The Dice game of “Pig” can be played with the following rules. Roll two six-side
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Question
The Dice game of “Pig” can be played with the following rules. Roll two six-sided dice. Add the face values together. Choose whether to roll the dice again or pass the dice to your opponent. If you pass, then you get to bank any points earned on your turn. Those points becom permanent. If you roll again, then add your result to your previous score, but you run the risk of losing all points earned since your opponent had rolled. Continue to roll as much as you want. However, once a “1” comes up on either die, your score is reduced to 0, leaving you only with point that you have previously "banked." Furthermore, you must pass the dice to your opponent. The first person to 100 points is the winner. To help with the probabilities and averages in this problem, you may want to write down all the possible outcomes. One way to organize that information would be a chart such as this (notice two example outcomes are given for you): Addition table for the game of pig. Let the event A be a failed roll (rolling a 1) and the event B is a successful roll (not rolling a 1).
Find P(A), rounded to four decimal places. To count the number of ways a failed roll can happen, determine how many outcomes are in the first row or first column. To count the number of total outcomes, count the total number of entries in the table above.
Find P(B), rounded to four decimal places. This should be the complement of the previous question. You can also directly find it by counting the number of outcomes that are NOT in the first row or first column and dividing by the total number of outcomes.
Looking at your table above, what is the mean value of a SUCCESSFUL roll? Add all the values that are not in the first row or first column, then divide by the number of outcomes you just added together. Note, the lowest successful roll is a 4 (2+2). The largest successful roll is a 12 (6+6). The mean should be somewhere between these values. Let's assume the number of points earned for a successful roll is the mean calculated in the previous problem. The number of points earned for a failed roll would be 0, since your score is reduced to zero when you roll a 1. What is the expected value if you were to only roll once? (Round to four decimal places.) Expected Value = P(success) * value(success) + P(failure) * value(failure) P(success) is your answer from #7 value(success) is your answer from #8 value(failure) is zero, so P(failure) * value(failure) is always going to be zero. Let's suppose you decide to roll two times. Considering the multiplication rule, what would be the probability of rolling successfully two times in a row? Round to four decimal places. (Make sure you don’t use your rounded values in calculations. Calculations should always be done using the exact value. We only report a rounded value in the end.) If your strategy was to roll two times, what would be your probabilty of failure? Hint - failure is the complement of success. Round your answer to four decimal places. (Make sure you don’t use your rounded values in calculations. Calculations should always be done using the exact value. We only report a rounded value in the end.) If you were to roll two times successfully, assume that your points earned would be two times the mean for a single roll. If you were to roll two times unsuccessfully, you would get 0 points. Find the expected value if you decided to roll twice. Round to four decimal places. (Make sure you don’t use your rounded values in calculations. Calculations should always be done using the exact value. We only report a rounded value in the end.) Using the logic discussed above, find the expected value for a strategy in which you roll three times before passing. Round to four decimal places. (Make sure you don’t use your rounded values in calculations. Calculations should always be done using the exact value. We only report a rounded value in the end.) Using the logic discussed above, find the expected value for a strategy in which you roll four times before passing. Round your answer to four decimal places. (Make sure you don’t use your rounded values in calculations. Calculations should always be done using the exact value. We only report a rounded value in the end.) Using the logic discussed above, find the expected value for a strategy in which you roll five times before passing. Round your answer to four decimal places. (Make sure you don’t use your rounded values in calculations. Calculations should always be done using the exact value. We only report a rounded value in the end.) Which strategy would you use? Why? Make sure your reasoning is statistical in nature, and not based off of gut feelings.
Explanation / Answer
Let us start by making the table for Dice game of PIG.
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A --> Rolling of 1 (failed event)
To count the number of ways a failed roll can happen, determine how many outcomes are in the first row or first column.
Number of ways we can get event A = 11 (Underlined outcomes of the table)
Number of total outcomes = 36
6) P(A) = 11/36 = 0.3055
B ---> Successful event
7) P(B) = 1 - P(A) = 1 - 0.3055 = 0.6945
8) Mean value of a SUCCESSFUL roll = Sum of successful values of table(BOLD outcomes of the table)/number of successful outcomes
= 200/(36-11) = 200/25 = 8
9) Expected Value = P(success) * value(success) + P(failure) * value(failure)
P(success) is your answer from #7 = P(B) = 0.6945
value(success) is your answer from #8 = Mean Value = 8
value(failure) is zero, so P(failure) * value(failure) is always going to be zero.
Expected Value = 0.6945*8 + 0 = 5.5560
10) P(rolling successfully two times) = P(B)*P(B) = 25/36*25/36 = 0.4822
11) P(failure) = 1- (1296-625)/1296 = 0.5177
12) 2 * mean = 16
Expected Value = P(success) * value(success) + P(failure) * value(failure)
P(success) is your answer from #7 = P(B) = 0.4822
value(success) is your answer from #8 = Mean Value = 16
value(failure) is zero, so P(failure) * value(failure) is always going to be zero.
Expected Value = 0.4822*16 = 7.7152
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