Reiner Knizia on Analysis Paralysis

Game Designer Hall of Fame

contact@recommend.games (Recommend.Games) — Fri, 17 May 2024 12:00:00 +0300

Spiel des Jahres 2024 nominations are just around the corner (watch this space for our annual predictions), but I wanted to take this opportunity for a look back over the history of the award. Specifically, I wanted to see which designers have been most successful at the Spiel, Kennerspiel and Kinderspiel des Jahres awards.

The first red meeple was given out in 1979. Since 1989, the jury also awarded the Kinderspiel des Jahres for children’s games, initially as an annual special award, and from 2001 as its own award with the blue meeple. (In what follows I count those special awards as regular Kinderspiel des Jahres winners.) Finally, in 2011, the Kennerspiel des Jahres was introduced for (slightly) more complex games with an anthracite meeple, again following a couple of special awards with a clearly similar intent, but less steady, so they just stand as special awards on their own in the statistics.

Games

Overall, the jury included 831 games on its various longlist across the three categories over its 45 year history, with (naturally) 45 Spiel, 35 Kinderspiel and 13 Kennerspiel des Jahres winners.¹ This is the overall number of games:

Award	Winner	Nominated	Recommended	Special award
Spiel	45	109	482	11
Kennerspiel	13	39	76	2
Kinderspiel	35	95	260	1

A quick word about the data: As usual, I rely on BoardGameGeek (BGG) for data about the games and their designers. If a game is not in the database or if the designer entry is wrong, the statistics here will necessarily be wrong too. I’ve collected data about the different Spiel des Jahres awards manually over the years, so that’s another potential source of errors. If you’re reading this from the future, I should point out that the statistics here should be complete up to and including Spiel des Jahres 2023.

The current format of having a longlist (which become the recommendations), out of which three games are selected for the shortlist (which are the nominated games), out of which one game will be given the award, has been in place since 2011 with the introduction of the Kennerspiel des Jahres. Formats have varied throughout the decades and I’ve tried to map them to the current situation as much as possible.

I’ve already mentioned that the first 12 Kinderspiel des Jahres winners were really special awards. There was another special award given during the first 18 years between 1979 and 1997 for “beautiful games”. Since this article is meant to focus on designers, not artists, it felt like those games clutter the statistics somewhat, so I treated them as recommendations rather than special award winners.

So without further ado, let’s take a look at the designers.

Designers

Over the course of 45 sessions, the jury has mentioned games by 566 different designers. By far the most successful designer is Wolfgang Kramer who has won Spiel des Jahres five times (four of those were co-designs) and Kinderspiel des Jahres once between 1986 and 2000. During that time period, Klaus Teuber won Spiel des Jahres four times, so during those 15 years Wolfgang & Klaus shared 11 awards between them. The record for most games on the longlists is held by Reiner Knizia with 36 games, which makes it even more surprising that he has “only” won two awards: one each of Spiel des Jahres and Kinderspiel des Jahres – both in 2008 (and both largely forgotten today, unlike his other timeless designs).

Overall, 94 designers have won at least one award (111 if we include special awards), with 21 of them winning more than one (22 including special awards). Only 8 of those have won their awards in more than one category: Wolfgang Kramer, Inka & Markus Brand, Alex Randolph, Reiner Knizia, Bruno Cathala, Antoine Bauza and Steffen Bogen. To this date, no designer has won in all three categories.

Looking at the shortlists (including winners), the jury nominated 217 different designers over the years. Of those, only 22 have been nominated in more than one category and just 4 designers can claim the distinction of being nominated in all three categories: Wolfgang Kramer, Antoine Bauza, Rüdiger Dorn and Wolfgang Warsch.

As mentioned earlier, the jury has recommended games by 566 different designers, with 68 of them appearing in more than one category and 9 in all three. In addition to the 4 mentioned above, those are Inka & Markus Brand, Reiner Knizia, Bruno Cathala and Roberto Fraga.

