Suppose you roll n dice.
In terms of n, what is the probability that the sum will be even? (n is a positive integer.) Or, if you don't speak math, roll a bunch of dice. Add up the results. What is the chance that number will be even?
This problem, slightly paraphrased, appeared on the Sprint round for the MathLeague* event I attended this weekend. The Sprint round was my least favorite, since the problems weren't all that interesting. Many of them were math at its worst**: if you understand a certain technique, you apply it in a straightforward manner, and you get the answer. Otherwise, you just can't solve the problem.
However, this problem stood out. After I stared at it for a little while, I figured it out. It's one of those problems that is very easy to solve provided you have the correct insight, but you can labor forever in vain. If you can't figure it out, the answer is in this footnote.*** (Although nothing will match the aha! you get from figuring it out yourself.)
Too often life is dominated by echoing something you have been trained to do. As we wake up, we yawn–an echo. We might stand up–an echo–and walk–an echo–towards the bathroom–an echo. In school, we echo the techniques and facts our teachers tell us, which they themselves are echoing. As we talk with other people, our speech patterns echo those of our mentors, which is to say, everyone. Even when I felt that aha! moment in MathLeague, it was still a form of echoing. I may have made an independent discovery, but someone else had found the answer before I did. I was also using the tools that someone else pioneered.
I can't speak for anyone else, but my greatest fear is living a life of echoes. That's essentially why you're reading this blog. Which still contains a fair amount of echoing, for sure. (And those are just the ones that are heavy echo.) I'm scared of living obediently by the techniques I've learned. I have a need to, at some point, make something new, that no one has ever found before. So far, the only time I've made something on that scale is probably the Four Fours problem, which will be its own blog post sometime.
In a nutshell: the Four Fours problem is a problem in math where you have to represent as many integers as possible with just four 4s and any amount of mathematical operations. I discovered a conclusive solution requiring only two 4s**** (or 3s or 5s, etc.) involving multifactorials that I won't detail here for space reasons (and since it would make a good post on its own.) However, even that is still an echo in some form, since I'm using someone else's techniques, notation, etc.
I hope that one day I can make something with no echo component, but it might be too late in the world for that. We can always hope.
Thanks for reading what I wrote. Maybe someday we will escape the echoes.
–beautifulthorns
Originally aired April 17, 2019
*MathLeague is essentially a worse-funded, worse-quality, worse-orchestrated, and worse-publicized version of MathCounts. It has five different rounds: Sprint, Target, Team, Relay, and Power. The Sprint round is the simplest: you get a sheet with 30 math problems and you have 1 hour to complete as many as possible.
**I'm a proud math nerd. Among the stereotypes about math nerds is that they like tedious problems of the sort found in school curriculum where you, say, multiply numbers with many digits. Real math nerds don't do that. In fact, my next math post is probably coming next week, and it's about a voting game which involves green vegetables. I'll explain why it is mathematical, so don't worry. And if you don't speak math, also don't worry. There's no tedious multiplication of numbers with many digits, or anything of the sort, involved.
***Hint: odd+odd=even, odd+even=odd, even+odd=odd, even+even=even.
The probability is 1/2, no matter the value of n. Think about it this way: imagine you save the last die for last, and roll the rest. The sum is either odd or even, but no matter which one it is, the last die will decide it. If it's odd, the last die must be odd, which has a 1/2 chance of happening. If it's even, the last die must be even, which also has a 1/2 chance of happening. So in all cases, the answer is 1/2.
****It's easy to burn any number, say n, by replacing an instance of n with √n*√n.