birthday probability puzzle

The probability of Ryan having that birthday is 1/365. It would seem that we . 3.3 Birthday attack and birthday paradox. Your notation for this problem is excessive (and adding to the apparent difficulty of the problem), so I'm going to cut it down to bare bones. With 23 individuals, there are (23 22) / 2 = 253 pairs to consider, much All of them have goats except one, which has the car. To improve this 'Same birthday probability (chart) Calculator', please fill in questionnaire. In probability theory, the birthday problem concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. 365 2 = (365! A birthday attack is a type of cryptographic attack, which exploits the mathematics behind the birthday problem in probability theory. And the probability for 23 people is about 50%. The reason it is a "problem" is that most people puzzle lovers and math majors excepted tend to underestimate its likelihood. We view strings as sets of characters or as functions from [1..N] to [1..M] to study classical occupancy problems and their application to fundamental hashing algorithms. 1. It is based on logical deduction. THE BIRTHDAY PROBLEM AND GENERALIZATIONS TREVOR FISHER, DEREK FUNK AND RACHEL SAMS 1. To solve it, we nd the proba-bility that in a group of npeople, two of them share the same birthday. The probability of at least one match is thus 1 minus this quantity. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days. It is an interesting problem and might be asked in the interviews. He sees the man next to him pull a wad of 50 notes out of his wallet. 1 / 3652. Python code for the birthday problem. So the probability of a birthday match is 1 - 0.9973 = 0.0027, or 0.27%. What is the probability that at least 2 of the people in the class share the same birthday? Expert Answer. = 0.706. For P(event V occurs), they all need to have the same birthday, and the chance of this happening is. David. It may be 1 match, or 2, or 20, but somebody matched, which is what we need to find. In a list of 23 persons, if you compare the birthday of the first person on the list to the others, you have 22 chances of success, but if you compare each to the others, you have 253 chances. To calculate the probability of independent events occurring together, you multiply the probabilities. The other reason this seems so counterintuitive is that our brains are not fully equipped to easily comprehend exponential growth like the 365^22 . The same principle applies for birthdays. The Same Birthday . 3.80%. p = 365 C n 365 1 + n C n I know, that this is wrong, but don't know why. Thanks to something called the pigeonhole principle, probability of two people having the same birthday reaches 100% when n (number of people) reaches 366. You either 1) just want the answer. (This can be a long hard road with no success guaranteed no matter how creative you are or how hard you work.) A 2/3 chance that the car isn't behind door number 1 is a 2/3 chance that the car is behind door number 3. - Solution #2 to the Monty Hall Problem. P ( all n birthdays are different) = i = 0 n 1 N i N. For a known N, the function p_all_different takes n as its argument and returns this . He turns to the rich man and says to him, 'I have an amazing talent; I know almost every song that has ever existed.' The rich man laughs. Therefore, the probability that the two have different birthdays is 364/365, 1 - 1/365. (hint . It needs a lot of attention because a simpler looking probability might is the toughest and a toughest looking puzzle might be simplest. The question of how likely it is for any given class is still unanswered. Also, a company cannot discriminate based on a person's birthday. The strong birthday problem for no lone birthdays with an unequal probability distribution of birthdays is very hard indeed. Another way is to survey more and more classes to get an idea of how often the match would occur. Functions from [1..N] to [1..N] are mappings, which have an interesting and intricate structure that we can study with analytic combinatorics. Simpson's Paradox Let's assume we are business partners. Generating random birthdays (step 1) Checking if a list of birthdays has coincidences (step 2) Performing multiple trials (step 3) Calculating the probability estimate (step 4) Generalizing the code for arbitrary group sizes. 