I’ll leave you with some more detailed statistics below. I hope you enjoy those and they help you bridge the time until the nominations are announced on June 11th. 🤩

Designer records

Overall

Most games on the longlist: 36
- Reiner Knizia
Most games on the shortlist: 13
- Wolfgang Kramer
- Reiner Knizia
Most wins: 6
- Wolfgang Kramer
Most games on the shortlist without win: 5
- Jürgen P. Grunau

Spiel des Jahres

Most games on the longlist: 25
- Reiner Knizia
Most games on the shortlist: 9
- Wolfgang Kramer
Most wins: 5
- Wolfgang Kramer
Most games on the shortlist without win: 5
- Rüdiger Dorn

Kennerspiel des Jahres

Most games on the longlist: 4
Most games on the shortlist: 3
- Stefan Feld
Most wins: 2
Most games on the shortlist without win: 3
- Stefan Feld

Kinderspiel des Jahres

Most games on the longlist: 13
- Kai Haferkamp
Most games on the shortlist: 5
- Reiner Knizia
- Heinz Meister
Most wins (incl special award): 3
- Marie Fort
- Wilfried Fort
Most wins (main award): 2
Most games on the shortlist without win: 3
- Jürgen P. Grunau

Designers in different categories

Longlist (incl shortlist and winners)

9 designers appear in all 3 categories
68 designers appear in at least 2 categories
566 designers appear in at least 1 category

Shortlist (incl winners)

4 designers appear in all 3 categories
22 designers appear in at least 2 categories
217 designers appear in at least 1 category

Winners

0 designers appear in all 3 categories
8 designers appear in at least 2 categories
94 designers appear in at least 1 category (111 incl special awards)

Winners per category

Spiel des Jahres

342 designers had a game on the longlist
105 designers had a game on the shortlist
58 designers won the award (incl special awards)
48 designers won the main award

Kennerspiel des Jahres

95 designers had a game on the longlist
52 designers had a game on the shortlist
20 designers won the award (incl special awards)
14 designers won the main award

Kinderspiel des Jahres

206 designers had a game on the longlist
86 designers had a game on the shortlist
41 designers won the award (incl special awards)
40 designers won the main award

Spiel & Kennerspiel des Jahres

22 designers had a game on both longlists
9 designers had a game on both shortlists
1 designer won both main awards

Spiel & Kinderspiel des Jahres

55 designers had a game on both longlists
15 designers had a game on both shortlists
5 designers won both main awards

Kennerspiel & Kinderspiel des Jahres

9 designers had a game on both longlists
6 designers had a game on both shortlists
2 designers won both main awards

Spiel, Kennerspiel & Kinderspiel des Jahres

9 designers had a game on all longlists
4 designers had a game on all shortlists
0 designers won all main awards

Designer Hall of Fame

This table includes designers with at least 2 wins or 5 nominations across all categories. Each entry shows the number of wins / nominations / recommendations for that designer in each category. Special awards are added in parentheses.

