Probability of a 45º Drunkard’s Walkstatic.squarespace.com/static/4ff39269c4aaf3fdb14faa05/t/...2.0 Developing a Method PAGE 8 David Kong – 2203 033 The probability of winning

Glenforest Secondary School

Extended Essay

Probability of a 45º Drunkard’s Walk

David Kong

2203-0033

Math

Word Count: 3814

ABSTRACT PAGE 2

David Kong – 2203 033

Abstract We take a glimpse into higher level probability with the ‚random walk,‛ a particle

moving in random directions on a plane. The focus, however, is very specific: a very

particular kind of random walk, as described by an ‚USA Mathematics Talent

Search‛ question,

3/4/20. A particle is currently at the point (0, 3.5) on the plane and is moving towards the origin. When the particle hits a lattice point (a point with integer coordinates), it turns with equal probability 45° to the left or to the right from its current course. Find the probability that the particle reaches the x-axis before hitting the line y = 6 USA (Mathematical Talent Search).

We begin by considering a smaller model in order to develop theorems based around

the symmetrical nature of the particle’s movement. These theorems are applied to

the original question and by focusing on particular points of the particle’s motion

and we solve a system of equations to reveal the answer to be

. We then check the

validity of our work by comparing the theoretical probability to the empirical

probability as found by an excel program we later create.

We then extend the problem to find that for a particle beginning downward at

(0,4.5), the probability of reaching the x-axis before reaching y=8 is

.

Finally, we use the patterns from these smaller models to create a system of

equations which, when solved, gives the answers to much larger models, including

the final assignment: finding the probability of reaching the x-axis before reaching

y=50. Assuming the particle, again, begins halfway (0,25.5), this was found by our

excel program to be around 0.555.

269 words

ABSTRACT PAGE 3


Table of Contents Abstract ................................................................................................................... 2

1.0 Introduction ................................................................................................... 4

2.0 Developing a Method ..................................................................................... 7

3.0 Creating an Excel Model .................................... Error! Bookmark not defined.

4.0 Solving the Problem ..................................................................................... 12

5.0 Extending the Problem ................................................................................ 19

6.0 Using Excel for Scalability ........................................................................... 29

7.0 Conclusion ................................................................................................... 43

8.0 Works Cited ................................................................................................. 46

9.0 Acknowledgements ....................................................................................... 46

10.0 Appendices ................................................................................................. 47

1.0 Introduction PAGE 4


1.0 Introduction Simple probability almost always involves an event space and sample space. Even the more difficult questions, such as those involving binomial distribution can still be represented in the number of times of success over the number of successes and failures.

However, in higher level probability, questions are elevated to a point where the sample space is infinite. As a direct approach becomes impossible, these questions are more difficult, and require original solutions. An integral part of higher level probability is the ‚random walk,‛ the study of a particle moving randomly in space. While applications of this field are numerous and significant, including the stock market and molecular movement (Brownian motion), it is outside the scope of this essay. Instead we will thoroughly investigate one particular type of random walk to demonstrate the sort of treatment a question of this calibre requires.

While our solution will include probability laws like that of ‚combined events,‛ it has a surprisingly high reliance on other pre-calculus disciplines.

From the USA Math Talent Search:

3/4/20. A particle is currently at the point (0, 3.5) on the plane and is moving towards the origin. When the particle hits a lattice point (a point with integer coordinates), it turns with equal probability 45° to the left or to the right from its current course. Find the probability that the particle reaches the x-axis before hitting the line y = 6 (Mathematical Talent Search).

The following diagram elucidates the motion described above.



The particle starts at (0,4),1 and begins its path down to (0,3). In this path, the particle first turns right of its current path at a 45° angle, towards the lattice point (-1,2), where it turns right 45° again. The particle continues to make left and right turns randomly until it hits y=6, where it terminates. This path is considered a failure.

1.1 Definitions and Notation Let us define the event of the particle reaching y=0 before y=6 a success, or winning. Let us refer to y=0 as the lower bound and y=6 as the upper bound. Let represent the probability of winning given that the particle is at point (x,y), and the upper bound is y=a while the lower bound is y=0. As the question specifies ‘lattice points,’ a, x and y and all other variables or parameters introduced can be expected to be either integers or a subset of .

