Casino Games  Red Dog
Author: YuiBin Chen, Ph.D.
The Copyright of this article is held by the author.
Introduction
First the dealer shows two cards. Aces are high and deuces are
low. You win if the third card falls between the first two
cards.
If the first two cards are consecutive, this is a tie.
If the first two cards are not consecutive and are not a pair,
then players can raise their bets up to the original bets and
the dealer later shows the third card. If its value falls
between the values of the first two cards, players win and
the payout is determined by the following chart.
If the first two cards are a pair, no raise is allowed and a
third card is drawn immediately. If they make threeofakind,
players receive 11 times of their bets. Otherwise, it is a
tie.
Rules
The dealer flips 2 cards: X, Y (assuming X <= Y).
Then, if necessary, he/she flips a third card (Z) after players' raise.
If (X = Y), then
(a) no raise, and
(b) if (Z = X), then players win, else Tie.
If (X+1 = Y), then Tie.
If (X+1 < Y), then
if (X < Z < Y), then players win,
else players lose.
Payout Schedule

Raise Allowed 
Raise Not Allowed 
Spread 
1 
2 
3 
4  11 
Consecutive 
Pair 
3ofakind 
Payout 
5:1 
4:1 
2:1 
Even 
Tie 
Tie 
11:1 
Strategy
In general, you should raise when the spread is greater than or equal
to 7. A perfect strategy indicates that you should raise when
Probability(win) × Return(win) > 1.
Expected Gain/Loss
Assume this game is dealt from n decks (n >= 1). We consider four
different cases and let Pr() and V() be the probability and winning
of each case.
(a) X = Y = Z :
Pr(a) = C(13,1)*C(4n,3)/C(52n,3)
V(a) = 11
(b) X = Y <> Z :
Pr(b) = C(13,1)*C(4n,2)*C(48n,1)
/ ( C(52n,3)*(3!/2!1!) )
V(b) = 0
(c) X + 1 = Y :
Pr(c) = C(12,1)*C(4n,1)*C(4n,1)/C(52n,2)
V(c) = 0
(d) X + 1 < Y : (let k = Y  X  1)
Pr(d,k) = C(12k,1)*C(4n,1)*C(4n,1)/C(52n,2)
Pr(d,k,win) = Pr(d,k)*C(4kn,1)/C(52n2,1)
Pr(d,k,lose) = Pr(d,k)*C(52n4kn2,1)/C(52n2,1)
V(d,1,win) = 5
V(d,2,win) = 4
V(d,3,win) = 2
V(d,4,win) = V(d,5,win) = V(d,6,win) = 1
V(d,7,win) = V(d,8,win) = V(d,9,win) = V(d,10,win) = V(d,11,win) = 2
(because of the raise bet)
V(d,i,lose) = 1, when i = 1,2,3,4,5,6
V(d,i,lose) = 2, when i = 7,8,9,10,11
The expected gain/loss per hand is
E(n) = Pr(a)*V(a)
+ sum(Pr(d,k,win)*V(d,k,win)
+ Pr(d,k,lose)*V(d,k,lose))
= (456n^2210n+143) / (13*(52n1)*(26n1)).
However, the expected bet size is higher than 1, because of the raise.
The expected bet size should be
B(n) = 1 + Pr(d,7) + Pr(d,8) + Pr(d,9)
+ Pr(d,10) + Pr(d,11)
= (796n13) / (13*(52n1)).
Then the expected gain/loss per dollor becomes
E(n)/B(n) = (456n^2210n+143)/((26n1)*(796n13)).
