Margolese 1
Rachel Margolese
Math 7
April 25, 2013
Essay 2
That is Logical
Often, our initial instinct upon hearing a puzzle for the first time is to remain focused on the words of the puzzle—to simply work with what is given. While this strategy works for many puzzles, it can lead to unnecessarily long and tedious processes to reach solutions. Instead, a good strategy to employ when confronted with relatively simple puzzles is to use logical deduction and what is not explicitly stated in the puzzle to arrive at a simple solution.
Take, for example, the puzzle of the hiking monk. Precisely at sunrise, a monk begins his climb up a winding path to a temple at the top of a mountain. He hikes at a variable speed and takes many rests as he ascends the mountain. He finally reaches the temple just before sunset. After spending several days meditating at the temple, he begins his return journey, again starting precisely at sunrise. On the way down he takes fewer breaks and his downward journey takes him much less time than his ascent. Was there a spot on the mountain that the monk pass at precisely the same time of day both going up the mountain and going down?
If using only the bare facts presented in the puzzle, the task seems nigh impossible. How can you figure out where the monk is at each time of day on each journey? You can’t. Consider instead an alternative scenario in which one monk is hiking up the mountain at the same time another monk is hiking down. Obviously, the two monks will pass each other on the path at some point and thus will both pass the same spot on the mountain at the same time of day.
A second simple puzzle that can be solved quite easily by considering alternative scenarios is the puzzle of four hands of cards. After a whole deck of cards is dealt to you and three other players, is the probability greater that you and the player on your left have none of the spades, or is the probability greater that you and the player on your left have all of the spades? This puzzle could be solved by writing out the probabilities of being dealt each card and computing the answer through a series of mathematical equations. Alternatively we could reach the answer more directly by considering the consequences of the scenario in which you and the person on your left hold none of the spades. If it is true that the two of you hold no spades, then the other two players must hold all of the spades. We can then see that the probability that you and the person on your left hold no spades is the same as the probability that the other two players hold all of the spades as the two scenarios are intrinsically linked. Therefore, there is equal probability that you and the player on your left hold all the spades and that you hold no spades.
For both of these relatively simple puzzles, consideration of alternative scenarios not stated in the puzzles themselves provides an easy to reach, and easy to agree with solution. This same method is useful in more challenging puzzles. One of my favorite puzzles for which this method is particularly useful is the puzzle of the set of chess games. A girl, her mother, and her father all love to play chess. One day, the girl asks her father for a raise in her allowance. Her father considers her request and decides that she must earn the raise in allowance by proving her skill at chess. He proposes a set of three games of chess with alternating opponents of himself and her mother. To earn the raise in allowance, she much win two games of chess in a row. As an added bonus, the girl gets to pick the order in which she plays her parents, either mother first then father then mother, or father first then mother then father. The girl considers the proposal. She knows that her father is a stronger chess player than her mother and that she has beaten both of them before, though she wins less often against her father. In which order should she play her parents in order to maximize her chance of winning two games in a row?
The girl has two choices: either she plays two games against a weaker opponent and one game against a stronger opponent, or she plays two games against a stronger opponent and one game against a weaker opponent. The first, more tedious and math intensive solution method, is to use brute force algebra and probability to find the answer. Take PM to be the probability that she wins against her mother and PF to be the probability that she wins against her father. It follows that the probabilities that she loses against her mother and father are (1-PM) and (1-PF) respectively. Also keep in mind that it is given that PF < PM. Then consider the first set of games she could play and all the possible outcomes that result in her earning the raise in allowance. For this set, she opts to play mother, father, mother.
First possible outcome: she wins all of her games. The probability of this outcome is PM x PF x PM = PM2 PF.
Second possible outcome: she wins the first two games and loses the third. The probability of this outcome is PM x PF x (1-PM) = PMPF - PM2 PF.
Third possible outcome: she loses the first games and wins the last two games. The probability of this outcome is (1-PM) x PF x PM = PMPF - PM2 PF.