Please note that our notation has a limitation in that it does not specify the direction. For simplicity, directional information will not be denoted in ‘ .’ We only consider the most direct path a particle can take to (x,y). For example, can only represent the probability of a particle moving south because the particle arrives in that situation in only one move whereas any other direction would require repositioning and thus many more moves. The reader will have access to

1 Although the question specifies that the particle starts at (0,3.5), we begin our paths on a lattice point without altering the results in any way.



graphs which accompany the mathematical equations which explicitly show the direction in question.

The answer to our problem is expressed as , since the particle necessarily goes through (0,3).

2.0 Developing a Method PAGE 7


2.0 Developing a Method To solve this problem, our strategy is to look at a smaller model.

Particle bound by y=0 and y=4

As the particle in the original problem began just above half, we will begin our smaller model at (0,3). The first move will be down to (0,2) where it can either turn left or right, and have these possible paths for the first five moves of the particle.

Theorem 2.1

where (d,e) and (f,g)are the two

possible progressions from (b,c).

Proof. From a tree diagram,

at(b,c)

turning right to (d,e) winning from (d,e)

losing from (d,e)

turning left to (f,g) wining from (f,g)

losing from (f,g)



The probability of winning at (b,c) is therefore:

Therefore the probability of winning on the first move,

(2.1)

However, notice from the diagram that the particle at (-1,1) is on a similar path as the one at (1,1), just in the opposite direction.

Theorem 2.2 Vertical Line of Symmetry. given that the particle at when flipped in the line x=d shares the same path as .

Proof. The particles at and have congruent paths, which is easily shown when one is flipped in the line x=d. Therefore, their probabilities of success must be equal.

Since we can simplify Equation 2.1,



This is useful because we can eliminate (-1,1) from our solution. In the diagram below, the orange path can be eliminated because for the same reason , we reason that .

One method of finding would be to apply theorem 2.1 over and over again as such:

(Theorem 2.2)

(

)

(

)

(2.2)

However, such a solution is unappealing and hard to follow. Instead, we will use a step counting method. Here is a graph for the interval . The endpoints, or the points in Equation 2.2, are marked.



Identifying the endpoints is important because each end point corresponds to a path, whose collective probabilities make up . In order for the particle to arrive at the point, must turn in the correct direction at every junction. For example, a particle at (4,4) arrives after turning left at each of the 3 junctions.2 The probability of passing the junction on the correct trajectory is 0.5 so

(

)

Since is the combination of the probabilities at all possible endpoints,

(

)

(

)

(

)

2 Please note that (0,2) and (3,3) are not counted as junctions. As Thorem 2.2 points out, no matter if the particle moves left or right at these point, the particle will follow congruent paths.



We reason that , since the probability of success from either point is definite. because it is a losing point.

(2.3)

Theorem 2.3 Horizontal at Half. (

)

for all b, given that the particle is

moving horizontally

Proof. Consider an horizontally east moving particle at (

). The two possible

progressions are therefore (

) and (

).

Let (

) . Reflecting the point (

) by

, we can

conclude that the probability of the reflected particle reaching is also . But

since it was reflected, is actually the probability of it reaching . Therefore,

the probability of success, (

) .

(

)

(

)

(

)

(

)

The case for a particle moving west is similar

Therefore from Equation 2.3,

(

)

The probability of a particle reaching the x-axis before y=4 is

.

4.0 Solving the Problem PAGE 12


4.0 Solving the Problem


Following a similar strategy as before, this diagram can help us find .

Based on the possible paths after the first 4 moves, employing the step counting method,

(

)

(

)

(

)

(

)

We can simplify since and are success points and is a horizontal at half (Theorem 2.3)

(

)

(

)

(

)

(

) (

)



(4.1)

Let us first deal with .

(Theorem 2.2)

(4.2)

One final theorem must be developed. Notice how the particle at (5,5) is making similar moves as the one at (4,1). These paths when translated horizontally are symmetrical by y=3, and their probabilities are related.



Theorem 4.1 Horizontal Line of Symmetry. for all b,c

and d, given that the particle at , when reflected in

shares the same path

at the particle at

Proof. By reflecting point (d,a-c) in the line

, we arrive at point (d, c), where

translated horizontally will have the same motion as the particle with .