Using these formulae, we know the house has a (E(n)/B(n)) edge, where
n is the number of decks used. We list the house edges for different
situations below:
Number of Decks 
House Edge 
1 
2.6718 % 
2 
2.6090 % 
4 
2.4472 % 
6 
2.3749 % 
8 
2.3353 % 
Infinite 
2.2033 % 
Counting Method
When a Duce or an Ace is removed from the shoe, players have a better
expected return. When an extra 6, 7, 8, 9, or 10 is added to the shoe,
players also have a better expected return. Therefore, after carefully
examining the change of expected return, we propose the following
counting method (called it "C" Counting System  when you plot the
CountCard in 2dimensional space, it looks like a "C"):
Card 
2 
3 
4 
5 
6 
7 
8 
9 
10 
J 
Q 
K 
A 
Count 
+5 
+1 
0 
1 
2 
2 
2 
2 
2 
1 
0 
+1 
+5 
The true count is defined as
Running count
True Count = 
Number of unused set
where a set is a quarter of deck (13 cards).
When the running count is positive, the game is more favorable to
players. However, in order to get a positive edge on Red Dog game,
the true count must be higher than +5. To support the theory, we
show a computer simulation of 200 million hands.
Simulation Results
Total number of rounds = 200,000,000 (some data are ignored because
of the lack of sampling points). It is simulated from an 8deck
game with 75% penetration.
True Count 
Number of Hands 
Total Bet Size 
Gain/Loss 
Percentage 
8 
21411 
26400 
1104 
4.182% 
7 
72182 
89084 
5764 
6.470% 
6 
243010 
298022 
12041 
4.040% 
5 
713879 
868765 
37757 
4.346% 
4 
1994749 
2412983 
93382 
3.870% 
3 
5295081 
6361955 
231228 
3.635% 
2 
13886986 
16573472 
523450 
3.158% 
1 
37704603 
44689349 
1247534 
2.792% 
0 
80092644 
94344958 
2204379 
2.337% 
1 
37373857 
43751974 
849609 
1.942% 
2 
14201195 
16509125 
234536 
1.421% 
3 
5431668 
6271514 
55569 
0.886% 
4 
2000302 
2292841 
6190 
0.270% 
5 
679187 
773411 
+2885 
+0.373% 
6 
205999 
232899 
2397 
+1.029% 
7 
59378 
66676 
925 
+1.387% 
8 
13309 
14802 
68 
+0.459% 
In the following table, we show the simulation results from 4, 6, and
8deck games (after 200 million hands):
True Count 
Gain/Loss Percentage 
4Deck 
6Deck 
8Deck 
8 
5.278% 
5.631% 
4.182% 
7 
5.392% 
4.707% 
6.470% 
6 
4.569% 
4.946% 
4.040% 
5 
4.627% 
4.396% 
4.346% 
4 
4.050% 
4.105% 
3.870% 
3 
3.704% 
3.657% 
3.635% 
2 
3.255% 
3.272% 
3.158% 
1 
2.848% 
2.827% 
2.792% 
0 
2.415% 
2.409% 
2.337% 
1 
1.959% 
1.912% 
1.942% 
2 
1.414% 
1.407% 
1.421% 
3 
1.023% 
0.995% 
0.886% 
4 
0.464% 
0.477% 
0.270% 
5 
+0.352% 
+0.316% 
+0.373% 
6 
+0.918% 
+0.712% 
+1.029% 
7 
+1.492% 
+1.600% 
+1.387% 
8 
+2.053% 
+1.469% 
+0.459% 
Expected Return 
2.4472% 
2.3749% 
2.3353% 
Expected Return 
2.4472% 
2.3749% 
2.3353% 
It seems that this counting method can be used at all kinds of games,
because when the true count reflects players advantage. However, in
order to make a profit while consistantly playing. You have to do a
1900 spreading. Or you can do the backcount and jump into the game
when the true count is +5 or higher (1 out of 200 chance, 0.48%). Since
each round uses only 2 or 3 cards, when the count reaches +5, it may
stay high for a while. This gives an advantage to back counters. But,
on the other hand, the chance of getting such a positive stream is far
below 0.48%.
Shuffle tracking is also useful to determine whether the incoming rounds
are favorable.