Designer	Spiel	Kennerspiel	Kinderspiel	Total
Wolfgang Kramer	5 / 4 / 12	0 / 1 / 0	1 / 2 / 3	28
Klaus Teuber	4 / 0 / 3	0 / 0 / 0	0 / 1 / 1	9
Michael Kiesling	3 / 3 / 4	0 / 2 / 0	0 / 0 / 0	12
Inka Brand	0 / 0 / 2	2 / 0 / 0	1 / 2 / 0	7
Markus Brand	0 / 0 / 2	2 / 0 / 0	1 / 2 / 0	7
Alex Randolph	1 / 0 / 13	0 / 0 / 0	2 / 1 / 0	17
Reiner Knizia	1 (1) / 6 / 17	0 / 0 / 1	1 / 4 / 5	36
Marie Fort	0 / 0 / 0	0 / 0 / 0	2 (1) / 1 / 0	4
Wilfried Fort	0 / 0 / 0	0 / 0 / 0	2 (1) / 1 / 0	4
Bruno Cathala	1 (1) / 0 / 7	0 / 0 / 1	1 / 0 / 0	11
Heinz Meister	0 / 1 / 2	0 / 0 / 0	2 / 3 / 5	13
Alan R. Moon	2 / 2 / 5	0 / 0 / 0	0 / 0 / 0	9
Manfred Ludwig	0 / 0 / 1	0 / 0 / 0	2 / 2 / 1	6
Antoine Bauza	1 / 0 / 2	1 / 0 / 1	0 / 1 / 0	6
Andreas Seyfarth	2 / 1 / 1	0 / 0 / 0	0 / 0 / 0	4
Klaus Zoch	0 / 1 / 1	0 / 0 / 0	2 / 0 / 0	4
Andreas Pelikan	0 / 1 / 0	2 / 0 / 0	0 / 0 / 0	3
Alexander Pfister	0 / 0 / 0	2 / 0 / 2	0 / 0 / 0	4
Steffen Bogen	1 / 0 / 0	0 / 0 / 0	1 / 0 / 1	3
Peter-Paul Joopen	0 / 0 / 0	0 / 0 / 0	2 / 0 / 1	3
Donald X. Vaccarino	2 / 0 / 0	0 / 0 / 0	0 / 0 / 0	2
Vlaada Chvátil	1 (1) / 0 / 2	0 / 0 / 0	0 / 0 / 0	4
Rüdiger Dorn	0 / 5 / 3	1 / 0 / 0	0 / 1 / 0	10
Wolfgang Warsch	0 / 1 / 0	1 / 1 / 0	0 / 2 / 0	5
Kai Haferkamp	0 / 0 / 0	0 / 0 / 0	1 / 3 / 9	13
Marco Teubner	0 / 0 / 1	0 / 0 / 0	1 / 3 / 6	11
Jens-Peter Schliemann	0 / 0 / 2	0 / 0 / 0	1 / 2 / 3	8
Roberto Fraga	0 / 0 / 1	0 / 0 / 1	1 / 2 / 1	6
Bernhard Weber	0 / 0 / 1	0 / 0 / 0	1 / 1 / 2	5
Michael Schacht	1 / 0 / 8	0 / 0 / 0	0 / 0 / 1	10
Sid Sackson	1 / 0 / 7	0 / 0 / 0	0 / 0 / 0	8
Rudi Hoffmann	1 / 0 / 6	0 / 0 / 0	0 / 0 / 0	7
Günter Burkhardt	0 / 0 / 5	0 / 0 / 0	1 / 0 / 1	7
Dirk Henn	1 / 0 / 5	0 / 0 / 0	0 / 0 / 0	6
Matt Leacock	0 / 3 / 0	0 (1) / 1 / 0	0 / 0 / 0	5
Uwe Rosenberg	0 (1) / 1 / 4	0 / 0 / 2	0 / 0 / 0	8
Jürgen P. Grunau	0 / 2 / 0	0 / 0 / 0	0 / 3 / 0	5
Stefan Dorra	0 / 2 / 6	0 / 0 / 0	0 / 1 / 1	10
Stefan Feld	0 / 0 / 5	0 / 3 / 1	0 / 0 / 0	9
Leo Colovini	0 / 2 / 5	0 / 0 / 0	0 / 1 / 0	8
Wolfgang Dirscherl	0 / 0 / 0	0 / 0 / 0	0 / 2 / 4	6
Reinhard Staupe	0 / 0 / 7	0 / 0 / 0	0 / 1 / 2	10
Friedemann Friese	0 / 1 / 3	0 / 0 / 1	0 / 0 / 0	5
Reinhold Wittig	0 / 0 / 10	0 / 0 / 0	0 / 0 / 1	11
Kris Burm	0 / 0 / 6	0 / 0 / 0	0 / 0 / 0	6
Roland Siegers	0 / 0 / 6	0 / 0 / 0	0 / 0 / 0	6
Haim Shafir	0 / 0 / 0	0 / 0 / 0	0 / 0 / 6	6
Jacques Zeimet	0 / 0 / 5	0 / 0 / 0	0 / 0 / 0	5

Download the full results here: designers and games. As always, you can find the code for this analysis from GitLab.