Therefore, the probability of hitting the line is for both particles. But

since one was reflected, is actually the probability of hitting the line

for the reflected particle:

, where P’ is the probability of failure (hitting y=a) therefore,

From Equation 4.2,

( )

(4.3)

Substituting this into Equation 4.1,

(

)

(4.4)



We will track our motion starting at (4,1):

in fact is a part of theorem 4.1. If the path of the particle from (5,2) to (5,3) were to be reflected in y=3, and translated horizontally, it would be the same path as the particle from (0,4) to (0,3). In such a way we know that

We also notice that since the particles are making the same move, just translated horizontally.



Therefore, from above,

( )

(4.5)

Now we have Equation 4.4 and Equation 4.5

We realize that we must solve a system. However we have 3 variable and only 2 equations. The solution is to write another one, ensuring it only has the three variables we’ve selected: , and . It seems reasonable that we begin the next equation with .



We take another look at the first diagram:

Counting the junctions from the endpoints to (2,2),

(

)

(

)

(

)

(

)

Knowing that 3,0)=1, (4,3) =

and from Equation 4.3,

(

)

(

)

(

)

(

) (

)

(

)

(4.6)



Solving the System Using equations 4.4, 4.5 and 4.6,

3 equations with 3 variables: solve using Microsoft Math:

Of course, the only value we need is

The decimal representation 0.7936 is very close to the predicted value at the end of section 3.0 (which was 79.45%).

5.0 Extending the Problem PAGE 19


5.0 Extending the Problem This problem lends itself to further investigation quite easily. The goal of this section will be to solve the probability for the particle bound by and . To see the general trend, the probability generator produced empirical probabilities for various upper bounds.3 The data is graphed:

3 Data for graph found in Appendix 10.7.



When these points are graphed on a scatter plot, a few interesting features are revealed. The points appear to form an asymptote to the line . This is easily explained. All probabilities are above 50% because all particles begin downwards, closer to the x axis closer, so that it is more likely that the particle reaches the x axis. However, as the bounds expand, the downward movements at the beginning grow increasingly irrelevant because there is a large area for the particle to change direction.

Seeing the increase in difficult from the problem to the problem, one can only assume the difficulty of solving a problem. Our initial method is therefore unscable. Solving a problem will be next to impossible.

The notation is also unscaleable. For a particle bound by y=8, becomes ambiguous.

Calculations might lead us to , which could refer to the blue path and the red path but have different probabilities. To add to the confusion, the purple path is shared by both.



Fortunately, our solution to the problem has created some hints with which we can continue our investigation. Here is a graph from before:

The large dots highlight the three crutial points which which we used to solve our system of equations: (0,3), (2,2), (4,1). With the exclusion of (0,3), which is the starting point, notice that both of the other two points lie on horizontal particle movements.

Notation Let represent the probability of success with upper bound when the particle is traveling on the line . For example, . Similarly, . However, does not necessarily equal .



We were only able to solve the y=6 problem after all probabilities of horizontal movements were calculated.

Note that and were calculated through a system at the end of section 4.0.


Using our observations concerning the smaller models, we employ the horizontal line method to solve for . To find the probabilites of all the horizontal lines, we would require . The probabilities will be stated in the terms of each other, creating a system of equations, which can be solved to find the the exact probabilities.



Begining the particle on a horizontal line of y=1, we will find the probability that the particle reaches the x axis.

This graph shows the possible paths of the particle, each terminating at either the bounds or at a horizontal line.

(

)

(

)

(

)

(

)

(

)

( )

(5.1)



This graph shows the particle beginning on a y=2 and having various end points.

Counting the number of junctions,

(

)

(

)

(

)

(

)

(

)

(

)

(5.2)



Similarly, for

(

)

(

)

( )

(5.3)



Solving the system using Equations 5.1, 5.2 and 5.3,

Therefore,

Now we can solve for .



The graph which maps particle motions for , shows three distinct end points located at horizontal movement and upper/lower bounds.