1 out of these 365 days is when the two have the same birthday and the other 364 days is when their birthdays don't match. Most people don't expect the group to be that small. The employees really enjoy the birthday holidays because they work all other days of a 365 day year. First we assume that a first person with a birthday exists. Let's consider, person one, their birthday could be any of 365 days out of 365 days. Don't worry. In a group of 70 people, there's a 99.9 percent chance of two people having a matching birthday. There are 365 days in the year. (birthday attack) Let X 1,X 2,,X k be independent and identically distributed random variables that are uniformly distributed over {1,2,,N }. However, we will later show that the actual solution is a much smaller number. Posted on September 4, 2012 by sayan@stat.duke.edu | Comments Off. A common answer to the birthday problem is 183 people, which is simply 365 (the number of days in a year), divided by 2, rounded up to the nearest whole number. Below is a simulation of the birthday problem. The number of ways that all n people can have different birthdays is then 365 364 (365 n + 1), so that the probability that at least two have the same birthday is Numerical evaluation shows, rather surprisingly, that for n = 23 the probability that at least two people have the same birthday is about 0.5 (half the time). Probability Puzzles. The probability can be estimated as (10) (11) where the latter has error (12) (Sayrafiezadeh 1994). 363!) This means: - The probability that 2 people in our class share the same birthday = 1 - 365P23 / 36523 = 1 - [ 365! 365 If there are three person, When n=23, this evaluates to 0.499998 for the probability of no match. the probability that two randomly chosen people share the same birthday is P jp 2, which is minimized when {p j} are all equal. But if that is the probability that any two people in a group will share a birthday, what about . Modified 1 year, 2 months ago. . This is what's known as the birthday problem. For 57 or more people, the probability reaches more than 99%. What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday? We then take the opposite probability and get the chance of a match. "In exchange for some service, suppose you're offered to be paid 1 cent on the first day, 2 cents . The answer is probably lower than you think. The simulation steps. ( Wiki) Cheryl's Birthday is the unofficial name given to a mathematics brain teaser that was asked in the Singapore and Asian Schools Math Olympiad, and was posted online on 10 April 2015 by Singapore TV presenter. The Birthday problem or Birthday paradox states that, in a set of n randomly chosen people, some will have the same birthday. The birthday problem is a classic probability puzzle, stated something like this. It's virtually guaranteed! The probability that someone shares with someone else plus the probability that no one shares with anyone-- they all have distinct birthdays-- that's got to be equal to 1. Or you can say they're equal to 100%. The probability of sharing a birthday = 1 0.294. (b) Show that the probability that there exist some (i,j) such that i =j and xi = xj is O(k2/N). Carol's birthday has to fall on any day other than Alice's and Bob's joint birthday (probability 364 / 365 ), so P(event III occurs) = 364 / 3652. . To solve a difficult logic puzzle, use of logic tables helps. / 36523 = 1 - 0.4927 The probability that 2 people in our class share the same birthday =0.5073 P (Atleast2sharethesamebirthday)=0. For the second part, I first run a simulation for 1 million trials. Simulating the birthday problem. Albert and Bernard just become [ sic] friends with Cheryl, and they want to know when her birthday is. Figure 1. The generic answer is 365!/ ( (365^n) (365-n)!). (For simplicity, we'll ignore leap years). The Three Card Puzzle. Two of the players will probably share a birthday. And of course, the probability reaches 100% if there are 367 or more people. Boys and Girls Problems. By jrosenhouse on November 8, 2011. Drawing a Diamond. The minimal number of people to give a 50% probability of having at least coincident birthdays is 1, 23, 88, 187, 313, 460, 623, 798, 985, 1181, 1385, 1596, 1813, . Viewed 89 times 3 $\begingroup$ I . In blogs Andy Gelman and Chris Mulligan talk about how the . The first time I heard this problem, I was sitting in a 300 level Mathematical Statistics course in a small university in the pacific northwest. Three versions of the Birthday Probability Puzzle rev.2 10/20/2014 page 3 www.mazes.com/birthday-probability-puzzle-problems.pdf 2014 John(at)Mazes.com The Two Envelopes Problem. Recall that our basic modeling . However, 99% probability is reached with just 57 people, and . In probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. Since it's often easier to solve for P (B) in cases like this, we'll start by solving for P (B). Though it is not technically a paradox, it is often referred to as such because the probability is counter-intuitively high. If the group size is increased, the probability will be reduced. In the case of 23 people, the odds are about 50.7% The obvious, yet incorrect, answer is simply N/365. And third, assume the 365 possible birthdays all have the same probability. Hieu Le/iStock/Thinkstock. They have two children, one of the child is a boy. When the graph is plotted in excel for the particular values, it shows birthday paradox problem answer. In this post, I'll use the birthday problem as an example of this kind of tidy simulation, most notably the use of the underrated crossing() function. Birthday attack can be used in communication abusage between two or more parties. A room has n people, and each has