4.2 Score Prediction Model
The score prediction model computes the expected
score distribution at the end of an end based on the
game situation without simulating all possible shots
until the end of the end. The model is constructed
by a simple fully-connected neural network. As input
data for the model, positioning information of stones
on the sheet and information of the game situation are
used. The information for each stone, as shown in
Table2, is inputted in order of proximity to the cen-
ter of the house for all fifteen stones. Information of
stones that do not exist on the sheet are all set to 0.
Information of the game shown in Table3 is also in-
put to the model. These are all one-hot vectors. The
output of the model is an 11-dimensional distribution
of expected scores. This indicates the probability of
scoring from -5 to 5 scores at the end of end. The
activation function of the output layer uses softmax.
Table 2: List of information of stones. Coordinates and dis-
tance from the center of house are represented as continuous
values, while other data are represented as discrete values.
Information Value
x coordinate [-2.375, 2.375]
y coordinate [32.004, 40.234]
Distance from center of house [0, 6.78]
Is there stone in the play area 0, 1
Is there stone in the house 0, 1
Owned player -1, 0, 1
Is there enemy stone on the inside 0, 1
Table 3: List of information of game. These information are
input in one-hot vectors.
Information Value(one-hot vector)
Score difference -2,-1,0,1,2
Remaining ends 0,1,2,3
4.3 Creation of Training Data
In this study, training data was created only for the
last shot of the end. Because the score of the end is
determined when the last shot is completed, so multi-
ple stages of simulation are not necessary, and the ex-
pected score distribution can be obtained with a small
number of simulations.
First, the number of remaining ends, score differ-
ence, and stone position data are given as input data
for the model. In this situation, all candidate shots
are simulated on the digital curling. Each candidate
shot is a vector of shots that reach each score on the
sheet divided on the grid. Grid is a square and its size
is the radius of the stone. No random fluctuation are
added to the simulation at this time. The score is cal-
culated from the stone position obtained after simula-
tion. Based on the points scored, the score difference
in the game situation and the number of remaining
ends, the expected winning probability is calculated
using the winning probability table. The formula for
this calculation is shown in Eq. 1.
E(x, y) =
3
∑
∆x=−3
5
∑
∆y=−5
p(x
0
, y
0
)w(r, d, s(x
0
, y
0
)) (1)
where x
0
= x + ∆x, y
0
= y + ∆y
p(x
0
, y
0
) is the probability of the shot stone reach-
ing the grid around the candidate shot after the simu-
lation when a random fluctuation is added to the can-
didate shot. ∆x and ∆y are the deviations from the grid
of the candidate shot. We consider 3 grids in the x di-
rection and 5 grids in the y direction centered on the
grid of the candidate shot. Since the random fluctu-
ation added to the shot are predetermined, the proba-
bility of the stones reaching the surrounding grids can
be determined in advance. w(r, d, s) is the expected
winning probability when the fluctuation of remain-
ing ends r, the score difference d, and the score at the
end of the end is s. This can be obtained from Table.
1. s(x
0
, y
0
) is the score at the end of the end when a
shot is made at x
0
and y
0
, obtained from the simulation
without random fluctuation described above.
Thus, we obtain the expected winning probability
for all candidate shots. The candidate shot with the
highest expected winning probability obtained is con-
sidered the best shot. Finally, we simulate this best
shot multiple times with the original stone arrange-
ment. Since random fluctuations are added to this
simulation, a probability distribution for each score
can be obtained. This probability distribution is the
expected score distribution as the target output, corre-
sponding to the game situation as the input.
As an example, the number of remaining ends is 1,
the score difference is 0, and the stone position is the
data shown in Fig. 3. Table. 1 shows that the winning
probability in this situation is about 67% if one point
is gained, 74% if a blank end is assumed, and about
4% if one point is lost. In other words, blank end
and scoring one point are high winning probability
actions, with blank end having the highest value. In
the situation, a shot as shown in the Fig. 4 is the best
shot in this situation because the expected winning
probability according to Eq. 1 is the highest. After
simulating the best shot multiple times, the expected
score distribution is as shown in Fig. 5. Fig. 5 shows
an example situation that can be blank end about 80%
of the time, but it is also a situation that a single point
be scored . It also shows that there is almost no risk
of steal. This means that the best shot in this situation
Creation of Training Data and Training for Prediction Model of Curling Scores Using Real Game Data
171