# How Much Can I Expect To Lose In Pinball?

This weekend, all going well, I’ll be going to the Michigan state pinball championship contest. There, I will lose in the first round.

I’m not trying to run myself down. But I know who I’m scheduled to play in the first round, and she’s quite a good player. She’s the state’s highest-ranked woman playing competitive pinball. So she starts off being better than me. And then the venue is one she gets to play in more than I do. Pinball, a physical thing, is idiosyncratic. The reflexes you build practicing on one table can betray you on a strange machine. She’s had more chance to practice on the games we have and that pretty well settles the question. I’m still showing up, of course, and doing my best. Stranger things have happened than my winning a game. But I’m going in with I hope realistic expectations.

That bit about having realistic expectations, though, makes me ask what are realistic expectations. The first round is a best-of-seven match. How many games should I expect to win? And that becomes a probability question. It’s a great question to learn on, too. Our match is straightforward to model: we play up to seven times. Each time we play one or the other wins.

So we can start calculating. There’s some probability I have of winning any particular game. Call that number ‘p’. It’s at least zero (I’m not sure to lose) but it’s less than one (I’m not sure to win). Let’s suppose the probability of my winning never changes over the course of seven games. I will come back to the card I palmed there. If we’re playing 7 games, and I have a chance ‘p’ of winning any one of them, then the number of games I can expect to win is 7 times ‘p’. This is the number of wins you might expect if you were called on in class and had no idea and bluffed the first thing that came to mind. Sometimes that works.

7 times p isn’t very enlightening. What number is ‘p’, after all? And I don’t know exactly. The International Flipper Pinball Association tracks how many times I’ve finished a tournament or league above her and vice-versa. We’ve played in 54 recorded events together, and I’ve won 23 and lost 29 of them. (We’ve tied twice.) But that isn’t all head-to-head play. It counts matches where I’m beaten by someone she goes on to beat as her beating me, and vice-versa. And it includes a lot of playing not at the venue. I lack statistics and must go with my feelings. I’d estimate my chance of beating her at about one in three. Let’s say ‘p’ is 1/3 until we get evidence to the contrary. It is “Flipper Pinball” because the earliest pinball machines had no flippers. You plunged the ball into play and nudged the machine a little to keep it going somewhere you wanted. (The game Simpsons Pinball Party has a moment where Grampa Simpson says, “back in my day we didn’t have flippers”. It’s the best kind of joke, the one that is factually correct.)

Seven times one-third is not a difficult problem. It comes out to two and a third, raising the question of how one wins one-third of a pinball game. Most games involve playing three rounds, called balls, is the obvious observation. But this one-third of a game is an average. Imagine the two of us playing three thousand seven-game matches, without either of us getting the least bit better or worse or collapsing of exhaustion. I would expect to win seven thousand of the games, or two and a third games per seven-game match.

Ah, but … that’s too high. I would expect to win two and a third games out of seven. But we probably won’t play seven. We’ll stop when she or I gets to four wins. This makes the problem hard. Hard is the wrong word. It makes the problem tedious. At least it threatens to. Things will get easy enough, but we have to go through some difficult parts first.

There are eight different ways that our best-of-seven match can end. She can win in four games. I can win in four games. She can win in five games. I can win in five games. She can win in six games. I can win in six games. She can win in seven games. I can win in seven games. There is some chance of each of those eight outcomes happening. And exactly one of those will happen; it’s not possible that she’ll win in four games and in five games, unless we lose track of how many games we’d played. They give us index cards to write results down. We won’t lose track.

It’s easy to calculate the probability that I win in four games, if the chance of my winning a game is the number ‘p’. The probability is p4. Similarly it’s easy to calculate the probability that she wins in four games. If I have the chance ‘p’ of winning, then she has the chance ‘1 – p’ of winning. So her probability of winning in four games is (1 – p)4.

The probability of my winning in five games is more tedious to work out. It’s going to be p4 times (1 – p) times 4. The 4 here is the number of different ways that she can win one of the first four games. Turns out there’s four ways to do that. She could win the first game, or the second, or the third, or the fourth. And in the same way the probability she wins in five games is p times (1 – p)4 times 4.

The probability of my winning in six games is going to be p4 times (1 – p)2 times 10. There are ten ways to scatter four wins by her among the first five games. The probability of her winning in six games is the strikingly parallel p2 times (1 – p)4 times 10.

The probability of my winning in seven games is going to be p4 times (1 – p)3 times 20, because there are 20 ways to scatter three wins among the first six games. And the probability of her winning in seven games is p3 times (1 – p)4 times 20.

Add all those probabilities up, no matter what ‘p’ is, and you should get 1. Exactly one of those four outcomes has to happen. And we can work out the probability that the series will end after four games: it’s the chance she wins in four games plus the chance I win in four games. The probability that the series goes to five games is the probability that she wins in five games plus the probability that I win in five games. And so on for six and for seven games.

