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 Meagan Summers
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1 In many situations in life, we are presented with opportunities to evaluate probabilities of events occurring and make judgments and decisions from this information. In this paper, we will explore four questions which have been asked by students of Sacramento State which can be evaluated using Statistical methods and Probability tools. These questions are: 1. Given that a student knows 2 questions of a possible 6 which can be on an exam, and on this exam only 3 of the 6 are chosen by the professor to be on the exam, and he or she is required to answer 1 question, what is the probability that the student will be able to answer 1 question? 2. Why does it seem that the team with the best chance of being chosen with the first draft pick in the NBA is being unfairly treated? Is this even the case? 3. Two people play a single round of Russian Roulette. A gun is used with six chambers, with a round randomly placed in one. a. If the cylinder is spun between trigger pulls, which player has the better probability of surviving after these two trigger pulls? Why? b. Does the probability change if the cylinder is not re spun after the first person shoots? If so, what is the probability of the second person dying? 4. Suppose one is given three fair dice; what is the probability of getting all dice to show the same number, a, if one is allowed to reroll any dice he or she wishes, and are allowed two rerolls are allowed? Question 1 This is a scenario often faced by students: given a limited time period to devote to many activities students will maximize the usefulness of the study time by only studying a limited set of the problems at hand. Given that of 6 questions could be on the exam, 3 are chosen by the professor, 1 of the 3 must be answered by the student, and the student knows the solutions to 2 of the 6 problems, what is the probability that 1 of the questions of the 3 chosen to be on the exam could be successfully answered by the student? Here you can just skip to the rule of the complement. You don t have to give so much background. From probability, we know that the probability of an event happening is 1 minus the probability the event does not happen, or in mathematical notation, P(A)=1 P(A C ). From probability, we know that the probability of some event occurring is 1 minus the probability that the event does not happen, or, in mathematical notation [1]: 1 Formatted: Superscript Formatted: Font: Do not check spelling or grammar Formatted: Centered Similarly, the probability that the student can answer at least one question is equal to 1 minus the probability that the student can answer none of the questions: 1 Formatted: Centered The probability that the student can answer at least one question as equal to the probability that, of the three questions chosen, two of them the student has studied, plus the probability that one of them the student has studied.
2 An axiom of probability states that the probability of all possible events occurring is equal to 1. (Not true as stated. If the events partition the sample space, then the last sentence is true.) Using this axiom, we can take the compleiment of this event the event that no questions are answerable and add it to the probability that at least one question is answerable. Because these two events are disjoint and completely cover all possible events in the sample space, their probabilities equal 1. In turn, knowing this, we can solve for the event that at least 1 question is answerable. Formatted: Font: Italic 1 1 To determine the probabilities of these events occurring, we must determine a way to describe the set of all possible outcomes, or the sample space. A useful sample space for this experiment consists of all possible sets of 3 questions the professor can choose to put on the exam. Assuming the professor randomly chooses the questions that will be on the exam, every possible choice of a set of 3 exam questions is equally likely. Thus, tthe probability that no questions are answerable is equal to the ratio of all outcomes in which no question is answerable against the size of the sample space. To calculate the size of the sample space, we must determine all the possible ways the professor can choose 3 questions out of the 6 available to put on the exam. Using counting methods [1] not explained here in DeGroot and Schervish (p.???), we can calculate the size of S, represented by ! 6 3! 3! 6! 3! 3! 20 There are 20 different ways to choose a set of 3 distinct items from 6, so there are 20 possible sets of questions the professor could put on the exam. The next item we must determine is the number of possible outcomes the event that the student does not know any of the questions on the exam. The set of questions can be split into two disjoint sets: two questions which the student knows, and four which the student does not know. The event that the student knows none of the questions on the exam must mean that the 3 questions have been chosen only from the 4 which the student does not know. Let K be the event that the student does not know any outcomes on the exam. The number of outcomes in K can be computed as follows: ! 4 3! 3! 2! 2 2! 2! 4! 1! 3! 2! 0! 2! Now that we know the sizes of the sample space and the number of outcomes in the event, the probability that the student knows no questions on the exam can be computed:
