Title: Discussion on Weiqi Ending in a Tie
Author/Editor: Jiang Qian 蒋牵
Publisher: Chengdu Shidai Publishing 成都时代出版社
Publisher Recommended Price: RMB25
The premise of this book is that is the thought that a perfect game of go ought to end in a tie. As a tie is possible in Xiangqi or in Chess, why is it that the rules of go makes it only possible for either a win or a loss? The author goes on to discuss a lot of ideas and thoughts on the game of go from the perspective of what the game is about rather than how to play it well. Don’t buy it if you are looking for a book for instant improvement to your rank; it’s a book more suitable for those with curious minds.
The content of the book (without translation)
I found this part quite interesting so I’ll translate/paraphrase it for your enjoyment.
The Probability of a Monkey Defeating Yi Chang-ho
Let’s assume there is a monkey who loves weiqi. With the size of his brain, the monkey only has capacity to play legal moves but not efficiently. Thus the monkey could only search randomly for legal moves and play them during his turn.
If we assume the lowest strength player who understands weiqi as 18k, then the monkey would probably rank 3 grades less at 21k.
Let’s say that at the zenith of weiqi strength stands Yi Chang-ho who plays extremely proper and efficient moves. We can assign him a rank 3 grades above a weak 9d. Of course this is just a theoretical ideal, so just think of the role of Yi Chang-ho as equivalent to someone you find as the greatest go player who has ever lived.
Now as the monkey makes random legal moves, it is possible that the monkey can beat Yi Chang-ho as the random moves selected could be the most devastating sequence of moves possible. Remember that all the moves the monkey selects are from the total number of possible moves on the board so theoretically there must be a combination of moves which could beat anyone who plays go.
Let’s say that every rank has only a 25% chance of beating someone 1 rank higher, then you can calculate the probabilities for each rank as follows and arrive at a number (P1) which shows the probability of a monkey beating Yi Chang-ho.
Strong 9d ——- 0.25
Normal 9d —– 0.0625
Weak 9d —– 0.015625
8d ——– 0.00393625
7d ——- 0.000976563
6d ——- 0.000244141
5d ——- 6.10352E_05
4d ——- 1.52588E_05
3d ——– 3.8147E_06
2d ——- 9.53674E_07
1d ——- 2.38419E_07
1k ——- 5.96046E_08
2k ——- 1.49012E_09
3k ——- 3.72529E_10
4k ——- 9.31323E_10
5k ——- 2.32831E_10
6k ——- 5.82077E_11
7k ——- 1.45519E_11
8k ——- 3.63798E_12
9k ——- 9.09495E_13
10k —— 2.27374E_13
11k —— 5.68434E_14
12k —— 1.42109E_14
13k —— 3.55271E_15
14k —— 8.88178E_16
15k —— 2.22045E_16
16k —— 5.55112E_17
17k —— 1.38778E_17
18k —— 3.46945E_18
19k —— 8.67362E_19
20k ——- 2.1684E_19
Monkey — 5.42101E_20
The probability of the monkey beating Yi Chang-ho is so tiny it would take more than billions upon billions of games before the monkey finally wins a game. But actually even this probability is an overestimate for the monkey’s ability. As the monkey can only play random legal moves, his actual weiqi strength is basically zero, so the probability of his winning is actually P0 and not P1.
Let’s think of another logical way to calculate P0.
Let’s assume that any move not played on the first or second line can be called a reasonable opening move. This gives a total of (19-2X2)^2 = 225 reasonable first moves. The odds of the monkey playing a reasonable move would be 225/361. Let’s say the probability of playing a reasonable move for each of the first 10 moves is basically the same as that of the first move.
Of course, every time a stone has been played, the number of proper reasonable moves is greatly reduced. When the board has 300 open space left, let’s assume that there will only be 6 moves which will not lose to Yi Chang-ho’s previous play. The probability of a monkey playing one of these moves would be 6/300 = 0.02 and we will assume that such a probability continues for the next 120 moves played.
Most games would probably end when a side has played 130 moves so the calculation of P0 is approximately (225/361)^10X0.02^120 = 1.1759E_206
As we can see, P0 is much much smaller than P1.
So just how small is the probability P0?
If the monkey and Yi Chang-ho can quickly play at a pace of one move per second, the monkey will not be able to win within the time period of the birth of the universe to the end of the universe. You could take millions of universal lifespans before the monkey finally wins a game.
So we can see that for expert players, the effort required to raise one rank is like reaching for the heavens. It takes herculean efforts and sacrifice for a normal 9d to become a strong 9d. But from the perspective of mathematics, all it takes is for the normal 9d to improve his win/loss percentage against the strong 9d from 25% to 50%.
For an absolute beginner who is learning weiqi at a natural pace, the journey is probably just a casual walk to get from the weak level probability P0 to a higher level probability P1. However, from the perspective of mathematics, we can see just how vast such a gap actually is.