How Much I Did Lose In Pinball

A follow-up for people curious how much I lost at the state pinball championships Saturday: I lost at the state pinball championships Saturday. As I expected I lost in the first round. I did beat my expectations, though. I’d figured I would win one, maybe two games in our best-of-seven contest. As it happened I won three games and I had a fighting chance in game seven.

I’d mentioned in the previous essay about how much contingency there is especially in a short series like this one. My opponent picked the game I expected she would to start out. And she got an awful bounce on the first ball, while I got a very lucky bounce that started multiball on the last. So I won, but not because I was playing better. The seventh game was one that I had figured she might pick if she needed to crush me, and if I had gotten a better bounce on the first ball I’d still have had an uphill struggle. Just less of one.

After the first round I got into a set of three “tie-breaking” rounds, used to sort out which of the sixteen players ranked as number 11 versus number 10. Each of those were a best-of-three series. I did win one series and lost two others, dropping me into 12th place. Over the three series I had four wins and four losses, so I can’t say that I mismatched there.

Where I might have been mismatched is the side tournament. This was a two-hour marathon of playing a lot of games one after the other. I finished with three wins and 13 losses, enough to make me wonder whether I somehow went from competent to incompetent in the hour or so between the main and the side tournament. Of course not, based on a record like that, but — can I prove it?

Meanwhile a friend pointed out The New York Times covering the New York State pinball championship:

ping @nebusj : the competition will be tough! https://t.co/OYdRqStBlZ

— harlem medic emeritu (@harlemedic) February 13, 2017

The article is (at least for now) at https://www.nytimes.com/2017/02/12/nyregion/pinball-state-championship.html. What my friend couldn’t have known, and what shows how networked people are, is that I know one of the people featured in the article, Sean “The Storm” Grant. Well, I knew him, back in college. He was an awesome pinball player even then. And he’s only got more awesome since.

How awesome? Let me give you some background. The International Flipper Pinball Association (IFPA) gives players ranking points. These points are gathered by playing in leagues and tournaments. Each league or tournament has a certain point value. That point value is divided up among the players, in descending order from how they finish. How many points do the events have? That depends on how many people play and what their ranking is. So, yes, how much someone’s IFPA score increases depends on the events they go to, and the events they go to depend on their score. This might sound to you like there’s a differential equation describing all this. You’re close: it’s a difference equation, because these rankings change with the discrete number of events players go to. But there’s an interesting and iterative system at work there.

(Points only expire with time. The system is designed to encourage people to play a lot of things and keep playing them. You can’t lose ranking points by playing, although it might hurt your player-versus-player rating. That’s calculated by a formula I don’t understand at all.)

Anyway, Sean Grant plays in the New York Superleague, a crime-fighting band of pinball players who figured out how to game the IFPA rankings system. They figured out how to turn the large number of people who might visit a Manhattan bar and casually play one or two games into a source of ranking points for the serious players. The IFPA, combatting this scheme, just this week recalculated the Superleague values and the rankings of everyone involved in it. It’s fascinating stuff, in that way a heated debate over an issue you aren’t emotionally invested in can be.

Anyway. Grant is such a skilled player that he lost more points in this nerfing than I have gathered in my whole competitive-pinball-playing career.

So while I knew I’d be knocked out in the first round of the Michigan State Championships I’ll admit I had fantasies of having an impossibly lucky run. In that case, I’d have gone to the nationals and been turned into a pale, silverball-covered paste by people like Grant.

Thanks again for all your good wishes, kind readers. Now we start the long road to the 2017 State Championships, to be held in February of next year. I’m already in 63rd place in the state for the year! (There haven’t been many events for the year yet, and the championship and side tournament haven’t posted their ranking scores yet.)

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:

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.

But How Interesting Is A Real Basketball Tournament?

When I wrote about how interesting the results of a basketball tournament were, and came to the conclusion that it was 63 (and filled in that I meant 63 bits of information), I was careful to say that the outcome of a basketball game between two evenly-matched opponents has an information content of 1 bit. If the game is a foregone conclusion, then the game hasn’t got so much information about it. If the game really is foregone, the information content is 0 bits; you already know what the result will be. If the game is an almost sure thing, there’s very little information to be had from actually seeing the game. An upset might be thrilling to watch, but you would hardly count on that, if you’re being rational. But most games aren’t sure things; we might expect the higher-seed to win, but it’s plausible they don’t. How does that affect how much information there is in the results of a tournament?

