Question

The Dice game of "Pig" can be played with the following rules. 1. Roll two six-sided dice. Add the face values together. 2. Choose whether to roll the dice again or pass the dice to your opponent. 3. If you pass, then you get to bank any points earned on your turn. Those points become 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. 4. 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 points that you have previously "banked." Furthermore, you must pass the dice to your opponent The first person to 100 points is the winner. When a player rolls two dice, the possible outcomes are as follows: Die #2 Roll Roll a 1 Roll a 2 Roll a 3 Roll a 4 Roll a 5 Roll 6 Roll a 1 0 0 0 Roll a 2 6 C 4 7 8 Roll a 3 7 Die #1 Roll Roll a 4 0 8 9 10 Roll a 5 C 10 11 Roll a 6 10 12 11 You can see that a player gets 0 points if they roll a 1 on either die. Otherwise their score comes from adding the results of the two dice. There are 25 ways to "successfully" roll the dice and 11 ways to "unsuccessfully" roll the dice, where "success" is not rolling a 1 on either die. Single Roll When we take the mean of the 25 successful rolls we get 8. We will assume that on a successful roll, the player earns 8 points. 25 P(success)= 36 11 P(failure)= 36 Value(success) = 8 Value(failure) 0 What is the expected value for a single roll? (Round to four decimal places) Two Rolls If we chose to pass the dice, then we would bank those points. If we chose to keep rolling, then in order to be successful we would need to roll no 1's twice. However, we would expect to get twice as many points IF we were to roll successfully. This would give us: 25 P(success)= 2 25 P(failure) 1- 36 Value(success) = 16 Value(failure) = 0 (Round to four decimal places) What is the expected value for rolling twice? More Rolls As you can see, if we chose to roll even more times, then we could calculate the probabilities and values in a similar way. If we were to roll n times, then we have: 25 P(success) = 36 25 P(failure) 1 36 Value(success) 8 - n Value(failure) = 0 What would the expected value be for: (Round to four decimal places) Three rolls? (Round to four decimal places) Four rolls? (Round to four decimal places) Five rolls? Strategy Having looked at the five different strategies, which would you choose? ORoll once before passing ORoll twice before passing ORoll three times before passing ORoll four times before passing ORoll five times before passing Why would you choose this strategy?

Answer 1

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