Then find the overall probability that the girl will earn the raise in allowance if she plays her parents in the order mother, father, mother, by adding all three probabilities: PMPF(2 - PM).
The other set option she has is to play father, mother, father. The probabilities for winning with this setup can be similarly calculated (either she wins all the games or just the first two or just the last) and yield the results of (PF2 PM), (PFPM - PF2 PM), (PFPM - PF2 PM) for winning all games, winning the first two, and winning the last two respectively. Once again, adding up all three probabilities gives the overall probability of the girl earning her raise in allowance, which, for this set, is PMPF(2 - PF).
We know that PF < PM, so it follows that PMPF(2 - PM) is less than PMPF(2 - PF). Therefore, the second set option (playing her parents in the order father, mother, father) gives the girl the greatest probability of winning two games in a row, and thus earning her raise in allowance.
This method of solving the puzzle is effective and provides an easy to follow and complete proof. An alternative way to find the solution simply involves flipping the puzzle on its head and considering logical deductions; no math required! However, as when this method was used to solve the previous two puzzles, this second method does not produce a complete proof; it merely provides a less mathematical way to reach a solution. To begin, there are three ways that the girl can earn her raise in allowance: she wins all the games, she wins the first two games, or she wins the last two games. Now instead of considering the games that the girl must win, consider the games that she can lose. In each scenario, she can only lose up to one game: either the first game or the last game. Once you see that she can only lose a single game and that that game cannot be the middle game, you can deduce that she must win that middle game. Therefore, to maximize her chance of earning her allowance, she needs to maximize her chance of winning the middle game which she can do by playing the middle game against her weaker opponent. Thus, she should play the games in the order father, mother, father, as her mother is her weaker opponent. From considering the games lost instead of considering the two games she must win, the solution to the puzzle becomes much easier to see. Although this is not a complete proof, knowing and understanding this solution creates a jumping off point for creating and understanding a full proof.
Although logical deduction is straightforward for some puzzles, the method can quickly become muddling and ineffective with longer puzzles. The challenge is to gauge at what point the logic and consideration of alternative solutions will become unnecessarily confusing. Say, for example, the father proposed five games of alternating opponents for the girl to play with the stipulation that she must win three in a row. Considering what games the girl can lose, as we did in the previous puzzle, we might deduce that since she must not lose the middle game she should play that game against her mother and thus play her parents in the order mother, father, mother, father, mother. But we might also consider the original scenario and deduce that she should play father first because she needs to beat him at least once, and as he is the stronger opponent she should have as many chances to beat him as she can. Thus using logical deduction we arrive at two opposite solutions.
Now if we compare algebraic probabilities in a complete proof, as done in method one of the previous problem, we can find the actual solution. Consider first the set of father, mother, father, mother, father. There are six combinations for which the girl will win: three ways to win three games, two ways to win four games, and one way to win five games. Those probabilities are PFPMPF(1-PM)(1-PF), (1-PF)PMPFPM(1-PF), (1-PF)(1-PM)PFPMPF, PFPMPFPM(1-PF), (1-PF)PMPFPMPF, and finally PFPMPFPMPF. Add those probabilities together and we get PFPM(2PF - PFPM - 2PF2 + PM - PF2PM). If we do the same for the set of mother, father, mother, father, mother, we obtain PMPF(2PM - PMPF - 2PM2 + PF - PM2PF). Knowing that PF < PM, we can compare these two summations and find that the former (the probability of winning the set beginning with father) is larger. Based upon this we conclude that the girl has a greater chance of earning her raise if she plays her father first. Even though the logical method was ineffective for this puzzle, it was quick to hash out and understand. Had it worked, it would have saved a lot of time and paper space. Logical deduction and consideration of alternative solutions do not work for every puzzle, but it is always worth a few minutes of attempts because, when it does work, it can save a lot of time and headache.