3 4 20 Plugging Substituting in this value into the original equation, we can calculate the probability that the student can answer at least one question: Therefore, the probability that at least one of the two questions known by the student will be chosen by the professor and placed on the exam is 0.8, or 80%. Question 2 The National Basketball Association determines the order by which the teams in the league are given chances to draft new talent each year using a weighted lottery. The lottery system is biased so that the lowest performing team in the previous season is given the greatest opportunity relative to any other team at being chosen as having the first draft pick of the year, and the worst team is guaranteed to not have any worst pick position than 4 th. The odds for the worst performing team getting the first draft pick position are as follows [2]: Draft Position Odds Total At first glance, iit is obvious that the team with the best chance at being chosen as the number 1first draft pick will more than likely not get the 1 st first pick, as clearly it is more probable that the team will be chosen as the 4 th fourth draft pick. Just because that single pick is the most likely, though, does not mean that it is the most likely event to occur. Though the chance of getting chosen 4 th is the highest at 0.357, it is much more likely that the team will get chosen first, second, or third, as the probability of getting one of those three is = (This doesn t seem relevant to me; did I not catch that in the original?) While this may have the appearance of being unfair, Tthe lottery only guarantees that the worst performing team gets the highest chance at winning compared with any other SINGLEsingle team, not that the worst performing team will more likely than not receive the first pick in the draft. To illustrate the difference, suppose there exists some game of chance played with four individuals, A, B, C, and D. Also, say for whatever reason, player Aone player has a 40% chance of winning, while the remaining individuals only have a 20% chance of winning. Player AThis single player has the best chance of winning, but it is still more likely (60% chance) that he will lose. Formatted: Font: Do not check spelling or grammar Comment [C1]: I got the impression from a note you wrote that considering multiple outcomes together like this wasn t a bad idea. However, I can see how it messes with the flow, and it makes sense to take it out. Formatted: Font: Italic Chris, now relate this to the lottery, i.e. The worst team has the highest chance at the first round draft pick (0.25) COMPARED TO ANY OTHER SINGLE TEAM. In fact, the second to worst team has a 19.9%
4 chance of winning first pick according to Wikipedia and the probabilities decrease as the teams rankings get better. However, when we combine the probabilities of getting the first round draft pick for all teams other than the worst, we obtain Thus, although the worst team has the best chance at the first round draft pick compared to any single team, they are more likely to lose the first round draft pick than to win it. (I think you are getting at this in the next paragraph, but I think it is better to make a parallel argument from your generic example to the NBA draft. Also, providing a concrete number like 19.9% can really help clarify the issue.)similarly, just because the worst team of the prior year has the greatest likelihood to get the first draft pick does not mean that they are more likely than not to get the first draft pick. The worst team has a 25% chance of getting picked first for the draft, and the remaining teams have a collective 75% chance of getting picked for the first position. In turn, the second worst team has the second best chance at being picked first, at 19.9% [2]. The remaining chance of getting the first pick is split in successively smaller amounts with successively better teams. Formatted: Font: Do not check spelling or grammar The same can be said of the worst performing NBA team and their bid for the first draft pick. Compared to any other team (this is the important point, good) in the league, the worst performing team has the highest chance of being chosen for the 1 st first position in the draft which is the purpose of the draft. However, Even though the worst team has the highest probability in being chosen first, it is three times more likely that they the team will not be chosen for the first position. Question 3 The game of Russian Roulette offers an opportunity to examine use (the final probabilities are total probabilities ) what is known in statistics as conditional probability. Conditional probability allows us to calculate the total probability probabilities that an event occurs, given that we know another one has previously occurredany other combination of related event has previously occurred. Russian Roulette, being a game which advances if a certain event does not occur, utilizes these conditional probabilities. In turn, the probability that either player perishes in the game is a total probability. For the purposes of the analysis of this question, let and represent the event that the gun goes off on the ith trigger pull for each participating individual, A and B. Part 1 The first part of the question asks if a game of Russian Roulette is played for a single round, and the cylinder is spun between each trigger pull, which turn is safer to take: the first or the second? Because the cylinder is being spun between each trigger pull, the probability that the gun fires on any trigger pull, given that the previous trigger pulls did not result in the gun going off, is independent of any other: (what if the bullet fires on the first pull? That changes prob of firing in second pull, so you have dependence. You can say Assuming the gun has not fired on any previous pull, the probability the gun fires is 1 in 6. ) 1 in 6. The only way the second person can fire is if the gun does not go off when the first person fired. Therefore, the probability for any trigger pull being one which fires is conditional upon the previous pull not firing.