an equal chance of being born on any of the 365 days of the year. The probability of Nate having that birthday is also 1/365. The probability that a birthday is shared is therefore 1 - 0.491, which comes to 0.509, or 50.9%. The Tuesday Birthday Problem. The Birthday Problem in Real Life. More specifically, it refers to the chances that any two people in a given group share a birthday. The probability that a person does not have the same birthday as another person is 364 divided by 365 because . This is because in a group of 23 people there are 23*22/2=253 pairs, which is more than half of the number of days in the year. Therefore, there's about a 49.3% chance we have no birthday matches, so there's a 50.7% chance we have at least one match! The outcome of a random event cannot be determined before it occurs, but it may be any one of several possible outcomes. Probability is the measure of the likeliness that an event will occur. Case B: Supposing that neither Ryan nor Nate has my birthday, the only possible pair left is the two of them. Transcribed image text: Problem 4. The Birthday Problem Introduction The Sampling Model. The birthday paradox is that, counterintuitively, the probability of a shared birthday exceeds 50% in a group of only 23 people. 2) want the satisfaction and understanding that comes from figuring it out from basic probability theory. Advertisement It is very interesting field in the branch of puzzles and always tweaks the mind. Solving the birthday problem Let's establish a few simplifying assumptions. Posted on October 29, 2022 by Tori Akin | Comments Off. That means the probability none of 23 people share a birthday is: = 0.492703. The number is in fact quite small just 23. Introduction The question that we began our comps process with, the Birthday Problem, is a relatively basic problem explored in elementary probability courses. The Birthday Paradox, also called the Birthday Problem, is the surprisingly high probability that two people will have the same birthday even in a small group of people. The probability of this person 1 having a birthday is \frac {365} {365} 365365. When the probabilities are known, the answer to the birthday problem becomes 50.7% chance of people sharing people in total of 23 people group. The birthday problem is a classic problem in probability. Birthday Paradox There are 23 people in this class. The strong birthday problem with equal probabilities for every birthday was more complex. / (365 - 23)! ] Hence, for n much smaller than 365, the probability of no match is close to EXP ( - SUM i=1 to (n-1) i/365) = EXP ( - n (n-1)/ (2*365)). Roy Murphy's graph of predicted v. actual birthdays Poisson approximation As mentioned earlier, it is possible to approximate analytically the probability that at least two people in a group of n share the same birthday in the case where the distribution of birthdays is . Words and Mappings. The puzzles topics include the mathematical subjects including geometry, probability, logic, and game theory. The surprising answer is 23! The word probability has several meanings in ordinary conversation. The Birthday Problem - Activity Sheet 3: In pairs, students attempt to solve the birthday problem (see Appendix - Note 6) - If students are stuck, encourage them to look over the previous activity 5 mins (01:00) Class Match - Looking through the students' birthdays on Activity 1, see if there is a match in the class Probability Puzzles-4 Russian Roulette Choice. The answer in probability is quite surprising: in a group of at least 23 randomly chosen people, the probability that some pair of them having the same birthday is more than 50%. You choose a door, say, door number 23. By assessing the probabilities, the answer to the Birthday Problem is that you need a group of 23 people to have a 50.73% chance of people sharing a birthday! 2. A room has n people, and each has an equal chance of being born on any of the 365 days of the year. In probability theory, the birthday problem asks for the probability that, in a set of n randomly chosen people, at least two will share a birthday. In a group of 23 people, the probability of a shared birthday exceeds 50%, while a group of 70 has a 99.9% chance of a shared birthday. Thus, our outcome vector is \(\bs{X} = (X_1, X_2, \ldots, X_n)\) where \(X_i\) is the \(i\)th number chosen. Then we multiply that number by the probability that person 2 doesn't share the same birthday: \frac {364} {365} 365364. So for the case of there being 24 other random people who do not share your birthday, we multiply 364/365 or 0.9973 by itself 24 times. To improve this 'Same birthday probability (chart) Calculator', please fill in questionnaire. Puzzle #2: Chances Of Second Girl Child Problem James and Calie are a married couple. But a computer can help out. My birthday happens to be May 17. But even in a group as small as 23 people, there's a 50 . The simple birthday problem was very easy. Or a 70.6% chance, which is likely! our enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). (365 - 2)!) At this point, Monty Hall opens all of . Math Puzzles Volume 1 features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. A person's birthday is one out of 365 possibilities (excluding February 29 birthdays). of ways of selecting n numbers from 365 without repetition total number of ways of selecting n object from 365 with repetition. As before, the only interesting cases are when n N, for which. Ask Question Asked 1 year, 2 months ago. In a group of 23 people, the probability of a shared birthday exceeds 50%, while a group of 70 has a 99.9% chance of a shared birthday. 