Video Games and Mathematics

Who says that you can't use anything at school for real things? They're all just #$(* jerks. Here are some problems presented in games that I just could not pass up.

Cure, Pure, and Cabins
In Final Fantasy (NES) I'm about to enter the Swamp Cave and I has 5000 GP to burn. I know I'll need a lot of Pure's and a lot of Cure's since my MP is worthless. I also want to to stock up on my Cabins at this point. Let's see I want three Cures for every two Pure's I have so how many Cure's and Pure's do I need (to keep track of my mindless button pressing shopping)?

GP
5000
Cure
60GP
Pure
75GP

Let C be the number of Cure's I need P be the number of Pure's I need i) 60C + 75P = 5000 ii) 3P = 2C P = 2C/3 ia) 60C + 75(2C/3) = 5000 C(60 + 50) = 5000 110C = 5000 C = 500/11 = 45.4545 iia) P = 1000/33 = 30.3030

Therefore I should buy 45 Cure's and 30 Pure's.

Cure, Pure, and Cabins
In Final Fantasy (NES) I'm about to enter the Swamp Cave and I has 5000 GP to burn. I know I'll need a lot of Pure's and a lot of Cure's since my MP is worthless. I also want to to stock up on my Cabins at this point. Let's see I want three Cures for every two Pure's I have so how many Cure's and Pure's do I need (to keep track of my mindless button pressing shopping)?
GP 5000 Cure 60GP Pure 75GP
Let C be the number of Cure's I need P be the number of Pure's I need i) 60C + 75P = 5000 ii) 3P = 2C P = 2C/3 ia) 60C + 75(2C/3) = 5000 C(60 + 50) = 5000 110C = 5000 C = 500/11 = 45.4545 iia) P = 1000/33 = 30.3030
Therefore I should buy 45 Cure's and 30 Pure's.

Kevin's Throws or Techs
I am about to settle the a plague and disputed question (somewhat) of Seiken Densetsu 3. Kevin of the Beast Kingdom is the only character with two techniques per level, a technical move and a throw. Many wonder whether each are equally likely or is there a bias towards one, or in other words what is the chance that Kevin will perform technique or a throw?
In a random sample of 100 techniques 61 were techniques and the rest throws. What is a the chance of a technique with a 95% confidence interval. (95% confidence means that the actual chance of a technique will 95% of the time be between these two numbers.)

π = p ± φ(95%)* SE_p p = K/n = 61/100 = .61 φ(95%) = 1.96 SE_p = (p(1-p)/n)^.5 = .0488 π = .61 ± 1.96*.0488 .5144 < π < .7056 51% < π < 71%

Using this small sample I can confidently say that the chances of Kevin performing a technique rather than a throw is between 51% and 71%. Had I more sample data I could narrow this down even more. But until then, there it is.
As you look at the evidence you probably have no clue to what it means. Here is a brief summary in laments terms. Ahem. A binomial random variable is a variable that counts whether a two outcomes can happens. For example the number of times heads appears in some amount of coin tosses can be represented with a binomial random variable. Thus the nature of Kevin's techniques is binomial. He either performs a move or a throw. Now if we were to guess the number of times Kevin would perform a technique it would be the average times a technique would show up, but we can't say that right out the bat. It is a random variable and we have a small sample, so to increase our confidence we can say that it is the average give or take a few standard errors. The more standard errors we take, the greater the confidence and vice versus. Thus we can create a confidence interval of average plus or minus X amount of standard errors, where X allows for the correct confidence interval. Well, for binomial the average is the chance of the event occurred which in this case is sixty-one divided by one humdred. The standard error for a binomial projecting to the actual variable (instead of the sample) is the sqaure of the the change times one minus the chance divided by the the number of trials. Now the correct number of standard errors is the number that covers ninety-five percent of the curve of the binomial when graphed. However since binomials approach a normal curve as the sample grows large, and we are estimating for the real value, we can figure that out real easy with a normal curve table. The value that gives 95% confidence on a normal curve is one and ninety-six hundredths. The rest is just plug and chug.

Kevin's Throws or Techs
I am about to settle the a plague and disputed question (somewhat) of Seiken Densetsu 3. Kevin of the Beast Kingdom is the only character with two techniques per level, a technical move and a throw. Many wonder whether each are equally likely or is there a bias towards one, or in other words what is the chance that Kevin will perform technique or a throw?
In a random sample of 100 techniques 61 were techniques and the rest throws. What is a the chance of a technique with a 95% confidence interval. (95% confidence means that the actual chance of a technique will 95% of the time be between these two numbers.)
π = p ± φ(95%)* SE_p p = K/n = 61/100 = .61 φ(95%) = 1.96 SE_p = (p(1-p)/n)^.5 = .0488 π = .61 ± 1.96*.0488 .5144 < π < .7056 51% < π < 71%
Using this small sample I can confidently say that the chances of Kevin performing a technique rather than a throw is between 51% and 71%. Had I more sample data I could narrow this down even more. But until then, there it is.
As you look at the evidence you probably have no clue to what it means. Here is a brief summary in laments terms. Ahem. A binomial random variable is a variable that counts whether a two outcomes can happens. For example the number of times heads appears in some amount of coin tosses can be represented with a binomial random variable. Thus the nature of Kevin's techniques is binomial. He either performs a move or a throw. Now if we were to guess the number of times Kevin would perform a technique it would be the average times a technique would show up, but we can't say that right out the bat. It is a random variable and we have a small sample, so to increase our confidence we can say that it is the average give or take a few standard errors. The more standard errors we take, the greater the confidence and vice versus. Thus we can create a confidence interval of average plus or minus X amount of standard errors, where X allows for the correct confidence interval. Well, for binomial the average is the chance of the event occurred which in this case is sixty-one divided by one humdred. The standard error for a binomial projecting to the actual variable (instead of the sample) is the sqaure of the the change times one minus the chance divided by the the number of trials. Now the correct number of standard errors is the number that covers ninety-five percent of the curve of the binomial when graphed. However since binomials approach a normal curve as the sample grows large, and we are estimating for the real value, we can figure that out real easy with a normal curve table. The value that gives 95% confidence on a normal curve is one and ninety-six hundredths. The rest is just plug and chug.

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