specified in Step 3. For n repetitions we will always youse the sequence np.arange(n)

3. Create a for loop. For each element of the repetitions sequence:

   1. Simulate one value by using the function you wrote in Step 2

   2. Augment the collection array with this simulated value

At this point, you can use the collection array as you would any other array. You can count how many simulated values fall into a particular category, place the array in the column of a table of visualize the distribution of the simulated values, and so on

Example: number of Heads in 100 Tosses

In this example we will simulate the number if heads in 100 tosses of a coin. The histogram of our results will give us some insight into how man heads are likely

Number of Heads in 100 Tosses : Step 1: What to simulate

The quantity we are going to simulate is the number if heads in 100 tosses

Number of Heads in 100 Tosses : Step 2: Simulating one Value

We have to figure out how to make one set of 100 tosses and count the number of heads. Let’s start by creating a coin

coin = make_array( ‘Heads’, ‘Tails’)

outcome = np.random.choice (coin,100)

num_of_heads = np.nonzero(outcome == ‘Heads’)

Output:

46

Since we want to do this multiple times, lets define a function that returns the simulated value of the number of heads. We can do this using the code developed in the code above:

def one_simulated_value():

outcome = np.random.choice(coin,100)

return np.non_zero(outcome == ‘Heads’)

Number of Heads in 100 Tosses : Step 3: Number of Repetitions

How many repetitions we will use is up to us. The more we use, the more reliable our simulations will be, but the longer it will take to run the code.

Let’s go for 20,000 repetitions. That means we are going to do the following 20,000 times:

• tossing a coin 100 times and count the number of heads

Number of Heads in 100 Tosses : Step 4: Simulating Multiple Values

We are ready to create create an array of 20,000 simulated values of the number of heads in 100 tosses of a coin

num_repetitions = 20,000 #number of repetitions

heads = make_array() #empty collection array

for i in np.arange(num_repetitions): # repeat the process num_repetitons times

heads = once_simulated_value() #simulated one value sing the function defined

heads = np.append(heads, new_value) # augment the collection array wit the simulated value

Check that the array heads contains 20,000 entries, one for each repetition of the experiment.

len(heads)

Output:

20,000

To get a sense of the variability in the number of heads in 100 tosses, we can collect the results in a table and draw a histogram

Simulating a table:

simulation_results = Table().with_columns(
    'Repetition', np.arange(1, num_repetitions + 1),
    'Number of Heads', heads
)

Simulating Histogram:

simulation_results.hist('Number of Heads', bins = np.arange(30.5, 69.6, 1))

Interpreting the Histogram

• Each bin has a width 1 and is centered at each value of the the number of heads

• The histogram looks symmetrical around 50 heads. The height of the bar at 50 is about 8%

• Since each bar is 1 unit wide, this is the same as saying that abut 8% of the receptions produced exactly 50 heads. That is not a huge percent, but it’s the largest compared to the percent at every other number of heads

• the histogram also shows that in almost all of the repetitions, the number of heads in 100 tosses was somewhere between 35 and 65. Indeed, the bulk of the repetitions produced numbers of heads in the range 45 o 55.

Example: Moves in Monopoly

Each move in the game Monopoly is determined by the total number of spots of two rolls of a die.If you play Monopoly, what should you expect to get when you roll the die two times?

We can explore this by simulating the sum of two rolls of a die. We will run the simulation 10,000 times. We’ve already completed Steps 1 and 3 of out simulation process s

Step 2 is the one in which we write a function to simulate the total number of spots in one pair of rolls. First, let’s plan our code, We will create an array containing number 1 through 6, draw two times from the array at random with replacement, and add the two numbers drawn

Step 2: define function

die = np. arange(1,7)

sum(np.random.choice(die, 2))

we can use the array die and the expression above to define that simulates one move in Monopoly

def one_simulation_move()

return sum(np.random.choice(die,2))

Step 4: Simulating Multiple Values

num_repetitions = 10,000

moves = make_array()

for i in np.arrange(num_repetitions):

new_moves = one_simulation_move()

moves = np. append( moves, new_move)

Visualization: Histogram

results = Table().with_columns(
    'Repetition', np.arange(1, num_repetitions + 1),
    'Sum of Two Rolls', moves
)

results.hist('Sum of Two Rolls', bins = np.arange(1.5, 12.6, 1))

Chance

Chance: Basics

• Lowest value: 0

  • chance of events that is impossible

• Highest value: 1 (or 100%)

  • chance of event that is certain

• Complement: if an event has a chance of 70%, then the chance that it doesn't

  • 100% - 70% = 30%

  • 1- 0.7 = 0.3

Equally likely Outcomes

Assuming: all outcomes are equally likely, the chance of an event A is:

P(A) = (number of outcomes that makes A happen) / (total number of outcomes)

• I have three cards: ace of hearts, king of diamonds, and a queen of spades

• I shuffle them and draw two cards at random without replacement

What is the chance that I get the Queens followed by the King?

