I was recently introduced to the traditional pub game Shut the Box by a friend. For those of you who aren't familiar, it is a singleplayer game where players bet on who can get the lowest score. Inside of a box, there are nine levers labeled with the numbers 1 through 9. Initially, the levers all start in the up position. The player rolls two standard dice and finds their sum. She then turns down levers whose values add to the same sum, removing those levers from the game. For example, if she rolled a 3 and a 5, then she could turn the levers $$1 \text{ and } 7\\ 1, 3, \text{ and } 4\\ \text{or just }8,$$ (the are other possibilities, as well). The game stops once the player is unable to match levers to the diceroll or all the levers are turned down. A player's score is the number created by reading all the up levers from small to large: if 1, 4, and 7 remain up, then her score is 147. If all the levers are turned down, a score of 0 is awarded. Clearly there is some strategy in the choice of which levers to flip down. In addition to choosing the correct levers, the player needs her rolls to add up to $45$ ($1+2+3+4+5+6+7+8+9$) exactly in order to achieve a perfect 0 score. Imagine a modified with the strategy removed: decompose each lever into the corresponding number of 1levers for a total of fortyfive 1levers. What is the probability that a player will win this modified game of no strategy? The problem can be rephrased as, "what is the probability that, in the course of rolling two standard dice repeatedly, the roller will reach the cumulative sum of exactly 45? Rather than focusing on two dice, we focus on one sixsided die and then halve the result. For $n\in \mathbf{Z}$, let $p_n$ be the probability that, in the course of rolling one sixsided die repeatedly, the cumulative sum of $n$ is achieved. Clearly $p_n=0$ for $n<0$ and $p_0=1$. Since each roll from 1 to 6 is equally likely, we have the recurrence relation $$p_n=\frac{1}{6}p_{n1}+\frac{1}{6}p_{n2}+\frac{1}{6}p_{n3}+\frac{1}{6}p_{n4}+\frac{1}{6}p_{n5}+\frac{1}{6}p_{n6}$$ for $n\ge 1$. Calculation has shown that $p_n$ approaches $0.2857...$ This number is suspiciously close to $\frac{2}{7}$, though I have not proven the result. When two dice are rolled at the same time, then the probability that any number is reached is halved. So as two dice are repeatedly rolled, the probability that any number is reached as a cumulative sum approaches $\frac{1}{7}$. The probability of winning our modified game is approximately $1$ in $7$.
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I started this blog to share my transformation from math nerd to math nerd who loves to share math with young people. I teach high school in Philadelphia, Pennsylvania. Your comments are always welcome. Archives
September 2017