Arguably, there’s 15 Kennerspiel des Jahres winners: In 2017, the jury gave the award to the first three games in the Exit series. In all statistics, I only count them as a single entry. Similarly, in 2019, the first three entries of the Sherlock series got recommended, which I also count as just a single game. ↩︎

How long is a game of High Society?

contact@recommend.games (Recommend.Games) — Tue, 27 Feb 2024 22:37:01 +0200

High Society is a classic bidding game by classic designer Dr Reiner Knizia, most recently released by Osprey Games with a wonderfully classic look:

The general concept is quite simple: the players are members of said “high society” and are trying to outdo each other in showing off their wealth. The game is played over a series of rounds, and in each round, a card is revealed from a deck of 16 cards. The players then take turns bidding on the card, and the player who wins the bid gets the card. The cards are worth points, and the player with the most points at the end of the game wins.

Here’s the twist though: The player with the least money at the end of the game is disqualified (that poor bastard can no longer be accepted as a member of this exclusive club!), so the players are trying to get the most points while also keeping enough money to stay in the game. This provides agonising decisions every turn – as well as a biting satire of the upper class.

One interesting detail is the way the game ends. There’s a total of 16 cards to be autioned off in the deck; 4 of these cards have a dark background. As soon as the 4th dark card is revealed, the game ends immediately. (That last card isn’t sold anymore.) This looming end adds extra tension to the game as players need to take gambles towards the end of the game and try to guess how many more chances they’ll get to bid.

This is where the maths comes in. While the strategy is way too deep to be easily analysed (and involves some interesting meta in pricing the cards), modelling the game length is pretty straightforward and we’ll do so in a couple of different ways. We’ll start with a simple simulation to understand what’s going on. Then we’ll develop exact formulae, which will culminate in meeting your new best friend: the negative hypergeometric distribution.

Let’s settle one important convention first though: In everything that follows, I’ll only count the cards that are actually auctioned off. The last card is never auctioned off, so it doesn’t count towards the game length. This means that the shortest possible game is 3 rounds long: this happens if all 4 dark cards are on top of the deck (very unlikely). The longest possible game on the other hand is 15 rounds long: this happens if the card at the bottom of the deck is dark and hence the 15 cards above it will be autioned off (actually pretty likely, as we shall see).

Simulation

It’s really simple to simulate this situation. Just create a bunch of (virtual) decks of 16 cards each, colour 4 of them dark, then shuffle them. Now we just need to count where in each of the decks the 4th dark card is. If our sample was big enough, this gives us a really good approximation of the distribution of game lengths.

If you know a little bit of Python and the ubiquitous numpy library, you can do this in just a few lines of code, so I thought I’d walk you through it. If you don’t care about coding, just skip ahead to the results. (Be warned that formulae are coming up next though, so the code might be the relaxing bit.)

import numpy as np

# Set up the parameters
num_cards = 16
num_dark = 4
num_games = 1_000_000
rng = np.random.default_rng(seed=13)

# Create the decks: num_games rows, num_cards columns
games = np.zeros((num_games, num_cards), dtype=bool)
# Colour the last four columns in each row dark
games[:, :num_dark] = True
# Shuffle each row individually
games = rng.permuted(games, axis=1, out=games)

# Count the game lengths: in each row,
# sum the number of dark cards revealed so far.
# Every column where this cumulative sum is
# less than the number of dark cards is a round of the game.
game_lengths = np.sum(np.cumsum(games, axis=1) < num_dark, axis=1)

