Monte Carlo Simulation
of the Craps Dice Game
Sencer Koç
Basics
• The player rolling the dice is the "shooter".
Shooters first throw in a round of Craps is called
the Come Out roll. If you roll a 7 or 11, you win
and the round is over before it started.
• If you roll a 2, 3, or 12 that's a Craps and you lose:
again, it's over before it started.
• Any other number becomes the Point. The purpose
of the Come Out roll is to set the Point, which can
be any of 4, 5, 6, 8, 9 or 10.
Objective
• The basic objective in Craps is for the shooter to
win by tossing the Point again before he tosses a
7. That 7 is called Out 7 to differentiate it from the
7 on the Come Out roll.
• If the Point is tossed, the shooter and his fellow
bettors win and the round is over. If the shooter
tosses Out 7, they lose and the round is over.
• If the toss is neither the Point nor Out 7, the round
continues and the dice keep rolling.
Craps Game
Come Out Roll
Win if 7, 11
Set Point
Lose if 2, 3, 12
Win if Point
Continue otherwise
Lose if 7 (Out 7)
Win if Point
Continue otherwise
Lose if 7 (Out 7)
Questions
• What is the probability that the roller wins?
– Note that this is not a simple problem.
• The probability of win at later rolls depends on the
point value, e.g., if the point is 8, P(8)=5/36
({2,6},{3,5},{4,4},{5,3},{6,2}) and if the point is
10, P(10)=1/36 ({5,5}).
• How many rolls (on the average) will the
game last?
Simulation and MATLAB
• We will simulate the craps game on a
computer using MATLAB.
• The command rand(1)
– Uniformly distributed between 0 and 1
– round(rand(1,2)*6+0.5)simulates a
single roll of a pair of dice.
• The following MATLAB code simulates M
games.
MATLAB Code
M = 10; % Number of simulations
lost = 0; win = 0; gamelength = zeros(1,M);
disp('
win
loss
gamelength')
data = zeros(10000,3);
for k=1:M
stop = 0;
curr_gamelength = 1;
x = sum(round(rand(1,2)*6+0.5));
if (x==7) | (x==11)
win = win+1; stop = 1;
elseif (x==2) | (x==3) | (x==12)
lost = lost+1; stop = 1;
else
point = x;
end
MATLAB Code (Continued)
while ~stop
curr_gamelength = curr_gamelength+1;
x = sum(round(rand(1,2)*6+0.5));
if (x==7)
lost=lost+1; stop = 1;
elseif (x==point)
win = win+1; stop = 1;
end
end
gamelength(k) = curr_gamelength;
disp(sprintf('%8.0f %8.0f %8.0f '...
,win, lost, curr_gamelength))
data(k,:) = [win lost curr_gamelength];
end
Sample Simulation
Program Output
Trial win
1
0
2
1
3
1
4
0
5
1
6
0
7
1
8
0
9
0
10
1
loss gamelength
1
2
0
7
0
3
1
2
0
5
1
10
0
8
1
2
1
4
0
1
Summary Data
Total over 10 simulations
win
loss gamelength
5
5
4.4
Generation of Useful Data
• Remember we are using a random number generator!
• We also need a larger sample size. We will run the
problem for many more games!
• Let us now try 1000 games and divide #wins and #loss by
M to determine the probabilities (comment out the disp
commands to suppress output of data).
• Also evaluate the mean of gamelengths.
• Our point estimators are
– P(win) = data(1000,1)/1000
– P(loss) = data(1000,2)/1000
– L = mean(data(:,3))
Sample Simulations (M=1000)
Trial
1
2
3
4
5
6
7
8
9
10
P(win)
0.479
0.496
0.497
0.467
0.512
0.485
0.480
0.494
0.492
0.495
P(loss)
0.521
0.504
0.503
0.533
0.488
0.515
0.520
0.506
0.508
0.505
L
3.646
3.774
3.351
3.450
3.386
3.435
3.476
3.591
3.466
3.312
Now make N=1000 such trials!
• This will take some time (about 5 minutes)
since the MATLAB code is written for
readability and not for speed!
• You may use the following command line:
– res=zeros(1000,3);for ii=1:1000;
mcscraps; res(ii,:)=[win/1000
lost/1000 mean(gamelength)];end;
Histograms
70
60
60
50
50
40
40
30
30
20
20
10
10
0
0.4
0.45
0.5
0.55
0.6
Histogram of P(win) for
1000 trials.
EPwin    p  0.4929
var Pwin    2p  2.483  10 -4
0
3
3.2
3.4
3.6
Histogram of L for 1000
trials.
EL   L  3.3748
var L   2L  0.0083
3.8
Confidence Intervals
P
P
t N 1,1 / 2  Pwin   0.4929 
t N 1,1 / 2
1000
1000
 1.6464
0.4929 
t999, 0.95
0.4921  Pwin   0.4937 95% confidence interval
L
L
t N 1,1 / 2  L  3.3748 
t N 1,1 / 2
1000
1000
 1.6464
3.3748 
t999, 0.95
3.3701  L  3.3795 95% confidence interval
Exact
The exact value of probability of win can be
calculated by using the theory of Markov Chains
0.4921  Pwin   0.4937 95% confidence interval
976
Pwin  
 0.4929
1980
Final Note
• The game length is an
exponentially
distributed random
variable, however its
mean over N=1000
trials (as shown
before) is
approximately
normally distributed
(Central Limit
Theorem)
3500
3000
2500
2000
1500
1000
500
0
0
10
20
30
40