Math team season is now over in my state, and other than the seniors' only state event coming up next month and Mu Alpha Theta nationals during the summer, competition for this year is over. But nationals are important and I still need to prepare, so I figured: hey! Why not actually put the blog I made to good use? Coach always says that teaching others helps to reinforce the ideas, so this should help me out, right?
Thus, this first post is born.
Oh, and for anyone reading an planning on subscribing because you think I'm going to teach stuff that'll help you do your homework: find another website. I can guarantee that my posts will be 50% math comics or something related, 25% funny math stories, 20% math that's kind of awesome but not really anything to help you in class, and 5% anything that will actually have relevance in pre-calculus or whatever the hell you're taking.
First mini-lesson in probability: You've got a 1/20 chance that this will be helpful for class, unless your teacher is actually doing a lesson focused on dice games and how to design them so you can take people's money. If you do have a teacher like that, then I just have to say you're one lucky S.O.B.
So come along with me on this magical journey as I teach you all how to figure out your odds at a basic dice game. I'll cover a couple of examples and show you how to design one that's in your favor.
Let's begin with an example, shall we?
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How does this deal sound right about now? It certainly seems like the odds could be in your favor, right? After all, $8 is over twice as much as $3. But don't just be stupid and just jump in. A good gambler needs to be a good mathematician.
The first step is to dissect problem to get the information we want. What are you looking for? Well, to figure out if this is a good idea or not, we need to figure out whether you or your friend has the higher chance of making money. In other words, we're looking for the expected win/loss per play.
5 Steps to find "expected win/loss per play" (for this problem):
• 1. First, figure out all of the possible combinations to get a winning number, in this case, a multiple of four. Since the maximum you can roll is 12 (6+6), that means the only winning rolls will have to add up to 4, 8, and 12.
• 2. Next, let's find all the possible combinations. (NOTE: Remember to treat the each dice roll as separate events, since the outcome of one roll doesn't directly depend on the other)
If your first roll is a 1, your next roll would need to be a 3 to equal 4.
If your first roll is a 2, your next roll would need to be a 2 to equal 4, or a 6 to equal 8.
If your first roll is a 3, your next roll would need to be a 1 to equal 4, or a 5 to equal 8.
If your first roll is a 4, your next roll would need to be a 4 to equal 8.
If your first roll is a 5, your next roll would need to be a 3 to equal 8.
If your first roll is a 6, your next roll would need to be a 2 to equal 8, or a 6 to equal 12.
These are the only possibilities. There are 9 of them. Remember that number.
• 3. Now find the total amount of all possible combinations. If you learned the basics of combinations and permutations in school, which you should have at one point, then you should know that to find the total of all possible combinations from two unrelated events (without any repeating elements within an event) you simply multiply the total possibilities from each event together. There are six sides on each of the dice.
6 x 6 = 36
If you didn't know that and you've at least made it to Alg II, then may the math gods smite both you and your Alg teachers because you should have damn well covered it by now. Hell, I learned this basic crap in elementary school.
...
Sorry, thinking about the school system not doing it's job makes me kinda swear-y.
• 4. Your winning chance is the number of winning possibilities over the number of total possibilities. In other words, 9 over 36, or, to simplify, 1/4.
So there is a 1/4 chance you will win $8. Since there are not other possibilities except losing $3, then you have a 3/4 chance of losing $3.
1/4 + 3/4 = 1
25% + 75% = 100%
There's no way to make my logic any clearer than this.
(Oh, and make sure that all your fractions have the same denominator for the next step.)
• 5. So now you know the chances for each. Next, to find the expected win/loss per play, multiply the numerator of each fraction to the corresponding money value (using negative numbers to represent the loss), add them together, and divide by the denominator.
( 1 x $8 ) + ( 3 x -$3 ) = $8 + (-$9) = -$1
-$1 / 4 = -$0.25
In other words, if you play this game with your friend, over time you will average to losing 25 cents per round.
• Answer 1: Tell your friend to piss off because you're not an idiot. If he were really your friend he wouldn't try to take your money, or at least he'd check his damn math first so that the game would be fair for the two of you, because you are going to LOSE 25 cents per round!
Friends don't let friends do math incorrectly.
Speaking of which, I will be cutting off this lesson for now, but in part two I will be covering how to design a dice game so that you can make it as fair (or unfair) as you'd like.
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Less than three,
The Artist