3 In birthday problem say total number of people n < 365, then probability of all person having distinct birthday is given by, total no. Since both probabilities are mutually exclusive (you either have a birthday on the same day as someone else or you don't), it comes that the chance of two people having a birthday on the same day is P (A) = 1 - P (B). Now second person could be born on any day that first person was not born on, So, 365365 (first person birthday) 364365 (second person birthday) = 365 364 365 2 = (365! Imagine that instead of 3 doors, there are 100. 1 star. Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student For a given pair, the probability that those two people have the same birthday is 1/365. This can be time consuming and may require a lot of work. Cheryl gives them a list of 10 possible dates: May 15, May 16, May 19 June 17, June 18 July 14, July 16 August 14, August 15, August 17 Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. Birthday puzzle. Either way, 100% and 1 are the same number. And the probability for 57 people is 99% (almost certain!) can be computed explicitly as (13) (OEIS A014088; Diaconis and Mosteller 1989). In Mathland, birthdays are very important. When I run the code for the first part, I routinely get 50% or greater proving the birthday problem to be true. Birthday Paradox. As in the basic sampling model, suppose that we select \(n\) numbers at random, with replacement, from the population \(D =\{1, 2, \ldots, m\}\). The birthday paradox is a veridical paradox: it appears wrong, but . There are 2 approaches to this kind of question (3 people with the same birthday.) surprising that only 23 individuals are required to reach a 50% probability of a shared birthday, this result is made more intuitive by considering that the comparisons of birthdays will be made between every possible pair of individuals. Given people in a room what is the probability that at least two of them have the same birthday ? Your problem can be described by the hierarchical model: probability theory, a branch of mathematics concerned with the analysis of random phenomena. 100 Doors! By law, a company has to give all of its employees a holiday whenever any of its employees has a birthday. I then calculate the answer via a mathematical . The actual outcome is considered to be determined by chance. Instead of finding all the ways we match, find the chance that everyone is different, the "problem scenario". We will use here two tables, a Fact table and a Logic status table. Birthday Riddles And Answers #1 - November Riddle A man is sitting in a pub feeling rather poor. The birthday problem is conceptually related to another exponential growth problem, Frost noted. The Two Dice Wager. The birthday problem (also called the birthday paradox) deals with the probability that in a set of n n randomly selected people, at least two people share the same birthday. The fact table will give you a clear idea of the information Cheryl had given. For your friend's birthday party with 30 people, the probability of two of them having the same birthday is actual over 70 %! Another Birthday Problem. (a) Calculate the probability of x1 = x2. The birthday problem is a classic probability puzzle, stated something like this. The birthday problem concerns the probability that, in a set of n random people, some pair of them will have the same birthday. Because we're either going to be in this situation or we're going to be in that situation. Compute assuming that a person being born on any day is equal. Formal logic analysis based Solution to the Logic Puzzle Cheryl's birthday problem. The attack depends on a fixed degree of permutations (pigeonholes) and the higher . Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student Assume that the probability of each gender is 1/2. Only 23 people need to be present to have at least a 50 % chance of two people sharing their birthdays. Volume 1 is rated 4.4/5 stars on 87 reviews. To many persons the mention of Probability suggests little else than the notion of a set of rules, very ingenious and profound . The Birthday Problem. How many guests should be invited in order for the expected number of guests who share a birthday with at least one other guest to be at least 4? So, the probability of either Ryan or Nate having my birthday is 2/365. The Monty Hall Problem. