#### An Interview with Math Professor Luke

**BB**: **X** = a player. **Z** = the number of games played. **Y** = the number of games needed to win. Can you write this out in math? Basically, for the best two out of three, three out of five, etc. This is for a two player pinball tournament article in Skill Shot.

**Luke**: Hey skipper! I can do my best. Let’s see here. You want to play an odd number of games, so things can’t end in a tie, so for some number **n**, **Z** **= 2n+1**. (That is, we can let **n** be any whole number we want, and then **Z** will be the **n**-th odd number; if **n=1** we get **Z=3**games, if **n=2** we get **Z=5** games, and so on.) So then it’ll take**Y=n+1** games to win. I introduced **n** as a convenience, but we can easily remove it now. As we see, **Z = 2Y-1**, or equivalently, **Y = (Z+1)/2**. Does that help?

**BB**: AWESOME!!!

**Luke**: I’m glad all those years I spent in grad school weren’t wasted after all! How’s it all going, mister?

**BB**: It’s going great. Working on Skill Shot **16**. So Gordon and I played a **2**-player **100** game pinball tournament. It took about a week, and he won **71**, I won **29**. Afterwards, we played a **10** game tournament, and I won **6**, he won **4**. So **100** games or **10**, we’re still close in skill. How’s your life?

**G2**: Hi Luke. We need another pinball equation for determining a fraction or percentage of a pinball tournament where it is possible to have a tie score. For example: If player **X** and player **Y** played **100**games of pinball, the equation would express the results as percentages. The equation could also be used in a tournament of **8**games or **7** or whatever. We’re thinking about calling the article “Percentages”. It would help people rank themselves.

**Luke**: Hey Gord! Let’s see here. Suppose you play a total of **T** games (**100**, or **8**, or **7**, or whatever) and win **X** of them. Then you’ve won**100*X/T** percent of your games. We can turn any fraction into a percentage this way, just by multiplying by **100: 0.6** or **6/10** is the same as **60%**, etc. Just type **100*X/T** into Google and press enter (replacing **X** and **T** with the actual numbers, of course) and it’ll even do the computation for you. I freaking love Google.

**BB**: Hey Luke, one more question. Do these equations apply to tournaments with more than **2** players?

**Luke**: Oh! Good question! The percentage one, definitely. In a tournament, if there are **T** games played in all, and a certain player wins **X** of them, he’s won **100X/T** percent of all games in the tournament. If a certain player wins **X** games and that same player plays a total of say **D** games (maybe he doesn’t play every single game in the tournament; if there are a total of say **30** games in the tournament but this player only plays in **10** of them, then **T=30** but**D=10**) then he’s won **100X/D** percent of the games he’s (or she’s) personally played.

The formula for the number of games required to win will still work in a multi-player tournament, *if* a single player has to win more than half of all games to win the tournament. If you just have to win more games than any other player it’d become a lot more complicated! I’ll have to think about that one.

**BB**: Anyway, this type of tournament isn’t really all about winning, it’s more of a way to rate your skill against other players. I mean, winning is great, but every game has to be played, even after one player has won the number of games required to beat the other(s). In the **100**game tournament, Gordon and I played all **100** games, long after he had already won the tournament by winning **51**.

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