kubikulann's Blog

CheminDeFer VS. Baccarat

My original curiosity was: where do the Baccarat rules of draw come from?
Natural hypothesis is that they derive from optimal strategies in Chemin de Fer.
So I computed these. (For ease of calculations, I took permanent probabilities, I.e. infinite # of decks.)


1) First, a consensus about notation
Two players, Punto (P) and Banco (B).
Punto has one decision set: knowing my hand, Draw or Stand?
Banco has several decision sets, according to Punto’s decision and the face-up drawn card: knowing that info and my hand, Draw or Stand?

Preliminary analysis shows that P’s strategy takes the form: draw if my hand is lower than N, stand if higher, randomize if equal (probability p of drawing)
N € {0 to 9} , p € ]0,1]
Notation X=N+p ———— Beware! This is pure notation as N and p do not represent the same measure. I write X for (N,p) actually. But as (N,1) is the same as (N+1,0), it is convenient and intuitive.

B’s strategy space is not so easily reduced a priori, but it will appear that the optimal responses to P do take the same form: for every info set c € { Card drawn= 0, 1,… ,9 ; P stood= 10}, draw if hand lower than M, prob q if equal to M.
Notation Y = {y(0), y(1), … , y(9) ; y(10) } with y(c)= M(c)+q(c)

So the Baccarat drawing rules are noted
X=6 ; Y={4,4,5,5,6,6,7,7,3,4 ; 6 }

Please note this is pure strategies (p=q=1):
N=5 ; M={3,3,4,4,5,5,6,6,2,3 ; 5 }

N,M is the highest value to draw. Here, X,Y is the lowest to stand.


2) Nash equilibrium calculates Best response strategies (aka Reaction Functions) X*(Y) and Y*(X) ,and solves the equation system
X = X*(Y)
Y = Y*(X)

[For economists, this game is defined as a Stackelberg equilibrium, because P has first move, hence he can influence B’s move, while the reverse is not true. That makes it easier to compute sequentially.]

Calculation of B’s reaction functions shows that B rarely resorts to mixed strategy (q<1): only on very specific values of X.
In particular:

X : N=5 p=	Y*(X)	label
0 to a	{4,4,5,5,5,6,7,7,3,3 ; 6 }	M0
a = 11/176	{4,4,5,5,mixed,6,7,7,3,3 ; 6 }	Q0
a to b	{4,4,5,5,6,6,7,7,3,3 ; 6 }	M1
b = 71/176	{4,4,5,5,6,6,7,7,3,mixed ; 6 }	Q1
b to c	{4,4,5,5,6,6,7,7,3,4 ; 6 }	M2 (this is the Baccarat rule)
c = 144/176	{4,4,5,5,6,6,7,7,3,4 ; mixed }	Q2
c to 1	{4,4,5,5,6,6,7,7,3,4 ; 7 }	M3

Now P’s Best response to M0 is to play p=1, and to M3, to play p=0. Classical Nash situation.
P must choose a mixed strategy. Here are his reactions (best response).

Y	X*(Y)
M0	6
Q0	6
M1	6
Q1	6
M2	6
Q2 q<q*	6
Q2 q=q*	mixed
Q2 q>q*	5
M3	5

Somewhere in Q2 - at q*= 1513/4888 = 0.3095 - X* switches from 6 to 5. Only at that q* is P indifferent, which means susceptible to use a mixed strategy. In equilibrium this strategy will be the value p* that causes B to select q*: p*=c=144/176= 9/11 = 0.8182

SUMMARY
PUNTO: 0 to 4 Draw / 5 Draw with a probability of .82 / 6,7 Stand.
BANCO: if P stands — 0 to 4 Draw / 5 Draw with a probability of .31 / 6,7 Stand. If P draws — { 4,4, 5,5, 6,6, 7,7, 3,4 }

3) Answers and new questions

• Baccarat rules are NOT the equilibrium strategies of Chemin de Fer players. Where do they come from, then?
• At équilibrium both players should use mixed strategies. There is no ‘bluff’ though (P trying to induce B in error.)
• Why is it that my solution is not the same as the Wizard’s? Maybe I made some mistake. Or maybe it is due to the infinite deck assumption. Or maybe he overlooked cases M1 and M2.