First solution:

AK, AQ, KQ, KA, QK, QA

Answer: 1/6

Alternative solution:

1. first draw one card, hope it is a Queen → 1/3

• if we got Queen in the fist step, proceed to the second

2. draw one card, hope it is a King

• chance of drawing a King GIVEN that we have already drawn the Queen → 1/2

Product Rules of probability: if these steps happen in order, then you can compute the probability:

(1/3) x (1/2) = 1/6

Multiplication Rule

chance that two events A and B both happen

= P(A happens) x P( B happens given that A has happened)

• The answer is less than or equal to each of the two chances being multiplied

• The more conditions you have to satisfy, the less likely you are to satisfy them all

Question: What is the chance that one of the cards I draw is a King and the other is Queen

AK, AQ, KQ, KA, QK, QA

Chance of QK or KQ: 2/6

Alternative:

KQ : 1/6

QK: 1/6

= 1/6 + 1/6 = 2/6 = 1/3

Addition Rule

If event A can happen in exactly one of two ways, then

P(A) = P(first way) + P(second way)

The answer is greater than or equal to thee chance of each individual way

Complement: At Least one Head

In 3 tosses:

• any outcome except TTT

• P(TTT) = (1/2) x (1/2) x (1/2) = 1/8

• P(at least one head) = 1 - P(TTT) = 1 - (1/8) = 87.5%

In 10 tosses:

• 1 - (1/2)**10 =0.999 = 99.9%

If there is a clear answer (e.g., “not a six”) whose probability you know, you can use…

Multiplication rule

If there is no clear answer (“could be K or Q, but then the next one would have to be Q or K) list all the distant ways your event could occur, use…

Addition Rule

If the sir above is log and complicated, look around the complement. If the complement is simple (e.g. the complement of “at least one” is “none”), you can find its chance and subtract that from 1. USE…

Compliment Rule

A population has 100 people, including Riya and Jordan. We sample two people at random without replacement

a. P(both Riya and Jordan are in the sample)

b. P(neither Riya nor Jordan is in the sample)

a.

P( first Riya, then Jordan) + P(Jordan first, then Riya)

(1/100) x (1/99) + (1/100) x (1/99) =0.000202

B.

P(neither Riya and Jordan is in the sample)

Step 1: Probability the first person is neither Riya nor Jordan

• There are 98 other people out of 100.

P(neither first) = 98/100

Step 2: Probability the second person is neither Riya nor Jordan

• Since the first person was someone else, 99 people remain. Of those, 97 are people other than Riya and Jordan.

• P(neither first | neither second) = 97/99

Multiply the probability:

(98/100) x (97/99) =0.96

When an Event Doesn’t Happen

P(an event doesn’t happen) = 1 - P(the event happens)

When All Events are Equally Likely

if you are rolling an ordinary die, a natural assumption is that all six faces are equally likely. Under this assumption, the probabilities of how one roll came out can be easily calculated as a ratio.

ex: the chat that the die shows an even number is

(number of even faces) / (number of all faces) =

#{ 2, 4, 6} / # {1,2,3,4,5,6} = 3/6

Similarly:

P( die shows a multiple of 3) =

#{3,6} / #{1,2,3,4,5,6} = 2/6

General formula:

P(an event happens) =

#{outcomes that make the event happen} / #{all outcomes}

When Two Events Must Both Happen

Scenerio:

Suppose you have a box that contains three tickets: one red, one blue, and one green. Suppose you draw two tickets at random without replacement; that is, you shuffle the three tickets, draw one, shuffle the remaining two, and draw another from those two. What is the chance you get the green ticket first, followed by the red one?

Consider:

There are six possible pairs of color: RB, BR, RG, GR, BG, GB. All of these re equally likely by the sampling scheme, and only one of them (GR) makes the event happen

P(Green first, then Red) =

#{GR} / # {RB, BR, RG, GR, BG, GB} = 1/6

Alternative: Thinking about the event in two stages

1. First the Green ticker has to be drawn → that has a chance of 1/3

2. The Red has to be drawn next → this happens in about ½ of those repetitions

P(green first, then red) = (1/3) x (1/2) = 1/6

The factor of ½ is called “the conditional chance that the red ticket appears second, given that the green ticket appears first

Multiplication Rule

P(two events both happen) = P(one event happens) x P(the other event happens)

Thus, when there are two conditions - one event must happen, as well as another — the chance is a fraction of a fraction, which is smaller than either of the two component fractions. the more conditions that have to be satisfied, the less likely they are all to be satisfied

When an Event Can Happen in Two Different Ways:

Suppose instead we want the chance that one of the two tickets is GREEN and the other is RED. This event doesn’t specify the order in which the colors must appear. So they can appear in either way

A good way to tackle problems like this is to partition the event so that it can happen in exactly one of several differnt ways. The natural partition of “one green and one rd” is: GR, RG

Each of GR and RG has chance 1/6 by the calculation above. So you can calculate the chancee of “one green and one red” by adding them up

P(one Green and one red) = P(GR) + P(RG) = 1/6 + 1/6 = 2/6

Addition Rule

P(an event happens) = P(first way it can happen) + P(second way it can happen)

hen an event can happen in one of two different ways, the chance that it happens is a sum of chances, and hence bigger than the chance of either of the individual ways

At Least One Success:

A question that sometimes arises is about the likelihood that a particular individual in the population is selected to be in the sample. To work out the chance, that individuals is called a ‘success’ and the problem is to find the chance that the sample contains a success

tossing a coin TWICE:

if you toss a coin twice, there are four equally likely outcomes: HH. HT, TH, and TT.

The chance of getting at least one head in two tosses is ¾

Another way of coming up with this answer is to work out what happens if you don’t get at least one head This is when both the tosses lands tails

P(at least one head in two tosses) = 1 - P(both tails) = 1 - ¼ = ¾

What is the probability of getting at least one six is:

For a fair six-sided die:

• Probability of rolling a six: 1/6

• Probability of not rolling a six: 1- 1/5 =5/6

If you roll the die twice, the probability of never getting a six is:

P(no six in 2 rolls) = (5/6)²

probability of getting at least one six is:

P( at least one six) = 1 - (5/6)²