Using the step-counting method, and knowing that

(5.4)

(

)

(

)



This result is extremely close to the 72.75%4 predicted by the excel program,

attesting to the accuracy of both methods. However, our method by hand is

arguable more useful because it provides the exact value of

.

Another benefit of the horizontal line method is how scalable it is. This method can handle a graph of any size because of its wrought formulaic approach. The original ‚point by point‛ method requires an immense amount of thinking on the mathematician’s part and is very difficult to complete for larger grids.

However, as good as the horizontal line approach is, it is limited by the amount of equations a mathematician can write out. A particle bound by y=20, for example, needs 9 equations with 9 variables, an extremely difficult task. So even the horizontal line approach is not scalable to the degree we want it to be.

The solution is the automatic generation of equations. The next section will create another excel program but this time, it will output exact answers instead of probabilities based on emperical data.

4 Refer to data found in Appendix 10.7

6.0 Using Excel for Scalability PAGE 29


6.0 Using Excel for Scalability In the last section, we solved the particle bound by y=0 and y=8 by finding the values of all of the probabilities of particles travelling horizontally, and then used those numbers in the final calculation of probability. Using a spreadsheet, we will find P(1), P(2), …, P(a-1) values automatically.

Finding a General Formula

The beauty of the horizontal line method is its recurring graphical representations.

Here is a graph we’re used to seeing, but without labels for the y axis. This is simply to show that the particle’s path is the same no matter where the bounds are. Assuming the particle begins y=b, we can write a general formula for P(b):



Using the step-counting method,

(6.1)



We realize that the coefficient is dependant on the distance from y=b (let this offset be denoted as n).

This table compares the highlighted values

A quick graph reveals that the relationship is a variation on the absolute value function . Since it was vertically stretched by a factor of 0.5 (as determined by the first differences) and translated up 1.5,

.

Offset from y=b, n Exponent, k(n)

-5 4 -3 3 -0.5 -1 2 -0.5 1 2 3 3 0.5 5 4 0.5



We can rewrite Equation 6.1 as:

This can be further simplified into:

∑(

)

Of course, since the equation never ends, we must eliminate some terms according to the defined bounds. Since the argument (b+n) is always within the bounds of the specified question, ( ). For example P(-1) does not exist and must be eliminated.

Therefore,5

∑ (

)

There is one part of the equation that is missing. In our horiontal line equations, we always have a constant. For example, in Equation 5.2, the constant was 0.25.

This constant is created because there is always a particle path that is ‘successful’ and terminates in the line y=0. The point at which this occurs is directly related with the value of b. The smaller b is, the closer the particle begins to y=0, and the likelihood of success is larger.

5 The following sigma notation is read: the sum of (

)

over all odd integers in

the specified range ( ).



The light blue lines on this graph represent possible x axes. When b=3, for example,

the particle begins 3 units above the x axis, allowing it to go through two junctions

before terimnating. Therefore, the constant value for b=3 is (

)

.



The constants for other values of b, determined by the graph, are as follows:

b Constan

t

Exponen

t

1

1

2 (

)

2

3 (

)

2

4 (

)

3

5 (

)

3

6 (

)

4

Expressing the exponent as a function is a little complicated because it resembles a floor/cieling function. Since b increases twice as fast as the exponent, the general formula is

⌊

⌋

Where ⌊ ⌋ is the largest integer smaller or equal to x. I.e. ⌊ ⌋ ⌊ ⌋ where x is any real number. In order to determine possible values for , we create a chart:

b

⌊

⌋

1 0.5+ 1

2 1+ 2

Using the definition ⌊

⌋

⌊

⌋

First row Second row

and

Since must render both and true, for simplicity. Therefore,

(

)⌊ ⌋



Now we can conclude that

∑ (

)

(

)⌊ ⌋

Using the General Formula

a and b are parameters when defined creates a system of equations. When a=8,

b=1,2,3,4,5,6,7. This creates 7 different equations that can be solved.

We will use a matrix to solve the system because it is the quickest way to do so on

excel. The previous equation rearranged for a matrix (isolate constants to the right

side) appears as:

∑ (

)

(

)⌊ ⌋

The matrices (for a=8) will look like:

(

)

(

)

(

)

We will arbitrarily define the row number as the starting point of the particle. ( ). Everything in Row 1 corresponds with events that occured when the particle began on line y=1.

