AP Statistics Curriculum 2007 Prob Simul
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| Outcome = Number of matching dice || None=loss || 1 || 2 || 3 | | Outcome = Number of matching dice || None=loss || 1 || 2 || 3 | ||
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- | | | + | | W (profit/payoff) || -1 || 1 || 2 || 3 |
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- | | P( | + | | P(W=w) || \( \left( \frac{5}{6} \right)^3 \) || \(\binom{3}{1} \left( \frac{5}{6} \right)^2 \frac{1}{6}\) || \(\binom{3}{2}\frac{5}{6} \left( \frac{1}{6} \right)^2 \) || \( \left(\frac{1}{6}\right)^3 \) |
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- | | | + | | w*P(W=w) || (-1)x0.5787 || 0.34722 || 2x0.06944 || 3x0.00463 |
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</center> | </center> | ||
- | Thus, if | + | Thus, if W is the [[AP_Statistics_Curriculum_2007_Distrib_RV |random variable]] denoting the return of this game (player's profit), its expected value is: |
- | : \( E[ | + | : \( E[W]=\sum_w{wP(W=w)} = -0.0787\). |
Similarly, we can compute the [[AP_Statistics_Curriculum_2007_Distrib_MeanVar#Variance|variance]] and [[AP_Statistics_Curriculum_2007_Distrib_MeanVar#Standard_Deviation|standard deviation]] of the outcome/payoff variable X: | Similarly, we can compute the [[AP_Statistics_Curriculum_2007_Distrib_MeanVar#Variance|variance]] and [[AP_Statistics_Curriculum_2007_Distrib_MeanVar#Standard_Deviation|standard deviation]] of the outcome/payoff variable X: | ||
- | : \( Var[ | + | : \( Var[W]= 1.23917\) and \( SD[W]= 1.11318 = \sqrt{Var[W]}\). |
: '''Question''': What is your interpretation of this quantitative analysis of the [http://socr.ucla.edu/htmls/exp/Chuck_A_Luck_Experiment.html Chuck A Luck Experiment]? Would you agree to play (or host) this game? | : '''Question''': What is your interpretation of this quantitative analysis of the [http://socr.ucla.edu/htmls/exp/Chuck_A_Luck_Experiment.html Chuck A Luck Experiment]? Would you agree to play (or host) this game? |
Revision as of 01:44, 17 April 2013
Contents |
General Advance-Placement (AP) Statistics Curriculum - Probability Theory Through Simulation
Motivation
Many practical examples require probability computations of complex events. Such calculations may be carried out exactly, using the proper probability rules, or approximately using estimation or simulations.
A very simple example is the case of trying to estimate the area of a region, A, embedded in a square of size 1. The area of the region depends on the demarcation of its boundary, as a simple closed curve shown on the picture. This problem relates to the problem of computing the probability of the event A as a subset of the sample-space S - square of size 1. In other words, if we were to throw a dart at the square, S, what would be the chance that the dart lands inside A, under certain conditions (e.g., the dart must land in S and each location of S is equally likely to be hit by the dart)?
This problem may be solved exactly by using integration, but an easier approximate solution would be throwing 100 darts at the board and recording the proportion of darts that landed inside A. This proportion will be a good simulation-based approximation to the real size (or probability) of the set (or event) A. For the instance of throwing 15 darts and having 7 land inside of A, the simulation-based estimate of the area (or probability) of A is .
SOCR simulations
There is a large number of SOCR Simulations that may be used to compute (approximately) probabilities of various processes and compare these empirical probabilities to their exact counterparts.
Ball and Urn Experiment
This experiment allows sampling with or without replacement from a virtual urn that has one of two types of balls - red (successes) and green (failures). The user may control the total number of balls in the urn (N), the number of red balls (R) and the number of balls sampled from the urn (n). Depending on whether we sample with or without replacement, the chance (or probability) of getting m red balls (successes) in a sample of n balls changes. Note that when sampling without replacement, , and when sampling with replacement, . The applet records numerically the empirical outcomes and compares these to the theoretical expected counts using distribution tables and graphs. Suppose we set N=50, R=25 and n=10. How many successes (red ball) do we expect to get in the sample of 10 balls, if we sample without replacement? Hypergeometric distribution provides the theoretical model for experiment and allows us to compute this probability exactly. We can also run the experiment once and approximate this answer by dividing the number of observed red balls by 10 (the sample size). You can try this experiment to gauge the accuracy of the simulation-based approximations relative to various setting, e.g., sample size. According to the Law of Large Numbers, the accuracy of the estimation will rapidly increase with the increase of the sample-size (n).