Last year, the NCAA College Men’s Basketball tournament inspired me to look up what the outcomes of various types of matches were, and which teams were more likely to win than others. If some person who wrote something for statistics.about.com is correct, based on 27 years of March Madness outcomes, the play between a number one and a number 16 seed is a foregone conclusion — the number one seed always wins — while number two versus number 15 is nearly sure. So while the first round of play will involve 32 games — four regions, each region having eight games — there’ll be something less than 32 bits of information in all these games, since many of them are so predictable.

If we take the results from that statistics.about.com page as accurate and reliable as a way of predicting the outcomes of various-seeded teams, then we can estimate the information content of the first round of play at least.

Here’s how I work it out, anyway:

Contest	Probability the Higher Seed Wins	Information Content of this Outcome
#1 seed vs #16 seed	100%	0 bits
#2 seed vs #15 seed	96%	0.2423 bits
#3 seed vs #14 seed	85%	0.6098 bits
#4 seed vs #13 seed	79%	0.7415 bits
#5 seed vs #12 seed	67%	0.9149 bits
#6 seed vs #11 seed	67%	0.9149 bits
#7 seed vs #10 seed	60%	0.9710 bits
#8 seed vs #9 seed	47%	0.9974 bits

So if the eight contests in a single region were all evenly matched, the information content of that region would be 8 bits. But there’s one sure and one nearly-sure game in there, and there’s only a couple games where the two teams are close to evenly matched. As a result, I make out the information content of a single region to be about 5.392 bits of information. Since there’s four regions, that means the first round of play — the first 32 games — have altogether about 21.567 bits of information.

Warning: I used three digits past the decimal point just because three is a nice comfortable number. Do not by hypnotized into thinking this is a more precise measure than it really is. I don’t know what the precise chance of, say, a number three seed beating a number fourteen seed is; all I know is that in a 27-year sample, it happened the higher-seed won 85 percent of the time, so the chance of the higher-seed winning is probably close to 85 percent. And I only know that if whoever it was wrote this article actually gathered and processed and reported the information correctly. I would not be at all surprised if the first round turned out to have only 21.565 bits of information, or as many as 21.568.

A statistical analysis of the tournaments which I dug up last year indicated that in the last three rounds — the Elite Eight, Final Four, and championship game — the higher- and lower-seeded teams are equally likely to win, and therefore those games have an information content of 1 bit per game. The last three rounds therefore have 7 bits of information total.

Unfortunately, experimental data seems to fall short for the second round — 16 games, where the 32 winners in the first round play, producing the Sweet Sixteen teams — and the third round — 8 games, producing the Elite Eight. If someone’s done a study of how often the higher-seeded team wins I haven’t run across it.

There are six of these games in each of the four regions, for 24 games total. Presumably the higher-seeded is more likely than the lower-seeded to win, but I don’t know how much more probable it is the higher-seed will win. I can come up with some bounds: the 24 games total in the second and third rounds can’t have an information content less than 0 bits, since they’re not all foregone conclusions. The higher-ranked seed won’t win all the time. And they can’t have an information content of more than 24 bits, since that’s how much there would be if the games were perfectly even matches.

So, then: the first round carries about 21.567 bits of information. The second and third rounds carry between 0 and 24 bits. The fourth through sixth rounds (the sixth round is the championship game) carry seven bits. Overall, the 63 games of the tournament carry between 28.567 and 52.567 bits of information. I would expect that many of the second-round and most of the third-round games are pretty close to even matches, so I would expect the higher end of that range to be closer to the true information content.

Let me make the assumption that in this second and third round the higher-seed has roughly a chance of 75 percent of beating the lower seed. That’s a number taken pretty arbitrarily as one that sounds like a plausible but not excessive advantage the higher-seeded teams might have. (It happens it’s close to the average you get of the higher-seed beating the lower-seed in the first round of play, something that I took as confirming my intuition about a plausible advantage the higher seed has.) If, in the second and third rounds, the higher-seed wins 75 percent of the time and the lower-seed 25 percent, then the outcome of each game is about 0.8113 bits of information. Since there are 24 games total in the second and third rounds, that suggests the second and third rounds carry about 19.471 bits of information.

Warning: Again, I went to three digits past the decimal just because three digits looks nice. Given that I do not actually know the chance a higher-seed beats a lower-seed in these rounds, and that I just made up a number that seems plausible you should not be surprised if the actual information content turns out to be 19.468 or even 19.472 bits of information.

Taking all these numbers, though — the first round with its something like 21.567 bits of information; the second and third rounds with something like 19.471 bits; the fourth through sixth rounds with 7 bits — the conclusion is that the win/loss results of the entire 63-game tournament are about 48 bits of information. It’s a bit higher the more unpredictable the games involving the final 32 and the Sweet 16 are; it’s a bit lower the more foregone those conclusions are. But 48 bits sounds like a plausible enough answer to me.