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 got to design a weeklong mathemagic curriculum as part of a youth summer camp. The campers were all elementary school aged. Each day we learned one new trick. At the end of the week, some of our young mathemagicians performed what they learned for the whole camp and the parents. Read all about it here. Image courtesy of Arvind Balaraman / FreeDigitalPhotos.net
When it comes to electing one person from a field of candidates, there are a number of different techniques for counting the votes. Two common approaches are the Condorcet Method and the Borda Count. The Condorcet winner is a candidate who would defeat every other candidate in a series of headtohead elections. The Borda Count asks voters to rank the candidates in order of preference. It then assigns 10 points, say, to a candidate for each firstplace vote, 9 points for every secondplace vote, and so on. The winner is the candidate with the highest point total.
Unfortunately, each system has its downsides. The Condorcet Method will sometimes fail to produce a unique winner. A tiebreak method is needed in those cases. The Borda Count sometimes leads to some counterintuitive results. If there are two favorites in a threecandidate race, the voters' ranking of the "irrelevant" third candidate can determine the outcome of the election. This third candidate acts as a spoiler. Based off of these observations, I wondered what would happen if I used the Condorcet Method as my primary votecounting technique, and used the Borda Count only as my tiebreaker. Could I use this method to always determine a winner and avoid the "spoiler effect" of the Borda Count? Imagine the following situation: three candidates A, B, and C are running in an election. Each voter must rank the candidates from most favored to least favored. In the first scenario, 17 votes are cast: Vote #1 ABC  6 votes BCA  5 votes CAB  6 votes Here, there is a Condorcet tie with A defeating B defeating C defeating A. By the Borda Count, A receives 18, B receives 16, and C receives 17 points. A wins! In a second vote, two voters changed their vote from BCA to CBA. Note that this does not affect A and C's relative rankings. Vote #2 ABC  6 voters BCA  3 voters CAB  6 voters CBA  2 voters Again, there is a Condorcet tie, but in this case, C steals two extra points from B. A receives 18, B receives 14, and C receives 19 points. C Wins! Even the Condorcet/Borda hybrid has its shortcomings, and in fact Arrow's Theorem tells us that there is no perfect votecounting system. If we concede defeat in looking for perfection, we may still seek a best possible votecounting technique. The above example notwithstanding, the paradoxes of the Borda Count are rare, and rarer still when Borda Count is only used as a tiebreaker. 
About Me
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 Hanoi, Vietnam. Your comments are always welcome. Archives
March 2018