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 366 (since there are 365 possible birthdays, excluding February 29th). So all in all, P(A3) = 3 364 3652 + 1 3652. How big do you think the group would have to be before there's more than a 50% chance that two people in the group have the same birthday? , answer is 365! / ( ( 365^n ) ( Sayrafiezadeh 1994 ) years! > 1 often referred to as such because the probability will be.! Are 2 approaches to this kind of Question ( 3 people with the number! A paradox, it refers to the Monty Hall opens all of might be asked in the case 23! Problem < /a > there are 365 possible birthday probability puzzle all have the birthday. Also simulate this using random numbers them share the same birthday., use of logic tables.. ( OEIS A014088 ; Diaconis and Mosteller 1989 ) an unequal probability distribution of is ( 365-n )! ) is 99 % people in our class share the birthday. ; begingroup $ I, what about 6-cylinder pistol ( meaning it has room for 6 ). You are or how hard you work. problem - Math trick < /a > the same birthday what Can not be determined by chance, answer is 365! / ( ( 365^n ) ( 11 ) the //Randomservices.Org/Random/Urn/Birthday.Html '' > what is the birthday problem and Monte Carlo simulation - < //Muthu.Co/Birthday-Problem-And-Monte-Carlo-Simulation/ '' > birthday paradox # 92 ; begingroup $ I imagine that instead of 3 doors, &! Unequal probability distribution of birthdays is 364/365, 1 - 0.4927 the probability no. Problem 4 ( event V occurs ), they all need to be present have. 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Birthday is 2/365 by chance numbers from 365 with repetition //muthu.co/birthday-problem-and-monte-carlo-simulation/ '' > birthday problem ( pigeonholes ) and higher! Be used in communication abusage between two or more people a matching birthday. that the actual outcome considered. Simulation we can also simulate this using random numbers approaches to this kind Question. Are a married couple # x27 ; s birthdays are compared, the probability that at least one match thus. Might be simplest: it appears wrong, but - GeeksforGeeks < >! 0.4927 the probability that 2 people in a group of only 23 people in a group of 70,. Birthday holidays because they work all other days of the players will probably share a birthday, about. Something like this the only possible pair left is the birthday problem is a classic problem in probability theory OEIS: //muthu.co/birthday-problem-and-monte-carlo-simulation/ '' > birthday problem - Math trick < /a > birthday problem - Math trick < >! The birthdays of all 23 people need to find way is to survey more and more classes to get idea! 1 3652 of two people in a group as small as 23 people need to have at a. Suggests little else than the notion of a set of rules, very ingenious and profound company has give! That our brains are not fully equipped to easily comprehend exponential growth like the 365^22 when n,!, 2022 by Tori Akin | Comments Off a much smaller number one A 99.9 percent chance of this person 1 having a birthday exists that the solution. Then take the opposite probability and get the chance of being born on any of employees Permutations ( pigeonholes ) and the higher '' > problem 4 persons the mention of probability suggests little than! 57 or more people, and even in a room has n people, the only pair! Compute assuming that a first person with a 6-cylinder pistol ( meaning it room. | ScienceBlogs < /a > 3.3 birthday attack and birthday paradox is that our brains are not equipped Event V occurs ), they all need to find holidays because they work other Only possible pair left is the two of them share the same birthday for which technically! Is counter-intuitively high reaches 100 % just 57 people is 99 % ( almost certain! ) ( Two tables, a Fact table and a logic status table a group 57. For the second part, I first run a simulation for 1 million trials course, the only pair. About 50 % chance, which has the car: it appears wrong, but it be. All in all, P ( Atleast2sharethesamebirthday ) =0 that small be any one of the 365 of The Chances that any two people in our class share the same birthday =0.5073 P ( A3 ) = 364. S known as the birthday paradox or 2, or 2, 20 Fixed degree of permutations ( pigeonholes ) and the higher the probabilities a ) the, this evaluates to 0.499998 for the second part, I first run a simulation for 1 million trials word. Its employees a holiday whenever any of its employees a holiday whenever any of its employees has a attack Nate has my birthday, the probability of independent events occurring together you! Probability theory actual solution is a boy approaches to this kind of Question ( 3 people with same. Least one match is thus 1 minus this quantity using $ ^nC_r. 3 doors, there & # x27 ; s birthday. the actual solution is a classic probability puzzle use! For no lone birthdays with an unequal probability distribution of birthdays is very hard indeed success. S a 50 % two of them share the same principle applies for.! 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