Problem 1. Yowie
Part 1
The first stage was actually taking all the required physical measurements of all the group participants. They can be found in Tab.1:
The only ratio we were familiar with was the one proposed by the assignment (height : arm span = 1 : 1), so we checked it. Based on our measurements we were able to make a conclusion that this ratio is actual but not precise. In our data, the ratios were: Courtney 167 : 160 = 1.04 : 1, Angela 169 : 170 = 1 : 1.01, Steev 176 : 180 = 1 : 1.02, Danielle 155 : 150 = 1.03 : 1. All the ratios are very close to 1 : 1.
As we found the average scale factors of body measurements to foot length, we could use them to find the presumable body measurements of Yowie based on his foot length. To do this, we multiplied Yowie’s foot length by the average scale factor of specific measurement to foot length. For example, Yowie’s height was found as height=height scale factor×foot length=6.56×43 cm=282.0 cm. All the other calculations were performed the same way. So, the Yowie’s presumable measurements are presented in Tab.3:
The stride length of Yowie can be estimated from his height using the scale factor 0.43 (Science Buddies, 2013), according to the information provided in Scientific American the length of the stride is 0.43 times height. So, stride length=height×0.43=282cm×0.43=121.3 cm.
If the footprints were re-measured and it was found that they are 44 cm long, then this would impact the Yowie’s estimated height. To find the difference that this 1 cm would produce, we multiplied it by the scale factor of height to foot length which we found previously – 1 cm×6.56=6.56 cm. Accordingly, the difference for some other measurements is the following: arm span 6.47 cm, shoulder width 1.64 cm, elbow to shoulder 1.22 cm, head circumference 2.23 cm. As we can see, the difference that this 1 cm makes is not groundbreaking but rather noticeable.
Part 2
Adding weight measurement to the problem can extend the problem. If I know my weight and height, I can find their ratio and assuming Yowie has the same weight to height ratio, his weight can be estimated. My weight is 70 kg and with the height of 176 cm, the ratio is 70 kg:176 cm=0.4 kgcm . For Yowie, the ratio would be weight :282 cm=0.4 kgcm, so weight=0.4 kgcm×282 cm=112.8 kg. Assuming that Yowie is nearly 3 m high, in my opinion, he would weigh more than 112.8 kg. So, I suppose Yowie has a different weight to height ratio than myself. This value can be compared to the average human ratio. The average male height is 174 cm, while average weight is 75 kg. The average ratio will then be 75 kg:174 cm=0.43 kgcm, which is very close to my own ratio.
Part 3
When solving this problem, we almost instantly agreed about using tables and ratios to solve this problem. Making measurements was the essential part of the task because it allowed getting the necessary data. The mathematical concepts and words we used were ratio, scale factor, average, centimeters, addition, division.
Part 1
The first, and by far the easiest, way to solve the problem was to simply fold the actual piece of paper. Prior to doing the folding, our group agreed that no more than 7 folds would be possible. The experiment actually supported our anticipations, as the 7th fold was already pretty hard to make, the 8th fold would be extremely hard if not impossible to make.
If a single sheet has a thickness of 0.1mm and each fold doubles the thickness, then the thickness of the folded paper could be found using guess and check approach. We doubled the thickness 7 times: 0.1 mm×2=0.2 mm;0.2 mm×2=0.4 mm;0.4 mm×2=0.8 mm;0.8 mm×2=1.6 mm;1.6 mm×2=3.2 mm;3.2 mm×2=6.4 mm;6.4 mm×2=12.8 mm.
The last question asked how many times we would need to fold a paper for a paper stack to reach the moon. The distance from the Earth to the Moon varies from 363 104 km to 406 696 km because its orbit is elliptic. The average distance is 384 403 km (Cain, 2013) and we decided to use this number for calculations. As we found out, the height of a paper stack increases very fast with the increase in the number of folds because the relationship between them is exponential. So, we made a guess that folding the paper 20 more times would be enough to cover the distance to the moon. The paper stack folded forty-two times would have the height of height=0.1 mm×242=4.398046511×1011 mm= 4.398046511×1011 mm×1 m1000 mm×1 km1000 m=439804.7 km. Apparently, our guess was right again, as forty-two folds are enough for a paper stack to reach the moon, while 41 folds would not be enough.
Part 2
The extension which can be devised for this problem can elaborate on the reverse task – what would be the size of the paper sheet when it is folded the specific number of times. For example, if we take one of the previous tasks in which the paper had to be folded to reach my height, what would be not the height but the size of such paper stack. The size of the sheet of paper can be described in terms of area. For example, the dimensions of a regular A4 sheet of paper as denoted on Paper Sizes Website (2016) are 210x297 mm. Accordingly, the area of this paper can be found by multiplying its height and width - area=height×width=210 mm×297 mm=62370 mm2. Each time the paper is folded its area is halved. As we can see, the pattern is the same as in finding the height of the paper stack. The difference is that when calculating the height of a paper stack the initial paper thickness is multiplied by 2 in the power of a number of folds, and when the area is calculated. The initial area should be divided by 2 by the power of the number of folds. As we already know, the paper should be folded 15 times to cover my height, so its area would be area=initial area2folds=62370 mm2 215=1.9 mm2. Such paper stack would be indeed tiny in terms of single layer area.
Part 3
In the course of solving this problem we used the concepts of powers, multiplication, division, unit conversions, exponential growth and finding the area. The power is a mathematical operation in which the number is multiplied by itself a certain amount of times. In this problem the power was, the number of folds made. We came to this formula by looking for patterns when examining the thickness of the folded paper. After having identified the formula, we adopted the guess and check strategy to find the specific number of folds for each task. The units we used in this problem and converted into one another were millimeters, centimeters, meters, and kilometers.