The column number refers to the argument (b+n), or where the particle ended up. (j=b+n). Column 2 contains all the coefficients of P(2).

The entry in Row 1 Column 2 corresponds with the coefficient of the term P(2) of the particle which began at y=1.



Right Hand Matrix

Since i=b,

(

(

)⌊

⌋

(

)⌊

⌋

(

)⌊

⌋

(

)⌊

⌋

(

)⌊

⌋

(

)⌊

⌋

)

Coefficient Matrix

Refering back to the general equation, note that the coefficient of is always one.

∑ (

)

(

)⌊ ⌋

When b=1, will have a coefficient of 1. When b=2, will have a coefficient of 1:

(

)

Also, since b+n=j, we can substitute b=i to create



Therefore, coefficient may be rewritten

(

)

(

)

This coefficient does not appear for every term. Instead it only appears for when n,

the offset is odd. Therefore, it appears to the left and right of P(b), and then

appears in every other entry.

(

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

)

All the empty entries, when the offset is even, are evidently given a value of 0.

This creates the scalability we need because this pattern can be expanded to whatever size we need it to be. (This matrix transfered into excel script can be found in appendix 10.5).

The matrices, for a particle bound by y=8, will look like

(

)

(

)

(

)



Solving for the variable matrix is simple:

(

)

(

)

(

)

(

)

(

)

This matrix method is evidently accurate as values for the first three match those

we already found

. is also

accurate because it is a horizontal line at half, and the final 3 values are simply

complements to the first three.



The beauty of this method is how simple it is to expand. For a particle bound by and , excel instantaneously calculates that the coefficient and right hand matrix to be

(

)

(

)

(

)

Please note how this matrix 9x9 matrix is simply an add-on to the 7x7 matrix seen in the y=8 example.



Excel then solves the matrix equation, giving

(

)

(

)



Again, a graph of the original particle’s path is needed:

(

)

(

)

The decimal approximation of 0.707 fits quite well with the predicted 70.6%.6





We end the investigation with a demonstration of scaleability. The excel program creates a 49x49 matrix and outputs values for . (Refer to appendix 10.6.)

Compared with the empirical value of 0.55587, we can reason that 0.555 is both accurate and the actual value (it has been converted to a decimal because the numerator/denominators were too large).


7.0 Conclusion PAGE 43


7.0 Conclusion As such, we find out the probability of a particle travelling randomly between y=0 and y=50, in less than a minute, when at the beginning of the investigation, it took dedication of time and thinking to produce even a particle travelling between y=0 and y=6.

And to highlight the gravity of , I illustrate a few of the possible paths of such a particle using a few graphs:



It is fulfilling to be able to predict the outcome of a particle’s path as diverse, random and interesting as those seen above.



In this investigation, we began with finding the probability that a particle randomly

travelling between and will hit the lower bound before hitting the upper

bound to be

. We extended the problem to find the same probability for a particle

between and . This was

. We also developed an excel program that

runs a set number of trials to find the empirical probability of a particle bound by

any two lines, as well as a method known as the ‚horizontal line method‛. This

horizontal line method uses a system of equations to quickly arrive at an answer

which would have taken much longer using the step-by-step approach that was

primarily adopted. This horizontal line method is limited by the number of

equations one would like to solve by hand. With technology, our second excel

program could find the exact probability of a particle bound by any two lines,

including the probability of hitting y=50 before y=0 (0.555).

However, our current method has a heavy reliance on technology. In the future, I’d

like to explore to find perhaps an even more direct method of attaining the answer.

8.0 Works Cited PAGE 46


8.0 Works Cited "USA Mathematical Talent Search Round 4 Problems Year 20 — Academic Year 2008–

2009." USA Mathematical Talent Search. March 9, 2009. 7 Sep 2009 .