# Mean and standard deviation of the game lengths
print(game_lengths.mean(), game_lengths.std())

# Histogram the game lengths
unique_lengths, length_counts = np.unique(game_lengths, return_counts=True)
print(dict(zip(unique_lengths, length_counts / num_games)))

Don’t worry if you can’t follow the code immediately, I’ll admit to playing a bit of code golf there. The important part are the results anyways, which should look something like this: the mean game length is 12.6 rounds with a standard deviation of 2.3. This is the full distribution of game lengths:

Length	Probability	Length	Probability
3	0.1%	10	6.6%
4	0.2%	11	9.1%
5	0.5%	12	12.1%
6	1.1%	13	15.8%
7	1.9%	14	20.0%
8	3.1%	15	25.0%
9	4.6%

If you prefer the visual representation, here’s a histogram of the game lengths:

As promised, very short games are extremely rare, whilst long games are in fact the most common. This isn’t necessarily what I would have expected when starting to think about this problem. I think there’s a pretty good intuition for why this is the case, but it’s instructive to examine the distribution from a more theoretic point of view first.

Hypergeometric distribution

If you’ve done your statistics 101, you’ve indubitably come across questions about poker hands. E.g., how likely is it to get a flush when drawing 5 cards out of a standard deck of 52? The answer to this question can be calculated using the hypergeometric distribution. If you find the name intimidating, wait till you see the formula:

\[ p_\text{HG}(N, K, n; k) = \frac{{K \choose k} {N - K \choose n - k}}{{N \choose n}}. \]

So let’s take this step by step. Applied to our situation, we have a total of \(N = 16\) cards, \(K = 4\) of which are “successes”, i.e., the dark ones. In the standard application of the hypergeometric distribution, we’d draw a fixed number of \(n\) cards and want to know the probability that exactly \(k\) of them are dark. Our problem however is a little different: we need to fix \(k = 4\) dark cards to be drawn and want to know the probability that this happens within the first \(n\) draws. This is almost what we want – it describes the probability that the game lasts less than \(n\) rounds. Let \(X\) be the random variable that describes the game length. Then what we just said can be expressed as

\[ P(X \lt n) = p_\text{HG}(16, 4, n; 4) = \frac{{4 \choose 4} {16 - 4 \choose n - 4}}{{16 \choose n}} = \frac{{12 \choose n - 4}}{{16 \choose n}}. \]

Now, it’s easy to recover the probability is that the game lasts exactly \(n\) rounds:

\[ P(X = n) = P(X \lt n + 1) - P(X \lt n) = \frac{{12 \choose n - 3}}{{16 \choose n + 1}} - \frac{{12 \choose n - 4}}{{16 \choose n}}, \]

where \(n \ge 4\) to make sure the binomial coefficients are defined (I’ll leave \(P(X = 3)\) as an exercise to the reader). If you’re bored, you can flex your algebraic muscles and simplify this expression:

\[ P(X = n) = \frac{12! \cdot n! \cdot 4}{(n - 3)! \cdot 16!}. \]

This is a nice closed form, but it’s a) tedious to calculate as those factorials really do blow up and b) not particularly insightful. Can we do better?

Pedestrian, but insightful formula

Did you notice above how \(P(X = 15) = 25\%\) is such a nice round number? Maybe it’s worth calculating this value in yet another way. So, what does it mean for the game to go over the full length? It means that the very last card has to be a dark one. What’s the probability of this happening? Well, 4 out of 16 cards are dark, so there’s a 1 in 4 chance of the last card being dark.