5 2 By the rules of conditional probability Multiplication Rule [1], we can rewrite the above as Formatted: Font: Do not check spelling or grammar Because each trigger pull is independent, Because the cylinder is being spun between each trigger pull, the conditional probability of the player B losing on the second trigger pull given that player A did not lose is equal to the probability of player B losing on any trigger pull, which is 1 in 6 because we are randomly choosing one of six chambers and only one has a bullet, or identical to the probabilities faced by A.. This allows us to solve for the probability of player B, losing: Again, Tthe probability that the first player loses on the first shot trigger pull is 1 in 6. Therefore, we can conclude: Therefore, it is marginally preferable to go second, rather than first, simply for the fact that there is a small probability that the second trigger pull will never occur, as the gun fired on the first trigger pull.. Part 2 The second part of the question asks a similar question as the first, but poses the condition that the cylinder is not spun between trigger pulls. The probability that the first player loses on the first trigger pull is the same: one in six. However, the probability that the second player losing on the second trigger pull has changed, since the cylinder is not being spun. If the second player has a chance to pull the trigger, the first player has verified that one of the chambers does not have the round, and there are five remaining chambers where the single round could be a probability of going off of 1 in 5. The symbolic description of the probabilities remains the same, but the values assigned to each probability are different: Therefore, the probability of either player dying the first player on the first trigger pull, the second on the second trigger pull is the same, and there is no advantage in going first or second. Question 4 The act of getting a set of thrown dice to all shown the same value is known colloquially as a. The probability of getting three dice to show a, with two allowed rerolls with the ability to
6 choose which dice to reroll can be calculated by summing the discrete events which result in a after one, two, or three throws. This is possible because the sets of events which result in a are disjoint. since these events are disjoint. There is only one possible way to get a on the first throw. However, there are multiple ways by which a could be achieved in two or three throws. The relationship between the possible throws allows one to construct a tree of possible outcomes from one, two, and three throws, to better comprehend what possible routes can be taken to achieve a. The results of the first throw dictate the choices for the following cases. If three unique values are thrown on the first toss, all three dice can be rerolled on the second throw. If a pair is rolled on the first throw, the single dice can be rerolled for an attempt at a. Lastly, it is possible to get a on the first throw. The relationship between the possible throws allows one to construct a tree of possible outcomes from one, two, and three throws, to better comprehend illustrate what possible routespaths can be taken to achieve a. (Before presenting diagram, explain what you will do after the first roll for each case. If the first throw consists of 3 distinct values, then all three dice will be rerolled on second throw and you could get. If the first throw consists of a pair and one different value, then you would reroll the one die that is different and you could get This will make the diagram make more sense.) Diagram would be improved if you highlight the branches that do lead to yahtzee. Nice layout with the background highlight for roll number.