Binomial Coin Toss Experiment
This experiment allows tossing of n independent coins, each with the probability of heads p. This may be a perfect model for any experiment that involves observing independent dichotomous measurement (e.g., success/failure, +/-, pro/con, up/down, presence/absence, etc.) where all measurements have the same fixed chance of success. Just like with the Ball and Urn experiment above, we can carry a simulation and estimate the probability of the coin being tosses by running n trials and computing the quotient of (number of successes)/n. Imagine that some one provided you only with the n outcomes (a sequence of Head and Tails), you will be able to estimate the P(H) of the process that generated these data. Using the graph and numerical table included in this applet, we can also compare the theoretical binomial probabilities against their empirical estimates.
Card Experiment
This experiment is more involved because the sample space is significantly larger. It demonstrates the basic properties of dealing n cards at random from a standard deck of 52 cards. At every trial, n cards are randomly drawn and their denomination and suit are recorded in the result table below. We can use this simulation to estimate the probabilities of various hands (e.g., the odds of getting a pair of cards with the same denomination). Run the experiment 100 times and count the number of 5-card hands that had at least one pair in them (at least one pair of cars in the 5-card had had a matching denomination, i.e., Y_{i} = Y_{j}, for ). Dividing this number by 100 gives the simulation-based estimate of the probability of the complex event of interest (at least one pair).
Roulette Experiment
The Roulette experiment presents another interesting example where we can draw direct synergies between exact theoretical probability and approximate simulation probability calculations. The Roulette wheel has 38 slots numbered 00, 0, and 1-36. Slots 00 and 0 are green. Half of the slots numbered 1-36 are red and half are black. Suppose we are interested in the odds of winning (i.e., the probability of the event) if we bet on a number between 1-18 turns up. Obviously, the chance of winning is given by the fraction 18/38 (which is less than 50%). However, we can run this experiment 20 times and the empirical number of time we win could be different each time, yet close to the theoretical value of 18/38. In the image below we won 10 out of the 20 trials and therefore, the empirical odds are 50-50.
Chuck A Luck Experiment
The Chuck A Luck Experiment shows how to compute the expectation of a real game, which indicated the odds of a player to win or loose money. In this game, a player/gambler chooses an integer from 1 to 6. Then three fair dice are rolled. If exactly k dice show the gambler's number, the payoff is k:1. The random variable W represents the gambler's net profit on a $1 bet.
Let's first look at the outcomes of each experiment and compute the corresponding outcome probability values. Then, we can use the expectation formula to compute the expected return (expected value) for this experiment. A zero, positive or negative expected return indicates if the game is fair, or biased for the player or the house, respectively.
The table below identifies the possible outcomes of this experiment and shows how to use the Binomial Distribution to compute their corresponding probabilities. Note that the probability of a single die turning up and matching (or not matching, the complementary event) the player betting number is \(\frac{1}{6}\) (or \(\frac{5}{6}\)).
Outcome = Number of matching dice | None=loss | 1 | 2 | 3 |
W (profit/payoff) | -1 | 1 | 2 | 3 |
P(W=w) | \( \left( \frac{5}{6} \right)^3 \) | \(\binom{3}{1} \left( \frac{5}{6} \right)^2 \frac{1}{6}\) | \(\binom{3}{2}\frac{5}{6} \left( \frac{1}{6} \right)^2 \) | \( \left(\frac{1}{6}\right)^3 \) |
w*P(W=w) | (-1)x0.5787 | 0.34722 | 2x0.06944 | 3x0.00463 |
Thus, if W is the random variable denoting the return of this game (player's profit), its expected value is:
- \( E[W]=\sum_w{wP(W=w)} = -0.0787\).
Similarly, we can compute the variance and standard deviation of the outcome/payoff variable X:
- \( Var[W]= 1.23917\) and \( SD[W]= 1.11318 = \sqrt{Var[W]}\).
- Question: What is your interpretation of this quantitative analysis of the Chuck A Luck Experiment? Would you agree to play (or host) this game?
Problems
- SOCR Home page: http://www.socr.ucla.edu
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