Similarly, we can check why \(P(X = 14) = 20\%\). This means the penultimate card will have to be the 4th dark card, i.e., the last card has to be light. In other words,

\[ P(X = 14) = \frac{12}{16} \cdot \frac{4}{15} = \frac{1}{5} = 20\%. \]

Once I started to examine this pattern, a lightbulb went on in my head. We need to flip the deck over! 💡 Now, we’re no longer asking where the 4th dark card is in the pile, but rather how many light cards are in a row at the bottom of it. This is pretty straightforward to write down, e.g., for the next probability in line:

\[ P(X = 13) = \frac{12}{16} \cdot \frac{11}{15} \cdot \frac{4}{14} = \frac{11}{70} \approx 15.7\%. \]

And in general:

\[ P(X = n) = \frac{4}{n+1} \cdot \prod_{k=n-2}^{12} \frac{k}{k+4}. \]

Again, if you feel bored, you can express this product in terms of the same factorials as above. But the big advantage of this formulation is that it’s pretty transparent what’s going on.

So, are we finally done? Not quite, there’s still another way of arriving at the same result, which is a bit more elegant and general.

Negative hypergeometric distribution

When playing with the hypergeometric distribution above, you might’ve stumbled across an important question we’ve ignored: is this even a proper distribution, i.e., do all those values sum up to 1? Those \(p_\text{HG}(N, K, n; k)\) assume you fix some parameters \(N\) (the number of cards), \(K\) (the number of dark cards), and \(n\) (the number of draws) and then run through all permissible values of \(k\) (the number of dark cards drawn). Summing \(p_\text{HG}(N, K, n; k)\) over all \(k\) for fixed parameters is guaranteed to give 1, but we’ve instead kept \(k\) fixed at 4 (i.e., draw all dark cards) and ranged \(n\) over all possible number of draws. Hasn’t someone already figured out the details of this distribution?

Of course they have – this is called the negative hypergeometric distribution. Whilst the hypergeometric distribution counts the number of successes in a fixed number of draws, the negative hypergeometric distribution counts the number of successes until a fixed number of failures occur. Translated to our problem, it’ll count how many light cards are drawn before the 4th dark card is drawn. Again, this is almost what we want – the game length will be the number of light cards drawn plus 3 for the first dark cards, which will be auctioned off as well.

The probability mass function of the negative hypergeometric distribution is given by

\[ p_\text{NHG}(N, K, r; k) = \frac{{k + r - 1 \choose k} {N - r - k \choose K - k}}{N \choose K}, \]

where \(N = 16\) is the total number of cards, \(K = 12\) is the number of light cards amongst them, \(r = 4\) is the number of dark cards (which will end the experiment / game), and \(k\) is the number of light cards drawn. Again, let \(X\) be the random variable that describes the game length. Then the probability that the game lasts \(n\) rounds is

\[ P(X = n) = p_\text{NHG}(16, 12, 4; n - 3) = \frac{{n - 3 + 4 - 1 \choose n - 3} {16 - 4 - n + 3 \choose 12 - n + 3}}{16 \choose 12} = \frac{{n \choose n - 3}}{16 \choose 12}, \]

which we can further simplify to

\[ P(X = n) = \frac{n!}{(n-3)! \cdot 3!} \cdot \frac{12! \cdot 4!}{16!} = \frac{12! \cdot n! \cdot 4}{(n - 3)! \cdot 16!}, \]

so thankfully, this checks out. 😌

Now, you might think it’s quite redundant to derive the same result over and over, but I find it immensely satisfying to obtain the same formula in 3 different ways – and have them verified by simulation. Probability theory can be very tricky and plausible calculations can go wrong in unexpected ways. This article could’ve just been a single paragraph if we had immediately given away the answer with the negative hypergeometric distribution, but I honestly wouldn’t have trusted the results just like that.

One final question remains: Did Reiner Knizia crunch the numbers when he designed the game? I can’t tell for sure, but he does hold a PhD in mathematics, so he would certainly have carefully considered the impact of the game length on the balance and gaming experience. When it comes to the “fun” in games, theory doesn’t matter and play testing is king, but calculations like these will provide a shortcut in the design process. Simulations are usually the fastest and most robust way to model a game, but calculations can lead you down really fun rabbit holes – I speak from experience… 🐇

PS: You can find the code for the simulation etc here.