So that’s neat. We can figure out the probability of the match ending after four games, after five, after six, or after seven. And from that we can figure out the expected length of the match. This is the expectation value. Take the product of ‘4’ and the chance the match ends at four games. Take the product of ‘5’ and the chance the match ends at five games. Take the product of ‘6’ and the chance the match ends at six games. Take the product of ‘7’ and the chance the match ends at seven games. Add all those up. That’ll be, wonder of wonders, the number of games a match like this can be expected to run.

Now it’s a matter of adding together all these combinations of all these different outcomes and you know what? I’m not doing that. I don’t know what the chance is I’d do all this arithmetic correctly is, but I know there’s no chance I’d do all this arithmetic correctly. This is the stuff we pirate Mathematica to do. (Mathematica is supernaturally good at working out mathematical expressions. A personal license costs all the money you will ever have in your life plus ten percent, which it will calculate for you.)

Happily I won’t have to work it out. A person appearing to be a high school teacher named B Kiggins has worked it out already. Kiggins put it and a bunch of other interesting worksheets on the web. (Look for the Voronoi Diagramas!)

There’s a lot of arithmetic involved. But it all simplifies out, somehow. Per Kiggins’ work, the expected number of games in a best-of-seven match, if one of the competitors has the chance ‘p’ of winning any given game, is:

$E(p) = 4 + 4\cdot p + 4\cdot p^2 + 4\cdot p^3 - 52\cdot p^4 + 60\cdot p^5 - 20\cdot p^6$

Whatever you want to say about that, it’s a polynomial. And it’s easy enough to evaluate it, especially if you let the computer evaluate it. Oh, I would say it seems like a shame all those coefficients of ‘4’ drop off and we get weird numbers like ’52’ after that. But there’s something beautiful in there being four 4’s, isn’t there? Good enough.

So. If the chance of my winning a game, ‘p’, is one-third, then we’d expect the series to go 5.5 games. This accords well with my intuition. I thought I would be likely to win one game. Winning two would be a moral victory akin to championship.

Let me go back to my palmed card. This whole analysis is based on the idea that I have some fixed probability of winning and that it isn’t going to change from one game to the next. If the probability of winning is entirely based on my and my opponents’ abilities this is fair enough. Neither of us is likely to get significantly more or less skilled over the course of even seven matches. We won’t even play long enough to get fatigued. But ability isn’t everything.

But our abilities aren’t everything. We’re going to be playing up to seven different tables. How each table reacts to our play is going to vary. Some tables may treat me better, some tables my opponent. Luck of the draw. And there’s an important psychological component. It’s easy to get thrown and to let a bad ball wreck the rest of one’s game. It’s hard to resist feeling nervous if you go into the last ball from way behind your opponent. And it seems as if a pinball knows you’re nervous and races out of play to help you calm down. (The best pinball players tend to have outstanding last balls, though. They don’t get rattled. And they spend the first several balls building up to high-value shots they can collect later on.) And there will be freak events. Last weekend I was saved from elimination in a tournament by the pinball machine spontaneously resetting. We had to replay the game. I did well in the tournament, but it was the freak event that kept me from being knocked out in the first round.

That’s some complicated stuff to fit together. I suppose with enough data we could possibly model how much the differences between pinball machines affects the outcome. That’s what sabermetrics is all about. Representing how severely I’ll build a little bad luck into a lot of bad luck? Oh, that’s hard.

Too hard to deal with, at least not without much more sports psychology and modelling of pinball players than we have data to do. The supposition that my chance of winning is fixed for the duration of the match may not be true. But we won’t be playing enough games to be able to tell the difference. The assumption that my chance of winning doesn’t change over the course of the match may be false. But it’s near enough, and it gets us some useful information. We have to know not to demand too much precision from our model.

And seven games isn’t statistically significant. Not when players are as closely matched as we are. I could be worse and still get a couple wins in when they count; I could play better than my average and still get creamed four games straight. I’ll be trying my best, of course. But I expect my best is one or two wins, then getting to the snack room and waiting for the side tournament to start. Shall let you know if something interesting happens.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

## 8 thoughts on “How Much Can I Expect To Lose In Pinball?”

1. Best of luck! I am loving these pinball posts! And there’s a pinball place in Alameda, CA that you’ve just inspired me to visit again.

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1. Thank you! I’m sorry I don’t find more excuses to write about pinball, since there’s so much about it I do like. And I’m glad you’re feeling inspired; I hope it’s a good visit.

The secrets are: plunge the ball softly, let the ball bounce back and forth on the flippers until it’s moving slowly, and hold the flipper up until the ball comes to a rest so you can aim. So much of pinball is about letting things calm down so you can understand what’s going on and what you want to do next.

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1. I honestly am not. The occasional lottery ticket is my limit. But probability questions are so hard to resist. They usually involve very little calculation but demand thoughtful analysis. It’s great.

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