Problem 3. Dice
Part 1
The game seems to be not fair from the start. To verify this, we decided to compose a table of all the possible outcomes of the game and calculate the probabilities of either player winning. The rows in the table stand for the score on the first dice, the columns – for the score on the second dice, the values in the cells are the differences between the dice scores.
Then, to obtain the probabilities of either player winning, we counted the number of outcomes with differences of 0, 1, 2 and 3, 4, 5. The total number of outcomes when the difference between the dice is 0, 1, 2 is 24, the number of outcomes when the difference is 3, 4, 5 is 12. To find the probability of each player winning, the number of outcomes, when the player wins, should be divided by the total possible amount of outcomes. So, for the Player A, the theoretical probability would be 2436=23, for Player B the probability would be 1236=13. The percentages of each player winning can be found by multiplying the probabilities by 100%: for Player A percentage=23×100%=66.6%, for Player B percentage=13×100%=33.3%. The experimental percentages can be found the same way. For Player A percentage=69100×100%=69%, for Player B percentage=31100×100%=31%. Both theoretical and experimental percentages show that the game is unfair as the players have different chances of winning. It is important to note that the theoretical and experimental percentages were really close, but still differed.
This game can be made fair if its rules are changed. The first way to make the game fair would be to play with four dice at the same time. And if Player A would get the points only from the first pair of dice and Player B – from both then they would have equal chances. The chances of Player A would not change and still be 66.6 %, while Player B will get his 33.3% chance from each pair of the dice and thus will have 33.3% + 33.3% = 66.6% which equals the chances of Player A.
The game can also be made fair if the initial rules are kept, but in the case of Player B winning, 2 points are awarded instead of one. Player A then has the probability 66.6% to win 1 point, while Player B has 33.3% to win 2 points which makes his/her chances 2×33.3%=66.6% making the game fair.
Another way to make the game fair is to award points to Player A if the different between the dice is 0, 3, 4 and Player B gets points if the difference is 1, 2, 5. The probabilities can be found using the table above. For Player A, the probability is 1836 and for Player B 1836.
This game can also be made fair if a point is awarded to Player A if the difference is 1 or 4, and to Player B if the difference is 2 or 3, and no points are awarded when the difference is 0 or 5. In this case, the probability for Player A is 1436 and for Player B - 1436. Since the chances are equal, the game is fair.
The last way to make it fair is to decide that Player 1 gets the point if the sum of the values on the dice is lower than 7, and Player B gets the point if the sum is greater than 7, if the sum equals 7 both Players get the point. To prove this the table of dice sums can be used:
Counting the probabilities the same way as in the previous task, for Player A, it is 2136, while for Player B it is 2136. As the probabilities are equal, the game is fair.
Part 2
Playing with 4-sided dice can extend this game. The table of differences will then look like:
In this case, the fair game rules would be to award a point to Player A if the difference is 1, and to Player B if the difference is 2 or 3, no points are awarded if the difference is 0. This would make the chances of winning 616 for Player A and 616 for Player B. Thus the game is fair.
Part 3
The strategies used for solving this problem were the table constructing, algebraic calculations, and pattern search. When solving it, we used the mathematical concepts and words such as probabilities, percentages, addition, subtraction and chances. To complete the problem, it was necessary to know how the probabilities and percentages should be calculated.
Reflection
Over the course of completing the three problems above, I have achieved multiple outcomes that I understand to a full potential. The most important aspect that was obtained by all the problems solving is to learn how to apply mathematical concepts and formulas to real-life objects and calculations.
In problem 1, we learned how the proportions, ratios, and scale factors could be used to calculate the unknown values from the known data. In the second problem, we learned about the exponential growth in real-life. We practiced in finding patterns and deriving formulas. In the third problem, we practiced in calculating probabilities and percentages, which is very important for real-life applications such as games and decision-making.
I found deriving five different ways to make the game fair in dice problem the most challenging. We managed to complete it by figuring out the patterns in the probabilities for different game setups using the tables we made up.
If I was doing the similar task in the future I would use tables in solving the folding problem. Recording and calculating values in tables is easier and more compact than doing it in line.
A skill that I still need to improve on is pattern seeking. I am the kind of learner who often makes mistakes in the first attempt but then reflects and corrects the mistakes.
I enjoyed this task because the algebraic calculations had to be combined with actual physical measurements and real-life tasks to complete the problems. I find it very important for our lives, as we have problems, which deal with real, not imaginary, objects, and thus the ability to combine measurements with calculations is really important.
References
A Paper Sizes - A0, A1, A2, A3, A4, A5, A6, A7, A8, A9, A10. (2016). Papersizes.org. Retrieved 10 August 2016, from http://www.papersizes.org/a-paper-sizes.htm
About Sydney Tower. (2016). Sydneytowereye.com.au. Retrieved 10 August 2016, from https://www.sydneytowereye.com.au/explore/about-sydney-tower/
Cain, F. (2013). What is the Distance to the Moon?. Universe Today. Retrieved 10 August 2016, from http://www.universetoday.com/103206/what-is-the-distance-to-the-moon/
Science Buddies. (2013). Stepping Science: Estimating Someone's Height from Their Walk. Scientific American. Retrieved 10 August 2016, from http://www.scientificamerican.com/article/bring-science-home-estimating-height-walk/