9.0 Acknowledgements I would like to thank websites like www.mrexcel.com for excellent instructions for all

of the formulas I have never used before. Creating the excel programs was extremely

enjoyable. Finally, I’d like to thank my extended essay mentor, Mr. De Bruyn for

digesting all the formulas and theorems in my essay. All the math teachers I

approached directed me to him because as they proclaimed, ‚his mind works in

wondrous ways.‛

10.0 Appendices PAGE 47


10.0 Appendices

10.1 Excel Program for Mapping Particles

A B C D

1 Step x y Movement

2

0 4 Down

3 1 0 3

=IF(RANDBETWEEN(0,1)=1,"Left","Right")

4

2

=IF(D4="Left",IF((B3-B2)=0,-(C3-C2),0)+IF((B3-B2)=(C3-C2),0,0)+IF((C3-C2)=0,(B3-B2),0)+IF((B3-B2)=-(C3-C2),(B3-B2),0)+B3, IF((B3-B2)=0,(C3-C2),0)+IF((B3-B2)=(C3-C2),(B3-B2),0)+IF((C3-C2)=0,(B3-B2),0)+IF((B3-B2)=-(C3-C2),0,0)+B3)

=IF(D4="Left", IF((C3-C2)=0, (B3-B2),0)+IF((B3-B2)=0,(C3-C2),0)+IF((B3-B2)=(C3-C2), (C3-C2),0)+IF((C3-C2)=-(B3-B2), 0,0)+C3, IF((C3-C2)=0, -(B3-B2),0)+IF((B3-B2)=0,(C3-C2),0)+IF((B3-B2)=(C3-C2),0,0)+IF((C3-C2)=-(B3-B2), (C3-C2),0)+C3)


5

3




6

4




Seen here is a truncated table used to produce a random path of the particle. While this only shows four steps, our program regularly has over 500.



10.2 Excel Program for Mapping Particles (2)

First x value for y= upper bound =MATCH(F2,C:C,0)

First x value for y= lower bound =MATCH(F3,C:C,0)

Where does particle reach first? =IF(F9



10.5 Coefficient Matrix and Right Hand Matrix

F I J K

1 1 2 3

2 1 =IF(I$1=$F2,1,0)+IF(MOD(I$1-$F2,2)=1,-1/2^(((ABS(I$1-$F2))+3)/2),0)

=IF(J$1=$F2,1,0)+IF(MOD(J$1-$F2,2)=1,-1/2^(((ABS(J$1-$F2))+3)/2),0)

=IF(K$1=$F2,1,0)+IF(MOD(K$1-$F2,2)=1,-1/2^(((ABS(K$1-$F2))+3)/2),0)







A translation to excel of the co-efficient matrix shown in the body of the essay.

C D

1 X

2 1 =0.5^(FLOOR(F2/2+1,1))

3 2 =0.5^(FLOOR(F3/2+1,1))

4 3 =0.5^(FLOOR(F4/2+1,1))

Right Hand Matrix



10.6 Particle Bound by y=50, Horizontal Line Probabilities

=

1 16/17

2 865/942

3 158/175

4 731/828

5 567/655

6 779/920

7 712/859

8 594/733

9 225/284

10 599/774

11 365/483

12 629/853

13 443/616

14 239/341

15 499/731

16 192/289

17 555/859

18 415/661

19 573/940

20 0.591312

21 0.57305

22 0.554787

23 0.536525

24 0.518262

25 0.5

26 0.481738

27 0.463475

28 0.445213

29 0.42695

30 0.408688

31 0.390426

32 0.372164

33 0.3539



34 0.335641

35 0.317374

36 0.299119

37 0.280844

38 0.262602

39 0.244306

40 0.226099

41 0.207747

42 0.189631

43 0.171128

44 0.153261

45 0.134351

46 0.117151

47 0.097146

48 0.081742

49 0.058791

Note that fractions greater than 3digits/3digits are represented as decimals



10.7 Empirical Probabilities

Upper bound y= % of success Trials

4 81.72 65000

6 79.12 65000

8 72.75 50000

10 70.6 50000

12 67 10000

14 66.6 1000

16 65.39 1000

18 64.33 1000

20 61.2 1000

24 61 1000

30 57.96 1000

40 56.1 1000

50 55.58 3341

Documents

Probability of a 45º Drunkard’s Walkstatic.squarespace.com/static/4ff39269c4aaf3fdb14faa05/t/...2.0 Developing a Method PAGE 8 David Kong – 2203 033 The probability of winning