7 Formatted: Font: Bold 1st Roll 2nd Roll 3rd Roll Formatted: Font: Bold 3 different 3 different 3 Different Start The possible series of events of which result in a can be described as follows, and summed together to get the overall probability of being achieved: (would be nice if order of terms in the equation followed order of diagram, reading from top of diagram to bottom) Though the possible outcomes of each throw are dependent on the previous throw, each throw is independent; (this needs rewording, it is a bit confusing. I think you need to say that the probabilities of outcomes on rerolls are indept of previous rolls, i.e. if you reroll two dice the prob of any outcomes is just like rolling a pair of dice from scratch ) this makes calculating the probabilities of the above events the product of the respective probabilities of each event in the chain occurring.though the possible results of the second and third throws are dependent on what has occurred on the previous throws, the
8 probabilities of these results occurring on the rerolls are independent of the previous rolls. That is to say, any time a die or dice are re rolled, the probabilities associated with the outcome of that particular throw are no different than if they were rolled at any other time. On the first roll, there are three possibilities: a, a two of a kind match, or all three dice are unique. The probability of getting a on the first throw is equal to the probability that all three dice share the same value, times the six possible values which the dice can take: The second possible outcome on the first throw is the event of getting two of a kind. This probability is equal to the probability that two dice show the same number while the third shows a different number, times the six possible numbers which can make up the two of a kind, times the 3 different ways one can choose two dice of the three available to make up the two of a kind: (Try to simplify this explanation, it can be made clearer. For example, this prob is equal to the probability of a pair on the first two dice, 6/36, times the prob of a different value on the third die, 5/6. However, we also need to account for the fact that the pair does not need to occur on dice 1 and 2 and could also occur on dice 1 and 3 or dice 2 and 3. Thus, there are 3 choose 2=3 ways we can select which of the 3 dice show the pair. The second possible outcome on the first throw is the event of getting two of a kind. This probability is equal to the probability that two dice show the same number, 6 in 36, while the third shows a different number, 5 in 6. However, we also need to take into account the matching pair can occur on any two of the three dice. The pair can occur on dice 1 and 2, dice 2 and 3, or dice 1 and 3 for a possible 3 ways to make the pair. Taking this into account, we can substitute in the probabilities previously calculated, and solve for the probability of getting two of a kind on the first roll: Formatted: Left (Chris, I think it is better to group the prob calcs. Thus, this part should go after prob of two of a kind on first throw, then you ll have another section after 3 unique values on first roll for yahtzee outcomes that correspond that that first roll)if a two of a kind is thrown on the first throw, then on the second throw, with one die remaining, there is a 1 in 6 chance for a, and a 5 in 6 chance that the die cast will not match. If the remaining die does not match, then again, on the final throw, there is a 1 in 6 chance for a.,, 1 6
9 , 5 6 The third possible outcome of the first throw is event that all three dice show unique values. In order to obtain an outcome with 3 distinct values, the first die can show 6 possible values, the second can show 5 possible values, and the third can show four possible values. In turn, there are 6, or 216 different ways the three dice can be arranged. In order to obtain an outcome with 3 distinct values, there are 6 choices for the first die, then only 5 choices of a different number for the second die This probability is equal to product of the number of ways each dice can take a unique value relative to the other dice over the number of possible ways to roll three dice: The probability of getting three unique values on the first roll is equal to the number of outcomes of which all the dice show unique values against the total number of possible outcomes: (Chris, I think it is better to group the prob calcs. Thus, this part should go after prob of two of a kind on first throw, then you ll have another section after 3 unique values on first roll for yahtzee outcomes that correspond that that first roll)if a two of a kind is thrown on the first throw, then on the second throw, with one die remaining, there is a 1 in 6 chance for a, and a 5 in 6 chance that the die cast will not match. If the remaining die does not match, then again, on the final throw, there is a 1 in 6 chance for a.,, 1 6, 5 6 If all three values are unique on the first throw, then if all three dice are re thrown the possible outcomes for the second throw are no different than the first: a, two of a kind, or three unique values will occur from the re casting of the dice. From this, we can calculate the probability: (Chris, relate this to your equation under the diagram, even rewrite it here for clarity)
10 Formatted: Font: +Body, Not Italic Formatted: Font: +Body Formatted: Font: +Body = Therefore, there is a 21.17% chance of getting a with three dice and two re throws.
11 Works Cited [1] M. H. DeGroot and M. J. Schervish, Probability and Statistics, Fourth Edition, Boston: Addison Wesley, [2] NBA, "NBA: Year by Year Lottery Probabilities," [Online]. Available: [Accessed 6 January 2012]. Formatted: Font:
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